Jim Randell

Updates from Jim Randell Toggle Comment Threads | Keyboard Shortcuts

Jim Randell 9:01 am on 16 July 2020 Permalink | Reply
Tags: by: Graham Smithers ( 82 )
Teaser 2682: 007
From The Sunday Times, 16th February 2014 [link] [link]

Eight villains Drax, Jaws, Krest, Largo, Morant, Moth, Sanguinette and Silva have been ordered to disrupt the orbits of the planets Earth, Jupiter, Mars, Mercury, Neptune, Saturn, Uranus and Venus, with each villain disrupting a different planet. Our agents Brosnan, Casenove, Connery, Craig, Dalton, Dench, Lazenby and Moore have each been assigned to thwart a different villain. For any villain/planet, villain/agent or planet/agent combinations just two different letters of the alphabet occur in both names.

List the eight agents in alphabetical order of their assigned villains.

[teaser2682]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 9:02 am on 16 July 2020 Permalink | Reply
  This is another puzzle that can be solved using [[ grouping ]] routines from the enigma.py library.
  
  Run: [ @repl.it ]
```
from enigma import grouping

# villains (in alphabetical order), planets, agents
V = ('Drax', 'Jaws', 'Krest', 'Largo', 'Morant', 'Moth', 'Sanguinette', 'Silva')
P = ('Earth', 'Jupiter', 'Mars', 'Mercury', 'Neptune', 'Saturn', 'Uranus', 'Venus')
A = ('Brosnan', 'Casenove', 'Connery', 'Craig', 'Dalton',  'Dench', 'Lazenby', 'Moore')

grouping.solve([V, P, A], grouping.share_letters(2))
```
  Solution: The ordering of the agents is: Dalton, Casenove, Connery, Lazenby, Craig, Moore, Dench, Brosnan.
  
  The groups are:
  
  Drax, Uranus, Dalton
  Jaws, Mars, Casenove
  Krest, Neptune, Connery
  Largo, Saturn, Lazenby
  Morant, Jupiter, Craig
  Moth, Earth, Moore
  Sanguinette, Mercury, Dench
  Silva, Venus, Brosnan
  
  LikeLike
Reply Cancel reply
Required fields are marked *

Name *

Email *

Website

Notify me of new comments via email.
Notify me of new posts via email.
Δ

This site uses Akismet to reduce spam. Learn how your comment data is processed.
Jim Randell 12:11 pm on 14 July 2020 Permalink | Reply
Tags: by: Victor Bryant ( 140 )
Teaser 2672: 11, 12, 13 …
From The Sunday Times, 6th December 2013 [link] [link]

On Wednesday the date will be 11.12.13. To celebrate this fact I have found three numbers with the lowest divisible by 11, another divisible by 12, and the remaining number divisible by 13. Furthermore, appropriately, the sum of the three numbers is 2013. Overall the three numbers use nine digits with no digit used more than once.

What (in increasing order) are my three numbers?

[teaser2672]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell and GeoffR are discussing. Toggle Comments
- Jim Randell 12:13 pm on 14 July 2020 Permalink | Reply
  The following Python program runs in 61ms.
  
  Run: [ @replit ]
```
from enigma import (irange, is_duplicate, printf)

# choose the first number (a multiple of 11)
for n1 in irange(11, 2013, step=11):
  # it has no repeated digits
  s1 = str(n1)
  if is_duplicate(s1): continue
  # and choose a larger second number (a multiple of 12)
  for n2 in irange(n1 + 12 - (n1 % 12), 2013 - n1, step=12):
    # n1 and n2 don't share digits
    s2 = str(n2)
    if is_duplicate(s1, s2): continue
    # and determine n3
    n3 = 2013 - (n1 + n2)
    # n3 is larger than n1
    if not (n1 < n3): continue
    # and a multiple of 13
    if n3 % 13: continue
    # and between them the numbers have 9 digits, all different
    s = s1 + s2 + str(n3)
    if len(s) != 9 or is_duplicate(s): continue
    # output solution
    printf("{ns}", ns=sorted([n1, n2, n3]))
```
  Solution: The three numbers are: 495, 702, 816.
  
  LikeLike
- GeoffR 5:44 pm on 14 July 2020 Permalink | Reply
```
% A Solution in MiniZinc
include "globals.mzn";
 
var 0..9:A; var 0..9:B; var 0..9:C; 
var 0..9:D; var 0..9:E; var 0..9:F; 
var 0..9:G; var 0..9:H; var 0..9:I; 

% Three numbers are ABC, DEF and GHI
var 100..999: ABC = 100*A + 10*B + C;
var 100..999: DEF = 100*D + 10*E + F;
var 100..999: GHI = 100*G + 10*H + I;

constraint A != 0 /\ D != 0 /\ G != 0;
constraint all_different ([A,B,C,D,E,F,G,H,I]);

constraint ABC mod 11 == 0 /\ DEF mod 12 == 0  
/\ GHI mod 13 == 0;

constraint ABC + DEF + GHI == 2013;

solve minimize (ABC);

output ["ABC = " ++ show (ABC) ++ ", DEF = " ++ show(DEF) 
++ ", GHI = " ++ show(GHI)];

% ABC = 495, DEF = 816, GHI = 702  
% or 495, 702, 816 in increasing order
% ----------
% ==========
```
  LikeLike
  - Jim Randell 5:12 pm on 15 July 2020 Permalink | Reply
    Providing we know the numbers all have 3 digits.
    
    But the smallest number is a multiple of 11, with no repeated digits, so must be at least 132 (unless it is 0). So the other two numbers must both be at least 3 digits, but between them they have only 9 digits, so they must indeed be all 3-digit numbers.
    
    (If the first number is 0, the other two must both be 4-digit numbers, but to add up to 2013, they would both have to start with 1, so this is disallowed).
```
#! python -m enigma -rr

SubstitutedExpression

"ABC % 11 = 0"
"DEF % 12 = 0"
"GHI % 13 = 0"

"ABC < DEF"
"ABC < GHI"

"ABC + DEF + GHI = 2013"

--answer="ordered(ABC, DEF, GHI)"
```
    LikeLike
Jim Randell 9:55 am on 12 July 2020 Permalink | Reply
Tags: by: John Feather ( 3 )
Brainteaser 1835: Dominic’s dominoes
From The Sunday Times, 16th November 1997 [link]

Dominic thought up a variation of the domino jigsaw game. He laid out a standard set of dominoes and produced the pattern below by copying the numbers of spots in each place. He put a 0 for all the blanks except for the one blank which stuck out of a side of the rectangle:

Find the positions of all the dominoes.

This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.

[teaser1835]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 9:56 am on 12 July 2020 Permalink | Reply
  I used the [[ DominoGrid() ]] solver from the enigma.py library (see: Teaser 1590). We pair up one of the edge squares with a 0 to make a domino, then attempt to fit the rest into the remainder of the grid. The following program runs in 88ms.
  
  Run: [ @repl.it ]
```
from enigma import DominoGrid, update, printf

# the grid
(X, Y, grid) = (5, 11, [
  6, 2, 6, 2, 0,
  6, 3, 1, 2, 0,
  3, 5, 1, 2, 0,
  3, 5, 3, 3, 1,
  6, 6, 5, 3, 0,
  4, 3, 0, 3, 2,
  2, 5, 4, 4, 4,
  4, 2, 4, 6, 6,
  5, 5, 4, 1, 1,
  2, 4, 1, 5, 0,
  1, 1, 0, 5, 6,
])

# remove one of the border squares and solve for the rest
for (i, b) in enumerate(grid):
  (y, x) = divmod(i, X)
  if x == 0 or x == X - 1 or y == 0 or y == Y - 1:
    # place the 0-b domino and solve for the remaining dominoes
    p = DominoGrid(X, Y, update(grid, [(i, None)]))
    for s in p.solve(used=[(0, b)]):
      printf("[0-{b} placed @ {x},{y}]\n")
      p.output_solution(s, prefix="  ")
```
  Solution: The layout of the dominoes is shown below:
  
  So the missing square is part of 0-2 domino.
  
  LikeLike
Jim Randell 5:11 pm on 10 July 2020 Permalink | Reply
Tags: by: Andrew Skidmore ( 85 ), flawed ( 53 )
Teaser 3016: Eureka
From The Sunday Times, 12th July 2020 [link] [link]

Archimedes’ Principle states that the upward force on a floating body equals the weight of water displaced by the body. In testing this, Liam floated a toy boat in a cylindrical drum of water. The boat had vertical sides and took up one tenth of the surface area of the water. He also had some identical ball bearings (enough to fit snugly across the diameter of the drum), which were made of a metal whose density is a single-figure number times that of water.

Liam loaded the boat with the ball bearings and noted the water level on the side of the drum. When he took the ball bearings out of the boat and dropped them in the water, the water level changed by an amount equal to the radius of a ball bearing.

How many ball bearings did Liam have?

This puzzle was not included in the book The Sunday Times Teasers Book 1 (2021).

[teaser3016]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell, Jim Randell, Dilwyn Jones, and 1 other are discussing. Toggle Comments
- Jim Randell 6:26 pm on 10 July 2020 Permalink | Reply
  Suppose there are n ball-bearings, each with a radius of r, then the radius of the cylindrical tank is nr.
  
  If we take the water level in the tank when the the unladen boat is floating in it to be “low water”, then when the balls are placed in the tank they displace an amount of water equal to their volume: n (4/3) 𝛑 r³.
  
  The boat still has the same draught, so the water level in the tank increases by:
  
  [1] (n(4/3)𝛑r³) / (𝛑(nr)²) = (4/3)(r/n)
  
  $\frac{n\left( \frac{4}{3}\pi r^{3} \right)}{\pi \left( nr \right)^{2}}\; =\; \frac{4}{3}\frac{r}{n}$
  
  [Note that the following step is flawed (although appears to be what the setter expects), see my follow-on comment below].
  
  When the balls are instead placed in the boat, then the weight of the boat is increased by: kn(4/3)𝛑r³ (where the metal of the balls is k times the density of water), and so this weight of water is displaced, so the level of water in the tank (above low water) is:
  
  [2] (kn(4/3)𝛑r³) / ((9/10)𝛑(nr)²) = (40/27)(kr/n)
  
  $\frac{kn\left( \frac{4}{3}\pi r^{3} \right)}{\frac{9}{10}\pi \left( nr \right)^{2}}\; =\; \frac{40}{27}\frac{kr}{n}$
  
  (the relative density of water is 1).
  
  The difference between these water levels is r:
  
  [2] − [1] (40/27)(kr/n) − (4/3)(r/n) = r
  n = (4/27)(10k − 9)
  
  Only one value of k = 1..9 has (10k − 9) divisible by 27.
  
  This Python program looks for values of k that give an integer value for n:
```
from enigma import (irange, div, printf)

for k in irange(1, 9):
  n = div(40 * k - 36, 27)
  if n is None: continue
  printf("k={k} n={n}")
```
  Solution: There are 12 ball bearings.
  
  And density of the ball bearings is 9 times that of water. So perhaps they are mostly made of copper (relative density = 8.96).
  
  Writing:
  
  40k − 27n = 36
  40k = 9(3n + 4)
  
  We see that k must be a multiple of 9, and as it must be a single digit the only possibility is k = 9, and so n = 12.
  
  [This seems to be the expected answer, but see my follow on comment for a more realistic approach].
  
  LikeLike
  - Jim Randell 9:22 am on 14 July 2020 Permalink | Reply
    
    In my original treatment, when the water was displaced from loading the boat, it was then added back in to the space around the boat. But unless the water is actually removed, and the boat drops anchor before the displaced water is added back in, the boat will rise as the water is added (we can think of the displaced water being added to the bottom of the tank, rather than the top). So the second water level calculated above is higher than it would be in “real life”.
    
    The correct calculation is:
    
    [2] = k×[1] (kn(4/3)𝛑r³) / (𝛑(nr)²) = (4/3)(kr/n)
    
    $\frac{kn\left( \frac{4}{3}\pi r^{3} \right)}{\pi \left( nr \right)^{2}}\; =\; \frac{4}{3}\frac{kr}{n}$
    
    (The cross-sectional area of the boat is immaterial, as the boat rises with the water).
    
    We then calculate the difference in water levels:
    
    [2] − [1] (4/3)(kr/n) − (4/3)(r/n) = r
    4(k − 1) = 3n
    
    So, n is a multiple of 4, and (k − 1) is a multiple of 3:
    
    n = 4, k = 4
    n = 8, k = 7
    n = 12, k = 10
    …
    
    For a single digit value of k there are two solutions (although k = 7 would be a more likely relative density for ball bearings [link], so the most reasonable solution would be that there are 8 ball bearings).
    
    But the relative cross-sectional areas of the boat and the tank don’t seem to matter, so perhaps the setter was expecting the same approach as that I originally gave above.
    
    LikeLike
  - Jim Randell 10:09 am on 25 March 2023 Permalink | Reply
    
    The published solution is:
    
    Teaser 3016
    We apologise that there was not a unique answer.
    4, 8 and 12 were all acceptable.
    
    LikeLike
- GeoffR 9:02 am on 14 July 2020 Permalink | Reply
  I did a Python programme for a range of ball bearing sizes and the result (as expected) was a constant number of ball bearings. The programme confirmed the difference in water height was the radius of the ball bearing in each case.
```
from math import pi

# Let rho be density factor of ball bearings c.f. water
# Let N = number of ball bearings

# ball bearing radius range considered is from 0.25 cm to 2.0 cm
for r in (0.25, 0.5, 1.0, 1.5, 2.0):  
  for rho in range(1, 10):   
    for N in range(1, 50):
      # calculate radius of tank, area of tank and boat
      tank_rad = N * r
      tank_area = round((pi * tank_rad**2), 3)
      boat_area = round((tank_area / 10), 3)
      
      # calculate weight and volume of N ball bearings 
      wt = rho * 4/3 * pi * r**3 * N
      vol = 4/3 * pi * r**3 * N

      # Let h1 be water height change after balls put in tank
      h1 = round((vol / tank_area), 3)

      # Let h2 = water height change after balls put in boat
      h2 = round ((wt / (tank_area - boat_area)), 3)
      # check radius of ball bearings = change in water height
      if r == h2 - h1:
        print(f"Ball Bearing Radius = {r} cm. Number of balls = {N} no.")
        print(f"1st height = {h1} cm., 2nd height = {h2} cm., Difference = {h2-h1} cm")
        print()

# Ball Bearing Radius = 0.25 cm. Number of balls = 12 no.
# 1st height = 0.028 cm., 2nd height = 0.278 cm., Difference = 0.25 cm

# Ball Bearing Radius = 0.5 cm. Number of balls = 12 no.
# 1st height = 0.056 cm., 2nd height = 0.556 cm., Difference = 0.5 cm

# Ball Bearing Radius = 1 cm. Number of balls = 12 no.
# 1st height = 0.111 cm., 2nd height = 1.111 cm., Difference = 1.0 cm

# Ball Bearing Radius = 1.5 cm. Number of balls = 12 no.
# 1st height = 0.167 cm., 2nd height = 1.667 cm., Difference = 1.5 cm

# Ball Bearing Radius = 2 cm. Number of balls = 12 no.
# 1st height = 0.222 cm., 2nd height = 2.222 cm., Difference = 2.0 cm
```
  LikeLike
- Dilwyn Jones 11:02 am on 17 July 2020 Permalink | Reply
  
  I agree with the corrected calculation and a set of possible answers. I wondered if there is an additional constraint – the cross sectional area of a ball bearing must be smaller than the area of the boat, but that is true for all the answers. Then I wondered if all the ball bearings must touch the flat bottom of the boat, which means there must be at least 10 of them ignoring the spaces in between. If instead each bearing fills a square of side 2r, there must be at least 13. However I am speculating, as this information is not given in the problem.
  
  LikeLike
  - Jim Randell 2:37 pm on 18 July 2020 Permalink | Reply
    
    I hadn’t given a great deal of thought to the shape of the boat, as I’d reasoned that if the boat was vertical tube with a hole down the middle to stack the balls into, then as long as n² ≥ 10 we could create a boat with the appropriate cross-sectional area. And this is true for all proposed values of n. Although there might be a problem with stability with this design of boat.
    
    However we know the balls will fit exactly across the diameter of the tank, so if we throw a convex hull around the balls (which would have to be infinitely thin for it to work), then we get a boat that holds all the balls in a line, and it has the added advantage that the boat would be unable to tip over.
    
    Then the cross-sectional area of the boat as a ratio of the cross-sectional area of the tank is:
    
    a = (𝛑 + 4(n − 1)) / (n²𝛑)
    
    Which has the following values:
    
    n = 4; a = 30.1%
    n = 8; a = 15.5%
    n = 12; a = 10.4%
    
    Which gets us close to the 10% stipulation at n = 12.
    
    The same idea, but stacking the balls two rows high gives a boat that will actually fit in the tank (but may tip over):
    
    a = (𝛑 + 2(n − 2)) / (n²𝛑)
    
    n = 4; a = 14.2%
    n = 8; a = 7.5%
    n = 12; a = 5.1%
    
    So with n=8 we can make a boat with a non-zero thickness hull around a cargo hold that can take the balls in a line stacked 2 high.
    
    LikeLike
Jim Randell 8:44 am on 9 July 2020 Permalink | Reply
Tags: by: Ptolemaeus ( 5 )
Brain-Teaser 2: Stop watch
From The Sunday Times, 5th March 1961 [link]

Setting one’s watch can be a tricky business, especially if it has a sweep second hand; for, unlike the hour and minute hands, the second hand is independent of the winder. The other day, when trying to set my watch by the midday time signal, I managed to get the hour hand and minute hand accurately aligned at 12 o’clock just as the pips signalled noon, but the second hand escaped me — in fact, on the pip of twelve it was just passing the 5-second mark. I can’t say that it was exactly on the 5-second mark, but it was within a second or two of it one way or the other. I didn’t bother to adjust it further in case I should finish up with the other hands wrong as well.

That night I forgot to wind my watch and, the next morning, found that (not unnaturally) it had stopped. I noticed that the hour hand, the minute hand and the second hand were all exactly aligned one above another.

(1) At exactly what time had my watch stopped?
(2) Where exactly was the second hand pointing (to the very fraction of a second) on the pip of twelve noon the previous day?

[teaser2]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 8:45 am on 9 July 2020 Permalink | Reply
  See also: Enigma 1761, Enigma 383, Enigma 404, Enigma 409.
  
  In each 12 hour period there are 11 times when the hour and minute hand coincide.
  
  This Python program looks at each of these eleven times, and works out far the second hand is ahead of its expected position if it is also coincident with the hour and minute hands. And if the second hand is between 3 and 7 seconds ahead of where it should be, then we have a solution to the puzzle.
  
  Run: [ @repl.it ]
```
from fractions import Fraction as F
from enigma import irange, sprintf as f, printf

# format fractional seconds
def fmt(s):
  (n, r) = divmod(s, 1)
  return f("{n:02d} + {r} s")

# number of seconds in a period of the clock (12 hours)
T = 12 * 60 * 60

# find times at which the hour and minute hands coincide
for i in irange(0, 10):
  t = F(T * i, 11)
  (h, ms) = divmod(t, 60 * 60)
  (m, s) = divmod(ms, 60)
  # the second hand should be at s, but is actually at ms/60
  # the difference should be between 3 and 7 seconds
  d = (F(ms, 60) - s) % 60
  if not(d < 3 or d > 7):
    printf("i={i} -> t={t} = {h:02d}:{m:02d}:{s}; diff = {d}", s=fmt(s), d=fmt(d))
```
  Solution: (1) The watch stopped at 8:43:38 + 2/11. (2) At exactly 12 noon the second hand was showing 5 + 5/11 seconds.
  
  LikeLike
Jim Randell 9:36 am on 7 July 2020 Permalink | Reply
Tags: by: Alan Bergson ( 4 )
Teaser 2663: Missed the plot
From The Sunday Times, 6th October 2013 [link] [link]

My friend has a triangular vegetable plot, all sides being whole numbers of metres. Coincidentally, the dimensions of the plot are such that its perimeter (in metres) is the same as its area (in square metres). Also, the length of one of the sides is the average of the lengths of the other two sides.

What are the lengths of the sides of the plot?

[teaser2663]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
GeoffR and Jim Randell are discussing. Toggle Comments
- Jim Randell 9:37 am on 7 July 2020 Permalink | Reply
  The triangle has sides of integer lengths (a, b, c) such that b = (a + c)/2.
  
  We can use Heron’s Formula [ @wikipedia ] to equate the area of the triangle with the perimeter, which gives us some restrictions on the sides of the triangle.
  
  However, we have encountered “equable” triangles before in Enigma 364, and generated a list of the 5 equable triangles, from which we can easily pick out the one that has one side that is an average of the other two.
  
  I also wrote code to generate k-equable triangles (integer sided triangles where the area is k times the perimeter), and that code can be reused here:
  
  Run: [ @repl.it ]
```
from enigma import irange, isqrt, divc, divf, printf

# find all triangles with integer sides where area = k . perimeter
def triangles(k=1):
  K = 4 * k * k
  for p in irange(1, isqrt(3 * K)):
    for q in irange(max(p, divc(K + 1, p)), divf(3 * K, p)):
      (r, z) = divmod(K * (p + q), p * q - K)
      if r < q: break
      if z == 0: yield (p + q, p + r, q + r)

# consider 1-equable triangles
for (a, b, c) in triangles(1):
  if b - a == c - b:
    printf("a={a} b={b} c={c}")
```
  Solution: The lengths of the sides of the plot are: 6m, 8m, 10m.
  
  LikeLike
- GeoffR 10:54 am on 8 July 2020 Permalink | Reply
```
for a in range (3, 25):
  for b in range (a+1, 25):
    for c in range(b+1, 25):
      # a < b < c and 2 * b = a + c
      if 2 * b != a + c: continue
      # Square of perimeter
      perim_sq = (a + b + c) ** 2
      # Area squared formula is based in Heron's formula
      area_sq = 1/16 *( 4*a*a*b*b - (a*a + b*b - c*c)**2)
      if perim_sq == area_sq:
        print(f"Length of sides are {a}, {b} and {c}m")

# Length of sides are 6, 8 and 10m
```
  A (5,12,13) triangle has the area equal to the perimeter numerically, but does not have one side as the average of the other two sides.
  
  LikeLike
Jim Randell 8:31 am on 5 July 2020 Permalink | Reply
Tags: by: Harold Reiter ( 5 )
Brainteaser 1823: The special seven
From The Sunday Times, 24th August 1997 [link]

A “special seven” fraction is one whose decimal expansion uses only 7s and/or 0s. So examples of special sevens are:

0.7707
0.70707070…

which are the decimal expansions of:

7707/10000
70/99

respectively.

You can set yourself all sorts of tasks with the special sevens. For example, is it possible to find some which add up to 1? In fact this is possible in many different ways, but…

… what is the smallest number of special sevens adding to 1?

This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.

[teaser1823]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 8:32 am on 5 July 2020 Permalink | Reply
  Each “special seven” number corresponds to a “special one” number, which can be derived by dividing by 7.
  
  So if we find a collection of special sevens that sum to 1, there is a corresponding set of special ones that sum to 1/7.
  
  The decimal expansion of 1/7 = 0.(142857)…
  
  Which if we write it as a sum of special ones will require at least 8 of them to construct the 8 digit.
  
  Solution: The smallest number of special sevens that sum to 1 is 8.
  
  We can construct a collection of 8 special ones that sum to 1/7 as follows:
```
0.(000100)...  [h]
0.(000101)...  [g] = [f]
0.(000101)...  [f]
0.(000111)...  [e]
0.(010111)...  [d] = [c]
0.(010111)...  [c]
0.(011111)...  [b]
0.(111111)...  [a]
-------------
0.(142857)...
```
  (This is not the only set, the 0’s and 1’s in each column can be re-ordered to give other sets, and there is no reason why each recurring section should use the same arrangement of columns).
  
  Writing these as fractions (from [a] to [h]) we have:
```
[a] = 0.(111111)... = 111111/999999 = 1/9
[b] = 0.(011111)... = 11111/999999
[c] = 0.(010111)... = 10111/999999
[d] = 0.(010111)... = 10111/999999
[e] = 0.(000111)... = 111/999999 = 1/9009
[f] = 0.(000101)... = 101/999999
[g] = 0.(000101)... = 101/999999
[h] = 0.(000100)... = 100/999999
```
  Which sum to: 142857/999999 = 1/7 as required.
  
  Multiplying these values by 7 gives us a set of 8 special sevens that sum to 1:
```
[a] = 0.(777777)... = 777777/999999 = 7/9
[b] = 0.(077777)... = 77777/999999
[c] = 0.(070777)... = 70777/999999
[d] = 0.(070777)... = 70777/999999
[e] = 0.(000777)... = 777/999999 = 7/9009
[f] = 0.(000707)... = 707/999999
[g] = 0.(000707)... = 707/999999
[h] = 0.(000700)... = 700/999999
```
  LikeLike
Jim Randell 5:14 pm on 3 July 2020 Permalink | Reply
Tags: by: Victor Bryant ( 140 )
Teaser 3015: Quid pro quo
From The Sunday Times, 5th July 2020 [link] [link]

In Readiland the unit of currency is the quid. Notes are available in two denominations and with these notes it is possible to make any three-figure number of quid. However, you need a mixture of the denominations to make exactly 100 quid. Furthermore, there is only one combination of denominations that will give a total of 230 quid.

What are the two denominations?

[teaser3015]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 5:32 pm on 3 July 2020 Permalink | Reply
  See also: Enigma 228, Enigma 1194, Enigma 1154.
  
  If we suppose the denominations are x and y (where gcd(x, y) = 1). Then the largest amount that cannot be made using the denominations is given by F(x, y) = xy − x − y.
  
  Both denominations are needed to make 100, so they must both be less than 100, and neither can be a divisor of 100.
  
  The following Python 3 program runs in 67ms.
  
  Run: [ @repl.it ]
```
from enigma import (coprime_pairs, express, printf)

# consider denominations x, y
for (x, y) in coprime_pairs(97):
  if 100 % x == 0 or 100 % y == 0: continue

  # largest inexpressible amount is < 100
  if not (x * y - x - y < 100): continue

  # there is exactly 1 way to express 230
  e230s = list(express(230, (x, y)))
  if len(e230s) != 1: continue

  # output solution
  printf("x={x} y={y}; 100 -> {e100s}, 230 -> {e230s}", e100s=list(express(100, (x, y))))
```
  Solution: The two denominations are 3 quid and 49 quid.
  
  The largest amount that cannot be made using these denominations is 95 quid. All larger amounts are possible.
  
  The only way to make 100 quid is 17× 3 quid + 1× 49 quid.
  
  The only way to make 230 quid is 44× 3 quid + 2× 49 quid.
  
  Manually:
  
  The largest integer that is not expressible in k different ways using denominations x, y is given by:
  
  F[k](x, y) = kxy − x − y
  
  So we need to find co-prime values for x, y, less than 100 that are not divisors of 100, and the following hold:
  
  F[1](x, y) < 100
  F[2](x, y) ≥ 230
  
  This will ensure that all 3-digit amounts are expressible, and that to make 100 requires both denominations. All that remains for each pair of candidate denominations is to determine if there is a unique expression for 230.
  
  Given a value for x we can use the inequalities to find a range of viable values for y.
  
  Starting with x = 3, gives y = 47 .. 51 and only y = 47, 49 are co-prime with x.
  
  230 can be expressed in two ways using (3, 47): 230 = 14× 3 + 4× 47 = 61× 3 + 1× 47.
  
  230 can be expressed in only one way using (3, 49): 230 = 44× 3 + 2× 49.
  
  So (3, 49) is a viable solution.
  
  For x = 4, we get only y = 34, but x divides 100, so this is not a candidate solution.
  
  And for larger values of x there are no possibilities for y, so we are done.
  
  LikeLike
Jim Randell 8:37 am on 2 July 2020 Permalink | Reply
Tags: by: Victor Bryant ( 140 )
Teaser 2748: I go up and up
From The Sunday Times, 24th May 2015 [link] [link]

I have in mind an arithmetic progression; ie, a sequence which goes up and up by a fixed amount (such as 1, 10, 19, 28, …) and — as in that example — one of the terms of my progression will eventually be 10000. But I have coded the numbers by consistently replacing digits by letters and in this way the arithmetic progression has become

I, GO, UP, AND, …

Find the value of POUND.

This puzzle is not included in the book The Sunday Times Brain Teasers Book 1 (2019).

[teaser2748]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
GeoffR and Jim Randell are discussing. Toggle Comments
- Jim Randell 8:38 am on 2 July 2020 Permalink | Reply
  We can use the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve this puzzle.
  
  The following run file executes in 93ms.
  
  Run: [ @replit ]
```
#! python -m enigma -rr

SubstitutedExpression

# the given terms must be evenly spaced
"GO - I == UP - GO"
"UP - GO == AND - UP"

# and eventually we reach 10000
"div(10000 - I, GO - I) > 0"

--answer="POUND"
```
  Solution: POUND = 28706.
  
  The first 4 terms in the progression are:
  
  4, 38, 72, 106, …
  
  Each term being derived from the previous term by adding 34. So the sequence is:
  
  S(n) = 4 + 34n
  S(294) = 10000
  
  LikeLike
- GeoffR 8:07 pm on 21 October 2021 Permalink | Reply
```
from itertools import permutations

for p1 in permutations('1234567890', 8):
  A, D, G, I, N, O, U, P = p1
  if '0' in (I, G, U, A, P): continue
  I = int(I)
  GO, UP, AND = int(G + O), int(U + P), int(A + N + D)
  if GO - I == UP - GO == AND - UP:
    POUND = int(P + O + U + N + D)
    # Look for a term with a value of 10000
    for n in range (1,500):
      # nth term of an arithmetic series
      if I + (n - 1)*(UP - GO) == 10000:
        print(f"POUND = {POUND}")
        print(f"I={I}, GO={GO}, UP={UP}, AND={AND}")
                     
# POUND = 28706
# I=4, GO=38, UP=72, AND=106   
```
  LikeLike
Jim Randell 9:25 am on 30 June 2020 Permalink | Reply
Tags: by: W Blackman ( 3 )
Brain-Teaser 524: [Digit sums]
From The Sunday Times, 27th June 1971 [link]

At a party at my friend Smith’s house recently, we were watching a rather attractive couple who were evidently enjoying each other’s company.

“I don’t know those two”, I said, “but I should be interested to know the lady’s age”.

“Well, I’m not the man to be explicit about that subject”, said Smith, “even though I know she’s rather proud of it. But her age is the sum of the digits in the sum of the digits in all the numbers from one to a hundred thousand inclusive”.

“Certainly not very explicit”, I said, “but then, how old is her companion?”

“Believe it or not”, said Smith, “but his age is the sum of the digits in the sum of the digits in all the numbers from one to a million inclusive”.

What was the difference between their ages?

This puzzle was originally published with no title.

[teaser524]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 9:26 am on 30 June 2020 Permalink | Reply
  It is not too onerous to tackle the problem directly using Python.
  
  This program counts the digits in each of the required sequences. It runs in 405ms.
  
  Run: [ @replit ]
```
from enigma import (fcompose, nsplit, irange, printf)

# sum the digits in a number
dsum = fcompose(nsplit, sum)

# sum the digits of the numbers in a sequence
dsums = lambda s: sum(map(dsum, s))

# age of the lady
a1 = dsum(dsums(irange(1, 100_000)))

# age of the gentleman
a2 = dsum(dsums(irange(1, 1_000_000)))

printf("a1={a1} a2={a2}, diff={d}", d=abs(a1 - a2))
```
  Solution: They are the same age, so the difference is 0.
  
  But it is also fairly straightforward so solve analytically:
  
  If we suppose DS(k) is the sum of the digits in the integers (represented in base 10), starting from 0, that are less than k, then:
  
  The sum of the digits in 0 .. 9 is 45.
  
  DS(10^1) = 45
  
  If we then consider numbers from 0 .. 99, the units digits comprise 10 lots of DS(10), and the tens digits comprise another 10 lots of DS(10):
  
  DS(10^2) = 20 DS(10^1) = 900
  
  For the digits from 0 .. 999, the tens and units digits together comprise 10 lots of DS(100), and the hundreds digits comprise 100 lots of DS(10).
  
  DS(10^3) = 10 DS(10^2) + 100 DS(10^1) = 13,500
  
  Carrying on this construction we find:
  
  DS(10^n) = 45n × 10^(n − 1)
  
  (See: OEIS A034967 [ @oeis ]).
  
  The sums we are interested in are therefore:
  
  DS(10^5) = 45×5×(10^4) = 2,250,000
  DS(10^6) = 45×6×(10^5) = 27,000,000
  
  The extra 0 at the beginning doesn’t make any difference, but we need to add in the digits for the end points, which are both powers of 10, so only contribute an extra 1.
  
  So we are interested in the digit sums of 2,250,001 and 27,000,001, both of which are 10.
  
  So both ages are 10, and their difference is 0.
  
  LikeLike

Jim Randell 8:10 am on 28 June 2020 Permalink | Reply
Tags: by: Colin

Brainteaser 1822: Teasing triangles

From The Sunday Times, 17th August 1997 [link]

Each of the ten line segments in the picture have been given a different whole number value from 0 to 9. A figure shown in a triangle equals the sum of the three values assigned to the sides of the triangle. The figures which have not been given from the other two triangles are the same as each other and the two add together to give a perfect square.

What are the numbers assigned to each side?

This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.

[teaser1822]

Jim Randell 8:11 am on 28 June 2020 Permalink | Reply
We can use the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve this puzzle.

The following run file executes in 96ms.

Run: [ @repl.it ]
```
#! python -m enigma -rr

SubstitutedExpression

# regions given
"A + F + G = 9"
"E + F + H = 7"
"C + I + J = 23"

# remaining regions have the same value...
"B + G + J == D + H + I"
# ... and sum to a square
"is_square(B + D + G + H + I + J)"

# suppress verbose output
--template=""
```
Solution: The numbers assigned to each line are shown below:

LikeLike
- Jim Randell 12:38 pm on 30 June 2020 Permalink | Reply
  The puzzle as originally published in The Sunday Times was worded as follows:
  
  Each of the ten line segments in the picture have been given a different whole number value from 0 to 9. A figure shown in a triangle equals the sum of the three values assigned to the sides of the triangle. The figures which have not been given from the other two triangles when added together to give 30. What is the value assigned to CD?
  
  CD being the external edge of the 23 triangle (which is the value C in my solution).
  
  We can adjust the run-file accordingly:
```
#! python -m enigma -rr

SubstitutedExpression

# regions given
"A + F + G = 9"
"E + F + H = 7"
"C + I + J = 23"

# the other two triangles added together to give 30
"B + D + G + H + I + J = 30"

--answer="C"

# suppress verbose output
--template=""
```
  And we find that there are 24 ways to fill out the numbers, but the value on the required edge is always 9.
  
  LikeLike

GeoffR 4:09 pm on 29 June 2020 Permalink | Reply


% A Solution in MiniZinc
include "globals.mzn";

% using the vertex labels A,B,C,D and E (as original diagram in link)
% plus F for the central point

% outside edges of the pentagon
var 0..9:AB; var 0..9:BC; var 0..9:CD; var 0..9:DE; var 0..9:EA;

% internal lines to central point F
var 0..9:AF; var 0..9:BF; var 0..9:CF; var 0..9:DF; var 0..9:EF;

constraint all_different ([AB, BC, CD, DE, EA, AF, BF, CF, DF, EF]);

% three given triangles are AEF, ABF and CDF
constraint AF + EF + EA == 7;
constraint AB + AF + BF == 9;
constraint CF + DF + CD == 23;

% the other two triangle sides (BCF and DEF) are the same total 
% and the two add together to give a perfect square

constraint DE + EF + DF == BF + CF + BC;

set of int: sq = {16, 25, 36, 49, 64, 81};
constraint (DE + EF + DF + BF + CF + BC) in sq;

solve satisfy;

% AB = 0; BC = 3; CD = 6; DE = 5; EA = 1;
% AF = 2; BF = 7; CF = 8; DF = 9; EF = 4;
% % time elapsed: 0.06 s
% ----------
% ==========

LikeLike

GeoffR 1:46 pm on 8 July 2020 Permalink | Reply

My solution below is updated for the original Sunday Times description, the solution above being the solution being based on the teaser in the book Brainteasers (2002, edited by Victor Bryant).


% A Solution in MiniZinc - original Sunday Times teaser
include "globals.mzn";
 
% using the vertex labels A,B,C,D and E (as original ST diagram in link)
% plus F for the central point
 
% outside edges of the pentagon
var 0..9:AB; var 0..9:BC; var 0..9:CD; var 0..9:DE; var 0..9:EA;
 
% internal lines to central point F
var 0..9:AF; var 0..9:BF; var 0..9:CF; var 0..9:DF; var 0..9:EF;
 
constraint all_different ([AB, BC, CD, DE, EA, AF, BF, CF, DF, EF]);
 
% three given triangles are AEF, ABF and CDF
constraint AF + EF + EA == 7;
constraint AB + AF + BF == 9;
constraint CF + DF + CD == 23;

% for figures in triangles not given ie triangles DEF and BCF
constraint DE + EF + DF + BF + CF + BC == 30; 
 
solve satisfy;
output ["CD = " ++ show(CD) ];

% CD = 9
% ----------
% ==========

LikeLike

Jim Randell 5:08 pm on 26 June 2020 Permalink | Reply
Tags: by: Danny Roth ( 84 )
Teaser 3014: Family business
From The Sunday Times, 28th June 2020 [link] [link]

George and Martha run a company with their five daughters. The telephone extensions all have four positive unequal digits and strangely only four digits appear in the seven extensions:

Andrea: ABCD
Bertha: ACDB
Caroline: BACD
Dorothy: DABC
Elizabeth: DBCA
George: CABD
Martha: CDAB

They noticed the following:

Andrea’s and Bertha’s add up to Dorothy’s.
Bertha’s and Elizabeth’s add up to George’s.
Caroline’s and Dorothy’s add up to Martha’s.

What is Andrea’s extension?

[teaser3014]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
GeoffR and Jim Randell are discussing. Toggle Comments
- Jim Randell 5:14 pm on 26 June 2020 Permalink | Reply
  Any one of the expressions is sufficient to determine the answer.
  
  I used the [[ SubstitutedExpression.split_sum() ]] solver from the enigma.py library to evaluate the alphametic expressions.
  
  The following run file executes in 77ms. (Internal runtime of the generated code is just 39µs).
  
  Run: [ @replit ]
```
#! python3 -m enigma -rr

SubstitutedExpression.split_sum

--invalid="0,ABCD"

"ABCD + ACDB = DABC"
"ACDB + DBCA = CABD"
"BACD + DABC = CDAB"

--answer="ABCD"
```
  Solution: Andrea’s extension is 2385.
  
  Here’s my manual solution:
  
  We can split each of the three sums into columns to give us partial equations, that may have a carry out or a carry in.
  
  B + D = C appears at both the left- and right-hand side of a sum, so can’t have a carry in or a carry out, so it is a complete equation.
  
  We then see that the partial sum BC + CD = AB cannot have a carry in, so the sum A + B = D is also a complete equation.
  
  The sum A + A = D cannot have a carry out, and if it does not have a carry in then we deduce that B = A, which is not possible. So it does have a carry in, and B = A + 1 is a complete equation.
  
  Using these three complete equations we derive:
  
  B = A + 1
  D = 2A + 1
  C = 3A + 2
  
  For digits in the range 1-9 the only viable values are:
  
  A = 1, B = 2, C = 5, D = 3
  A = 2, B = 3, C = 8, D = 5
  
  Only one of these assignment works in the original three sums, so we have the answer.
  
  Or, we can use the partial equation BC + CD = AB, which we now know must have a carry out, to give a fourth complete equation: 9B + 11C + D = 10A + 100. We can then substitute in the other equations to get the value of A, and from that the remaining values:
  
  9B + 11C + D = 10A + 100
  9(A + 1) + 11(3A + 2) + (2A + 1) = 10A + 100
  44A + 32 = 10A + 100
  34A = 68
  ⇒
  A = 2, B = 3, C = 8, D = 5
  
  LikeLike
- GeoffR 9:19 am on 27 June 2020 Permalink | Reply
  I simulated a normal addition sum in MiniZinc for the first equation to find an answer.
  
  The values obtained confirmed the other two equations.
```
% A Solution in MiniZinc
include "globals.mzn";
 
% Using the first equation only ie Andrea + Bertha = Dorothy
% 
%  A B C D
%  A C D B
%  -------
%  D A B C
%  -------

var 1..9:A; var 1..9:B; var 1..9:C; var 1..9:D;
var 0..1:carry1; var 0..1:carry2; var 0..1:carry3; 

constraint C == (B + D) mod 10 
/\ carry1 = (B + D) div 10;

constraint B = (carry1 + C + D) mod 10 /\ 
carry2 = (carry1 + C + D) div 10;

constraint A = (carry2 + B + C) mod 10
/\ carry3 = (carry2 + B + C) div 10;

constraint D = carry3 + 2*A;

solve satisfy;

output ["A,B,C,D = " ++ show([A,B,C,D]) ];
```
  LikeLike
- GeoffR 2:41 pm on 27 June 2020 Permalink | Reply
  A short Python programme confirms that the second and third equations give the same value for Andrea’s extension as my MiniZinc programme for the first equation:
```
from itertools import permutations

# Bertha’s and Elizabeth’s extensions add up to George’s extn.
# 2nd Equation: ACDB + DBCA = CABD

for p in permutations((1,2,3,4,5,6,7,8,9),4):
    a, b, c, d = p
    acdb = 1000*a + 100*c + 10*d + b
    dbca = 1000*d + 100*b + 10*c + a
    cabd = 1000*c + 100*a + 10*b + d
    if acdb + dbca == cabd:
        # Find Andrea's extension
        abcd = 1000*a + 100*b + 10*c + d
        print("2. Andrea's extension =", abcd)

# Caroline’s and Dorothy’s extensions add up to Martha’s extn.
# 3rd Equation : BACD + DABC = CDAB

for p in permutations((1,2,3,4,5,6,7,8,9),4):
    a, b, c, d = p
    bacd = 1000*b + 100*a + 10*c + d
    dabc = 1000*d + 100*a + 10*b + c
    cdab = 1000*c + 100*d + 10*a + b
    if bacd + dabc == cdab:
        # Find Andrea's extension
        abcd = 1000*a + 100*b + 10*c + d
        print("3. Andrea's extension =", abcd)
```
  LikeLike

Jim Randell 7:16 am on 25 June 2020 Permalink | Reply
Tags: by: H. Wright ( 2 )

Brain-Teaser 1: Tall story

From The Sunday Times, 26th February 1961 [link]

“We’re a biggish family in more ways than one”, said Jeremy. “Five, including triplets, and three of us the same height. I’m 6 ft. and aged 72; Jim is three inches taller than John and three years older; and John and Julian’s combined heights are 5 ft. 11 in. more than Joe’s and their combined ages 71 years more than his. Our aggregate height in inches equals our aggregate age in years, but no one’s age in years equals Joe’s height in inches”.

What were the name, age and height of the tallest of the triplets?

This was the first of the regular Teaser puzzles published in The Sunday Times. It was accompanied by the following introduction:

Brain-Teasers: a Weekly Feature

Great numbers of our readers have found entertainment and interest in the Brain-Teasers, or mathematical problems, which we have published from time to time, usually at holiday week-ends. Now we intend to make them a weekly feature of The Sunday Times.

Readers themselves have supplied many of the best of the problems in the past, and we invite them to continue to do so. A fee of £10 will be paid for each problem accepted.

Problems should fulfil the following conditions: Both the problem and the solution must be expressible in reasonably brief compass. No advanced or specialist mathematical techniques should be necessary. Solutions should be unique, or otherwise admit of no uncertainty in judging. Problems must be original. Diagrams are admissible if they are not too complicated.

The problem below is the invention of a reader who observes: “Any who are stumped by it after the expiry of an hour should feel cause for concern”.

A prize of £3 was offered.

[teaser1]

Jim Randell 7:17 am on 25 June 2020 Permalink | Reply

We can associate values (A, B, C, D, E) with (Jez, Jim, Jon, Jul, Joe), that correspond to either their heights in inches or their ages in years, both give rise to the same set of equations:

A = 72
B = C + 3
C + D = E + 71
x = y = z

where (x, y, z) are some three of (A, B, C, D, E).

This gives us a set of 5 simultaneous equations in 5 variables.

This Python program finds solutions to these equations (using the [[ Matrix.linear() ]] solver from the enigma.py library), groups the solutions by the sum of the set of values, and then looks for two sets with the same sum (one set for the ages, and another set for the height) that satisfy the condition that Joe’s height does not appear in the set of ages.

It runs in 55ms.

Run: [ @replit ]

from enigma import (Matrix, subsets, as_int, multiset, group, printf)

# labels for the people involved
# A = Jez, B = Jim, C = Jon, D = Jul, E = Joe
labels = (A, B, C, D, E) = (0, 1, 2, 3, 4)

# solve equations specified as: (ps, ms, k)
# for integer solutions
def solve(eqs):
  (A, B) = (list(), list())
  for (ps, ms, k) in eqs:
    eq = [0] * 5
    for (i, n) in multiset.from_seq(ps).items(): eq[i] += n
    for (i, n) in multiset.from_seq(ms).items(): eq[i] -= n
    A.append(eq)
    B.append(k)
  try:
    # find values that are positive integers
    return Matrix.linear(A, B, valid=(lambda x: as_int(x, include='+')))
  except ValueError:
    return

# find values that satisfy the equations
def generate():
  # choose the three with the same value
  for (x, y, z) in subsets(labels, size=3):
    # solve the equations
    r = solve([
      ([A], [], 72), # A = 72
      ([B], [C], 3), # B = C + 3
      ([C, D], [E], 71), # C + D = E + 71
      ([x], [y], 0), # x = y
      ([y], [z], 0), # y = z
    ])
    if r is None: continue
    # check there is a set of triplets
    if not (3 in multiset.from_seq(r).values()): continue
    # return candidate solutions
    yield r

# group sets of solutions by the sum of the values
d = group(generate(), by=sum)

# consider values for the sum
for (t, rs) in d.items():
  # choose two sets of values for the ages and heights
  for (age, height) in subsets(rs, size=2, select="M"):
    # no one's age is the same as Joe's height
    if height[E] in age: continue
    for (n, a, h) in zip("Jez Jim Jon Jul Joe".split(), age, height):
      printf("{n}: age {a} yr, height {h} in")
    printf()

Solution: Jim is the tallest of the triplets. His age is 72, his height is 6ft 2in.

The complete characteristics are:

Jez: age 72 yr, height 72 in
Jim: age 72 yr, height 74 in
Jon: age 69 yr, height 71 in
Jul: age 74 yr, height 71 in
Joe: age 72 yr, height 71 in

So: Jeremy, Jim, Joe are the triplets (same age), and: John, Julian, Joe are the same height.

There are 7 ways to solve the equations, but there is only one pair that has the same sum, and this gives rise to the solution.

LikeLike

GeoffR 4:10 pm on 25 June 2020 Permalink | Reply

I found three solutions in MinZinc.

The first solution was invaiid as there was no single maximum height.

The second solution was the same as Jim’s solution above.

There appears to be another solution, which gives the maximum height of Jim as 6ft 3 in, with an age of 75 years. If the puzzle had stated that age and height could not be the same, then this last solution could be ruled out – it appears to conform to the stated conditions in the puzzle.


% A Solution in MiniZinc
include "globals.mzn";
 
var 10..99: JezAge; var 10..99: JimAge; var 10..99: JonAge; 
var 10..99: JulAge; var 10..99: JoeAge; 

var 50..80: JezHt; var 50..80: JimHt; var 50..80: JonHt; 
var 50..80: JulHt; var 50..80: JoeHt; 

% Three of us the same height
% (A,B,C,D,E) = (Jez, Jim, Jon, Jul, Joe)
constraint (JezHt == JimHt /\ JimHt == JonHt)  % ABC
\/ (JezHt == JimHt /\ JimHt == JulHt)   % ABD
\/ (JezHt == JimHt /\ JimHt == JoeHt)   % ABE
\/ (JezHt == JonHt /\ JonHt == JulHt)   % ACD
\/ (JezHt == JonHt /\ JonHt == JoeHt)   % ACE
\/ (JezHt == JulHt /\ JulHt == JoeHt)   % ADE
\/ (JimHt == JonHt /\ JonHt == JulHt)   % BCD
\/ (JimHt == JonHt /\ JonHt == JoeHt)   % BCE
\/ (JimHt == JulHt /\ JulHt == JoeHt)   % BDE
\/ (JonHt == JulHt /\ JulHt == JoeHt);  % CDE

% Three of us the same age(triplets)
constraint (JezAge == JimAge /\ JimAge == JonAge)  % ABC
\/ (JezAge == JimAge /\ JimAge == JulAge)   % ABD
\/ (JezAge == JimAge /\ JimAge == JoeAge)   % ABE
\/ (JezAge == JonAge /\ JonAge == JulAge)   % ACD
\/ (JezAge == JonAge /\ JonAge == JoeAge)   % ACE
\/ (JezAge == JulAge /\ JulAge == JoeAge)   % ADE
\/ (JimAge == JonAge /\ JonAge == JulAge)   % BCD
\/ (JimAge == JonAge /\ JonAge == JoeAge)   % BCE
\/ (JimAge == JulAge /\ JulAge == JoeAge)   % BDE
\/ (JonAge == JulAge /\ JulAge == JoeAge);  % CDE

% Jez is 6 ft. and aged 72
constraint JezHt == 72 /\ JezAge == 72;

% Jim is three inches taller than John and three years older
constraint JimHt - 3 == JonHt /\ JimAge - 3 == JonAge;

%  John and Julian’s combined heights are 5 ft. 11 in. more than Joe’s
%  and their combined ages 71 years more than his
constraint JonHt + JulHt - 71 == JoeHt;  
constraint JonAge + JulAge - 71 == JoeAge;

% Our aggregate height in inches equals our aggregate age in years
constraint  JimHt + JezHt + JulHt + JonHt + JoeHt 
== JimAge + JezAge + JulAge  + JonAge + JoeAge;

% No one’s age in years equals Joe’s height in inches
constraint JoeHt != JezAge /\ JoeHt != JimAge
/\ JoeHt != JonAge /\ JoeHt != JulAge;

solve satisfy;

output ["Ages: " ++ show([JezAge, JimAge, JonAge, JulAge, JoeAge])]++
["\nHeights: " ++ show([JezHt, JimHt, JonHt, JulHt, JoeHt])];

%           Jez Jim Jon Jul Joe
% Ages:    [72, 72, 69, 72, 70]  << no single tallest member 
% Heights: [72, 72, 69, 72, 70]  << Therefore, an invalid solution
% ----------
% Ages:    [72, 72, 69, 74, 72]  <<  Jim is tallest member at 6 ft. 2 in.
% Heights: [72, 74, 71, 71, 71]  <<  with an age of 72 years
% ----------
% Ages:    [72, 75, 72, 72, 73]  << Jim is tallest member at 6 ft. 3 in.
% Heights: [72, 75, 72, 72, 73]  <<  with an age of 75 years
% ----------
% ==========

LikeLike

Jim Randell 4:15 pm on 25 June 2020 Permalink | Reply

@Geoff: The condition “no one’s age in years equals Joe’s height in inches”, means that Joe’s age is not equal to Joe’s height. So this disposes of two of your solutions. (And effectively means that the same set of solutions cannot be used for both ages and heights).

LikeLike

GeoffR 4:38 pm on 25 June 2020 Permalink | Reply

@Jim: Fair point. I was interpreting only other people’s ages only could not be equal to Joe’s height. It could have been more clearly stated that this condition also applied to Joe himself!

LikeLike

Frits 3:51 pm on 28 July 2020 Permalink | Reply

from enigma import SubstitutedExpression, icount #, multiset

# suppose Jez not part of three --> Also either Jon or Jim not part of three 
#   --> Joe and Jul part of three --> Jez = 72, Jon = 71, Jim 74, Joe = 71, Jul = 71 or
#       (Jon has to be 71)            Jez = 72, Jon = 71, Jim 74, Joe = 74, Jul = 74 
# suppose Jez part of three 
#   --> if both Jim and Jon not part of three --> Jon = 71, Jim = 74
#       else Jim or Jon is 72 --> Jez = 72, Jon = 72, Jim 75, Jul == Joe - 1 or
#                                 Jez = 72, Jon = 69, Jim 72, Jul == Joe + 2
# --> Joe 70-74
# --> Jim 72,74,75
# --> Jul 71,72,74
# --> Jon 69,71,72
    

# the alphametic puzzle
p = SubstitutedExpression(
  [
    #  Jez Age    Joe Age    Jim Age    Jul Age    Jon Age
    #    AB         CD         EF         GH         IJ     
    #  Jez Hght   Joe Hght   Jim Hght   Jul Hght   Jon Hght
    #    KL         MN         OP         QR         ST   
    
    # Jez is 6 ft. and aged 72
    "AB = 72", "KL = 72",
  
    # Jim is three inches taller than John and three years older
    "IJ + 3 == EF" ,"ST + 3 = OP",
   
    # Jul and Jon's combined heights are 5 ft 11 in more than Joe's
    # and their combined ages 71 years more than his
    "QR + ST == MN + 71", "GH + IJ == CD + 71",
    
    # aggregate height in inches equals aggregate age in years
    "AB + CD + EF + GH + IJ == KL + MN + OP + QR + ST",
    
    # No one's age in years equals Joe's height in inches
    "AB != MN", "CD != MN", "EF != MN", "GH != MN", "IJ != MN",
    
    # Three have the same age, three have same height
    "icount([AB, CD, EF, GH, IJ], (lambda d: v([AB, CD, EF, GH, IJ]) == d)) == 3",
    "icount([KL, MN, OP, QR, ST], (lambda d: v([KL, MN, OP, QR, ST]) == d)) == 3",
    #"sorted(multiset.from_seq([AB, CD, EF, GH, IJ]).values()) == [1, 1, 3]",
    #"sorted(multiset.from_seq([KL, MN, OP, QR, ST]).values()) == [1, 1, 3]",
    
    # ----------------------------
    # try to speed up 
    # ----------------------------
    
    #"CD in ([70,71,72,73,74])",
    #"EF in ([72,74,75])",
    #"GH in ([71,72,74])",
    #"IJ in ([69,71,72])",
    
    #"MN in ([70,71,72,73,74])",
    #"OP in ([72,74,75])",
    #"QR in ([71,72,74])",
    #"ST in ([69,71,72])",
  ],
  distinct="", 
  code="v = lambda x: max(x,key=x.count)",
  answer="AB, CD, EF, GH, IJ, KL, MN, OP, QR, ST"
)      
                          
# Print answers

for (_, ans) in p.solve(verbose=0): 
   print("        Age        |       Height") 
   print("Jez Joe Jim Jul Jon Jez Joe Jim Jul Jon")
   print(ans)
   
# Output:   
#         Age        |       Height
# Jez Joe Jim Jul Jon Jez Joe Jim Jul Jon
# (72, 72, 72, 74, 69, 72, 71, 74, 71, 71)

Remarks regarding speed:

"IJ + 3 == EF" ,"ST + 3 = OP",  # Approx 1.3 secs
"IJ + 3 = EF" ,"ST + 3 == OP",  # Approx 14 secs
"IJ + 3 = EF" ,"ST + 3 = OP",   # Approx 14 secs
"IJ + 3 == EF" ,"ST + 3 == OP", # SyntaxError: too many statically nested blocks

The speed up logic I had to invent before I discovered the “IJ + 3 == EF” ,”ST + 3 = OP” equations.
In theory the multiset equations are too restrictive

LikeLike

Jim Randell 8:52 am on 23 June 2020 Permalink | Reply
Tags: by: DJT Hogg ( 3 )
Teaser 2525: [Sandwich numbers]
From The Sunday Times, 13th February 2011 [link] [link]

“Sandwich numbers” are those of the form BFB, where B is the bread and is a single digit, and F is the filling of any length: BFB has to be divisible by F. For example, 11371 is a sandwich with filling 137 since 11371 = 83 × 137. Incidentally, all of 21372, 31373, … , 91379 are also sandwich numbers, and these nine are said to make up a “sandwich box”.

The filling in my sandwich box today is the smallest beginning with a 3.

What is my filling?

This puzzle was originally published with no title.

[teaser2525]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 8:58 am on 23 June 2020 Permalink | Reply
  My first thought was to generate integers that start with a 3, and then check to see if they make a “sandwich box”. But this was taking a long time, so here’s some analysis to produce a faster program:
  
  If f is an n digit “filling” number, then the corresponding sandwich numbers in a sandwich box are:
  
  b.10^(n + 1) + 10.f + b
  
  for b = 1 .. 9. And each of these must be divisible by f.
  
  Clearly f will divide the 10.f part, which leaves b(10^(n + 1) + 1), when b = 1, we must have f divides (10^(n + 1) + 1), and if this is the case f will divide the remaining sandwich numbers with b > 1.
  
  So, f must be a divisor of (10^(n + 1) + 1), and we are looking for the smallest f that starts with a 3.
  
  f.g = 10^(n + 1) + 1
  
  and f is an n digit number that starts with 3, so g = 26 .. 33.
  
  This Python program runs in 49ms. (Internal runtime is 46µs).
  
  Run: [ @replit ]
```
from enigma import (irange, inf, printf)

def solve():
  # consider n digit fillings
  x = 101
  for n in irange(1, inf):
    # f is an n digit divisor of x
    for g in irange(33, 26, step=-1):
      (f, r) = divmod(x, g)
      if r == 0:
        yield f
    x *= 10
    x -= 9

for f in solve():
  printf("f = {f}")
  break
```
  Solution: The smallest filling beginning with 3 is: f = 3448275862069.
  
  Where: 29f = 10^14 + 1.
  
  So we see why the simple program was taking so long. (In the end it took 32 minutes).
  
  In fact there is a family of filling numbers that start with 3 that take the form:
  
  344827586206 [8965517241379310344827586206] 9
  
  where the section in brackets can be included 0 or more times, to give fillings with length (13 + 28k) digits.
  
  We can adapt the program to generate all filling numbers without the restriction that the leading digit is 3.
```
from enigma import (irange, inf, match, printf)

# generate "filling" numbers
def filling():
  # consider n digit fillings
  x = 101
  for n in irange(1, inf):
    # f is an n digit divisor of x
    M = x // 10
    for g in irange(101, 11, step=-1):
      (f, r) = divmod(x, g)
      if r == 0 and f < M <= 10 * f:
        yield f
    x *= 10
    x -= 9

for f in filling():
  printf("f = {f}")
  if match(f, '3*'): break
```
  See: OIES A116436 [ @oeis.org ]
  
  LikeLike
Jim Randell 9:48 am on 21 June 2020 Permalink | Reply
Tags: by: R Postill ( 20 )
Brain-Teaser 523: [Alphabet cricket]
From The Sunday Times, 20th June 1971 [link]

The Alphabet Cricket Club (26 playing members, every surname having a different initial letter) ran into trouble last season. The selected 1st XI consisted of those with initials A to K; the 2nd XI was to be picked from the remainder, but personal feeling had crept in. (To save space, players are referred to by initials).

U refused to play in the same team as Z. Q said that he would not be available if either L or M was picked. P announced that he would not turn out (a) unless M was in the side, (b) if either O or W played. Z declined to play in an XI containing more than one of R, S, T. Finally, V agreed to play only with Q.

The secretary managed to pick an XI but then a further complication arose: two of the 1st XI fell ill and two of the selected 2nd XI were taken to fill the gaps. Fortunately the secretary was still able to field a 2nd XI (which, incidentally, won, M making a century).

Who were the two promoted to the 1st XI, and who were omitted altogether?

This puzzle was originally published with no title.

[teaser523]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
John Crabtree and Jim Randell are discussing. Toggle Comments
- Jim Randell 9:49 am on 21 June 2020 Permalink | Reply
  This Python program runs in 52ms.
```
from enigma import (subsets, intersect, diff, join, printf)

players = "LMNOPQRSTUVWXYZ"

# generate possible 2nd XI's
def generate():
  for ps in subsets(players, size=11):

    # verify conditions:

    # "U refused to play in the same team as Z"
    if 'U' in ps and 'Z' in ps: continue

    # "Q said that he would not be available if either L or M was picked"
    if 'Q' in ps and ('L' in ps or 'M' in ps): continue

    # "P announced that he would not turn out (a) unless M was in the
    # side, (b) if either O or W played"
    if 'P' in ps and ('M' not in ps or 'O' in ps or 'W' in ps): continue

    # "Z declined to play in an XI containing more than one of R, S, T"
    if 'Z' in ps and len(intersect((ps, "RST"))) > 1: continue

    # "V agreed to play only with Q"
    if 'V' in ps and 'Q' not in ps: continue

    yield ps

# choose two sets for the second eleven
for (s1, s2) in subsets(generate(), size=2, select="P"):

  # M is in the final set
  if 'M' not in s2: continue

  # the two promoted are in the first but not the second
  ps = diff(s1, s2)
  if len(ps) != 2: continue

  # the ones who were never picked
  ns = diff(players, s1 + s2)
  printf("promoted={ps} omitted={ns} [{s1} -> {s2}]", ps=join(ps), ns=join(ns), s1=join(s1), s2=join(s2))
```
  Solution: Q and V were promoted. P and Z were omitted.
  
  The original 2nd XI was: N O Q R S T U V W X Y.
  
  And then Q and V were promoted to the 1st XI. L and M stepped in to take their places.
  
  So the actual 2nd XI was: L M N O R S T U W X Y.
  
  P and Z were omitted from both line-ups.
  
  V will play only with Q, so it makes sense that Q and V were promoted together. (Although I didn’t check that the requirements for the 2nd XI were satisfied by the new 1st XI, but in this case they would be).
  
  These are the only two possible line-ups allowed for the 2nd XI for the given conditions, so the one with M in must be the actual line-up.
  
  LikeLike
- John Crabtree 8:06 pm on 25 June 2020 Permalink | Reply
  
  Four players were initially not picked.
  From the two statements involving Z, if Z plays, U does not play and at least two of R, S and T do not play.
  
  The two statements involving Q may be written as:
  Q + ~Q.~V = 1
  Q.~L.~M + ~Q = 1
  Combining them gives Q.~L.~M + ~Q.~V = 1
  Either L and M do not play, or Q and V do not play, and so Z does not play,
  The statements involving P may be written as P.M.~O.~W + ~P = 1
  If P plays, Q, V, O, W and Z do not play. And so P does not play.
  
  And so Q and V were promoted. P and Z did not play.
  L and M were brought in to the 2nd XI as replacements.
  
  LikeLike
Jim Randell 7:06 pm on 19 June 2020 Permalink | Reply
Tags: by: Howard Williams ( 43 )
Teaser 3013: Arian Pen-blwydd
From The Sunday Times, 21st June 2020 [link] [link]

When I thought that my daughter was old enough to be responsible with money I gave her on her next, and all subsequent birthdays, cash amounts (in pounds) which were equal to her birthday age squared.

On her last birthday her age was twice the number of years for which she received no such presents. I calculated at this birthday that if I had made these gifts on all of her birthdays then she would have received 15% more than she had actually received. I then decided that I would stop making the payments after her birthday when she would have received only 7.5% more if the payments had been made on all of her birthdays.

What was the amount of the final birthday payment?

There are now 300 puzzles available on the site.

[teaser3013]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 9:43 pm on 19 June 2020 Permalink | Reply
  If she didn’t receive gifts for the first k years, then the “missing” gifts are the sum of the squares of 1 .. k. The amounts actually received are the sum of the squares of (k + 1) .. 2k.
  
  This Python program finds the value of k when the amount of the missing gifts is 15% of the actual amount, and then continues looking at future gifts until the amount becomes 7.5%.
  
  It runs in 53ms.
  
  Run: [ @repl.it ]
```
from enigma import irange, inf, printf

# solve the puzzle
def solve():
  # consider the k years before the gifts started
  for k in irange(1, inf):
    # total before amount
    before = sum(x * x for x in irange(1, k))
    # total after amount
    after = sum(x * x for x in irange(k + 1, 2 * k))
    # before = 0.15 * after
    if not(100 * before == 15 * after): continue
    printf("k = {k}; before = {before}, after = {after}")

    # now add in future gifts
    future = after
    for n in irange(2 * k + 1, inf):
      future += n * n
      # before = 0.075 * future
      if not(1000 * before == 75 * future): continue
      printf("-> n = {n}; n^2 = {n2}, future = {future}", n2=n * n)
      return

solve()
```
  Solution: The final gift was £ 1,764.
  
  The gifts started on the 18th birthday, so the “missing” gifts (years 1 – 17) would amount to £ 1,785.
  
  The actual gifts between ages 18 and 34 amount to £ 11,900, and 15% of £ 11,900 is £ 1,785.
  
  The gifts are to continue to age 42, making the total amount £ 23,800, and 7.5% of £ 23,800 is also £ 1,785.
  
  Which means the final gift is made on the 25th (silver) anniversary of the first gift.
  
  Analytically:
  
  (See: Enigma 1086).
  
  The sum of the first n squares is given by the square pyramidal numbers:
  
  SP(n) = n (n + 1) (2n + 1) / 6
  
  So the first part of the puzzle is to solve:
  
  SP(k) = 0.15 (SP(2k) − SP(k))
  20 SP(k) = 3 (SP(2k) − SP(k))
  23 SP(k) = 3 SP(2k)
  23k (k + 1) (2k + 1) = 6k (2k + 1) (4k + 1)
  23k + 23 = 24k + 6
  k = 17
  
  The second part is to solve, for n > 34:
  
  SP(k) = 0.075 (SP(n) − SP(k))
  3 SP(n) = 43 SP(k)
  SP(n) = 25585
  n = 42
  
  The required answer is then: n² = 42² = 1764.
  
  LikeLike
  - Jim Randell 11:19 pm on 20 June 2020 Permalink | Reply
    Or, more simply:
    
    We are looking for values of n, k where:
    
    SP(k) = (3/43) SP(n)
    SP(2k) = (23/43) SP(n)
    
    This Python program runs in 51ms.
    
    Run: [ @repl.it ]
```
from enigma import irange, inf, div, printf

# accumulate the square pyramidal numbers, map: SP[n] -> n
d = dict()
x = 0
for n in irange(1, inf):
  # calculate: x = SP(n)
  x += n * n
  d[x] = n
  # can we find a corresponding y = SP(k)?
  y = div(x * 3, 43)
  k = d.get(y, None)
  if k is None: continue
  # and verify z = SP(2k)
  z = div(x * 23, 43)
  k2 = d.get(z, None)
  if k2 is None or k2 != k * 2: continue
  printf("n^2={n2} [n={n} k={k}; SP[{k}]={y} SP[{k2}]={z} SP[{n}]={x}]", n2=n * n)
  break
```
    LikeLike
Jim Randell 7:12 am on 18 June 2020 Permalink | Reply
Tags: by: Nick MacKinnon ( 52 )
Brainteaser 1819: Early bath
From The Sunday Times, 27th July 1997 [link]

There are 20 teams in one country’s premier league. They each play each other once in the first half of the season, and then they each play each other a second time in the rest of the season. Each team plays each Saturday of the season, earning three points for a win and one point for a draw. At the end of the season the bottom three teams are relegated and the top team wins the league championship.

Last season was the most boring ever. It was possible to determine the relegated teams well before the end of the season. In fact it would be impossible in any season to be able to determine the three relegated teams in fewer weeks.

Three further Saturdays after the relegations were determined the league championship was also determined when the league leaders were in a 0-0 draw and then found that they were unassailable. There were great celebrations that night.

At that time, how many points did the current top two teams have?

This puzzle is included in the book Brainteasers (2002). The puzzle text above is taken from the book.

[teaser1819]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
John Crabtree and Jim Randell are discussing. Toggle Comments
- Jim Randell 7:12 am on 18 June 2020 Permalink | Reply
  
  After a tortuous chain of reasoning we arrive at the answer:
  
  In the first half of the season there are C(20, 2) = 190 matches, and then another 190 in the second half of the season. Each team plays 19 matches in the first half of the season, and another 19 matches in the second half of the season. The season lasts 38 weeks, with 10 matches played each week.
  
  At some point there are 3 teams doomed to relegation, as they cannot possibly catch up with any of the remaining 17 teams.
  
  Suppose in the first half of the tournament there are three teams that lose all their matches, except the matches they play amongst themselves, which are drawn.
  
  Each of the three doomed teams has only 2 points (from their draws with the other two doomed teams). If all the other 187 matches were won outright there are 187×3 = 561 points to distribute between the remaining 17 teams. This gives an average of 33 points per team. But, if one of the doomed teams were to win all their matches in the second half of the tournament they would end up with 2 + 3×19 = 59 points, so their relegation is not guaranteed by the end of the first half of the tournament.
  
  If we carry on for k weeks into the second half of the tournament, with each doomed team losing each of their matches, and each of the other matches being won, then how many weeks is it before enough points are accumulated, so that each of the 17 remaining teams are out of reach?
  
  There are (19 − k) weeks remaining and a doomed team could win all their remaining matches, so they would end with 2 + 3(19 − k) = 59 − 3k points.
  
  And the total number of points to be shared between the 17 remaining teams would be: 561 + 30k.
  
  This can happen when:
  
  (561 + 30k) / 17 > 59 − 3k
  561 + 30k > 1003 − 51k
  81k > 442
  k > 5.46…
  
  i.e. after the 6th week of the second half of the tournament, at the earliest.
  
  There would then be 13 weeks remaining, so a doomed team that had a sudden change of fortune could finish with 2 + 3×13 = 41 points. But by that stage in the tournament 561 + 30×6 = 741 points could have been scored by the remaining 17 teams, which is enough for each of them to have accumulated at 43 points each. So the doomed teams can indeed be doomed to relegation.
  
  Now after three more weeks, i.e. after the 9th week of the second half of the tournament, with 10 weeks remaining, the championship was determined. So the team with the most points must be more than 30 points ahead of the next highest team.
  
  If we go back to 6th week of the second half of the tournament, then 16 of the 17 non-doomed teams could have 42 points, and the other team could have 69 points. So they are 27 points ahead.
  
  If they win the following 2 weeks, and the week after that they draw (as we are told in the puzzle text), then after the 9th week they have 76 points, and in order to be unassailable the next highest team can have no more than 45.
  
  So now we need to make sure none of the other 16 teams can get more than 45 points by the end of week 9. An easy way is to suppose all the matches apart from the ones involving the future champions are drawn.
  
  Then at the end of the 7th week, the best any of the 16 challengers can do is 43 points. The future champions have 72 points, and there are 12 weeks remaining, so they are assailable.
  
  At the end of the 8th week, the best any of the challengers can do is 44 points. The future champions have 75 points, and there are 11 weeks remaining, so they are still assailable.
  
  At the end of the 9th week, the best any of the challengers can do is 45 points. The future champions have 76 points, and there are 10 weeks remaining, so they are now unassailable.
  
  So this is a viable scenario for the puzzle.
  
  Solution: The top team had 76 points, and the next highest team had 45 points.
  
  LikeLike
  - John Crabtree 5:14 pm on 19 June 2020 Permalink | Reply
    
    This teaser just works. After 6 weeks of the second half, 16 teams have 42 points and one has 69, ie a total of 741 for those teams and the maximum possible.
    
    LikeLike

Jim Randell 8:42 am on 16 June 2020 Permalink | Reply
Tags: by: Angela Newing ( 37 )

Teaser 2662: Number please

From The Sunday Times, 29th September 2013 [link] [link]

I finally got through to the operator of a large company and asked to be connected to an appropriate department. He tried eight different extension numbers before then finding the correct one. The eight numbers were 1933, 2829, 3133, 4630, 5089, 5705, 6358 and 6542. Curiously, each of these wrong numbers did have at least one correct digit of the correct extension in the correct position!

What was the correct extension number?

[teaser2662]

Jim Randell 8:42 am on 16 June 2020 Permalink | Reply

Here is a recursive program that starts with unassigned digits in all positions, and then at each step looks for a number that doesn’t already match and uses it to fill out one of the unassigned digits. It runs in 53ms.

Run: [ @repl.it ]

from enigma import (update, join, printf)

# find sequences that match ns in at least one position
# ds - candidate digits ('?' for unassigned)
# ns - remaining numbers to check
def solve(ds, ns):
  # are we done?
  if not ns:
    yield join(ds)
  else:
    # consider the digits of the next number
    ms = list(zip(ds, ns[0]))
    # if any digits already match move on to the next one
    if any(a == b for (a, b) in ms):
      yield from solve(ds, ns[1:])
    else:
      # set one of the unassigned digits
      for (i, (a, b)) in enumerate(ms):
        if a == '?':
          yield from solve(update(ds, [(i, b)]), ns[1:])

# numbers provided
numbers = "1933 2829 3133 4630 5089 5705 6358 6542".split()

# solve the puzzle, starting with 4 ?'s
for n in solve("????", numbers):
  printf("number = {n}")

Solution: The correct extension number is 6739.

A brute force search of all possible extension numbers is shorter, and only slightly slower:

from enigma import (subsets, join, printf)

# the numbers tried (each has at least one correct digit in the correct position)
numbers = "1933 2829 3133 4630 5089 5705 6358 6542".split()

# consider digits for the actual number
for ds in subsets("0123456789", size=4, select="M"):
  if all(any(a == b for (a, b) in zip(ds, n)) for n in numbers):
    printf("number = {n}", n=join(ds))

LikeLike

GeoffR 10:29 am on 6 October 2020 Permalink | Reply


% A Solution in MiniZinc 
include "globals.mzn";

var 0..9: A; var 0..9: B; var 0..9: C; var 0..9: D;

constraint all_different ([A,B,C,D]);

% Each of these wrong numbers did have at least one correct
% digit of the correct extension in the correct position

% Extn Try 1 - 1933 
constraint sum ([A==1,B==9,C==3,D==3]) > 0;

% Extn Try 2 - 2829 
constraint sum ([A==2,B==8,C==2,D==9]) > 0;

% Extn Try 3 - 3133 
constraint sum ([A==3,B==1,C==3,D==3]) > 0;

% Extn Try 4 - 4630 
constraint sum ([A==4,B==6,C==3,D==0]) > 0;

% Extn Try 5 - 5089 
constraint sum ([A==5,B==0,C==8,D==9]) > 0;

% Extn Try 6 - 5705 
constraint sum ([A==5,B==7,C==0,D==5]) > 0;

% Extn Try 7 - 6358  
constraint sum ([A==6,B==3,C==5,D==8]) > 0;

% Extn Try 8 - 6542 
constraint sum ([A==6,B==5,C==4,D==2]) > 0;

solve satisfy;

output ["The correct extension number is " ++
show(A),show(B),show(C),show(D) ];

% The correct extension number is 6739
% % time elapsed: 0.03 s
% ----------
% ==========

LikeLike

Jim Randell 12:18 pm on 14 June 2020 Permalink | Reply
Tags: by: Peter G Chamberlain ( 11 )

Teaser 2655: Sudoprime

From The Sunday Times, 11th August 2013 [link] [link]

The grid shows a cross-figure with two of its digits given. The eleven answers (five of them being “across” and the other six “down”) are all different prime numbers with no leading zeros. No digit appears more than once in any row, column or main diagonal.

What are the two largest of the eleven primes?

[teaser2655]

Jim Randell 12:19 pm on 14 June 2020 Permalink | Reply

See also: Enigma 1730, Enigma 1740, Enigma 1755.

I used the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve this puzzle. The condition that there are no repeated digits in the rows, columns or diagonals can be expressed using the [[ --distinct ]] parameter, which lets us specify multiple groups of symbols, where none of the groups may contain repeated digits.

The following run file executes in 93ms.

Run: [ @replit ]

#! python -m enigma -rr

SubstitutedExpression

#
#  A # # # B
#  C D # E F
#  # G H I #
#  J K # L M
#  N # # # P
#

# the two values we are given
--assign="G,8"
--assign="N,7"

# the 11 answers are all different prime numbers
"is_prime(AC)"
"is_prime(BF)"
"is_prime(CD)"
"is_prime(EF)"
"is_prime(JK)"
"is_prime(LM)"
"is_prime(JN)"
"is_prime(MP)"
"all_different(AC, BF, CD, EF, JK, LM, JN, MP)"

"is_prime(DGK)"
"is_prime(EIL)"
"is_prime(GHI)"
"all_different(DGK, EIL, GHI)"

# no digit appears more than once in any row, column, or main diagonal
--distinct="AB,CDEF,GHI,JKLM,NP,ACJN,DGK,EIL,BFMP,ADHLP,NKHEB"

# answer is the two largest primes
--answer="last(ordered(DGK, EIL, GHI), 2)"

# suppress verbose output
--template=""

Solution: The two largest primes are: 821, 983.

The completed grid looks like this:

LikeLike

GeoffR 3:21 pm on 14 June 2020 Permalink | Reply

% A Solution in MiniZinc - same grid as Jim
include "globals.mzn";

var 1..9:A; var 1..9:B; var 1..9:C; var 1..9:D; var 1..9:E; 
var 1..9:F; var 1..9:G; var 1..9:H; var 1..9:I; var 1..9:J; 
var 1..9:K; var 1..9:L; var 1..9:M; var 1..9:N; var 1..9:P; 

% Given values
constraint G == 8 /\ N == 7;

% Rows
var 11..97:CD = 10*C + D; 
var 11..97:EF = 10*E + F; 
var 11..97:JK = 10*J + K; 
var 11..97:LM = 10*L + M; 
var 101..997:GHI = 100*G + 10*H + I;
% Columns
var 11..97:AC = 10*A + C; 
var 11..97:JN = 10*J + N; 
var 101..997:DGK = 100*D + 10*G + K; 
var 101..997:EIL = 100*E + 10*I + L; 
var 11..97:BF = 10*B + F; 
var 11..97:MP = 10*M + P; 

predicate is_prime(var int: x) = 
x > 1 /\ forall(i in 2..1 + ceil(sqrt(int2float(ub(x))))) 
((i < x) -> (x mod i > 0));

% The eleven answers are all prime numbers
constraint is_prime(CD) /\ is_prime(EF) /\ is_prime(JK) /\ is_prime(LM)
/\ is_prime(GHI) /\ is_prime(AC) /\ is_prime(JN) /\ is_prime(DGK)
/\ is_prime(EIL) /\ is_prime(BF) /\ is_prime(MP);

% All the prime numbers are different
constraint all_different([CD,EF,JK,LM,GHI,AC,JN,DGK,EIL,BF,MP]);

% Top and bottom rows have different digits
constraint A != B /\ N != P;

% Other rows have different digits
constraint all_different([C,D,E,F]) /\ all_different([G,H,I])
/\ all_different([J,K,L,M]);

% All the columns have different digits
constraint all_different([A,C,J,N]) /\ all_different([D,G,K])
/\ all_different([E,I,L]) /\ all_different([B,F,M,P]);

% The diagonals have different digits
constraint all_different([A,D,H,L,P]) /\ all_different([B,E,H,K,N]);

solve satisfy;

output ["Three digit primes are : " ++ show([DGK,GHI,EIL])
++ "\nA, B, C, D, E, F = " ++ show([A,B,C,D,E,F])
++ "\nG, H, I, J, K, L = " ++ show([G,H,I,J,K,L])
++ "\nM, N, P = " ++ show([M,N,P]) ];

% Three digit primes are : [983, 821, 617]
% A, B, C, D, E, F = [6, 9, 1, 9, 6, 7]
% G, H, I, J, K, L = [8, 2, 1, 4, 3, 7]
% M, N, P = [1, 7, 3]
% ----------
% ==========
% Grid solution
% 6 # # # 9
% 1 9 # 6 7
% # 8 2 1 #
% 4 3 # 7 1
% 7 # # # 3

LikeLike

Jim Randell 4:48 pm on 12 June 2020 Permalink | Reply
Tags: by: Stephen Hogg ( 62 )
Teaser 3012: Number blind rage
From The Sunday Times, 14th June 2020 [link] [link]

After school, angry at getting “50 lines”, I kicked my satchel around. Impacts made my 11-digit calculator switch on. An 11-digit number was also entered and the display was damaged. Strangely, I found “dYSCALCULIA” displayed and saved this to memory (as shown).

After various tests I confirmed that all arithmetic operations were correct and the decimal point would appear correctly if needed. No segments were permanently “on”, two digits were undamaged, and for the other digits, overall, several segments were permanently “off”.

Retrieving “dYSCALCULIA”, I divided it by 9, then the result by 8, then that result by 7, then that result by 6. No decimal point appeared and the last result (at the right-hand side of the display) had three digits appearing as numerals.

What number was “dYSCALCULIA”?

[teaser3012]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 9:07 pm on 12 June 2020 Permalink | Reply
  I used the standard set of digits (as illustrated in Enigma 1701).
  
  This Python program runs in 90ms.
```
from itertools import product
from enigma import join, div, printf

# normal digits
normal = "0123456789"

# map digits to illuminated segments, arranged as:
#
#   0
# 1   2
#   3
# 4   5
#   6
#
# map digits to segments
f = lambda *ss: frozenset(ss)
seg = {
  # normal digits
  '0': f(0, 1, 2, 4, 5, 6),
  '1': f(2, 5),
  '2': f(0, 2, 3, 4, 6),
  '3': f(0, 2, 3, 5, 6),
  '4': f(1, 2, 3, 5),
  '5': f(0, 1, 3, 5, 6),
  '6': f(0, 1, 3, 4, 5, 6), # or could be (1, 3, 4, 5, 6)
  '7': f(0, 2, 5), # or could be (0, 1, 2, 5)
  '8': f(0, 1, 2, 3, 4, 5, 6),
  '9': f(0, 1, 2, 3, 5, 6), # or could be (0, 1, 2, 3, 5)
  # malformed digits
  'a': f(0, 1, 2, 3, 4, 5),
  'c': f(0, 1, 4, 6),
  'd': f(2, 3, 4, 5, 6),
  'l': f(1, 4, 6),
  'u': f(1, 2, 4, 5, 6),
}
# map segments to normal digits
norm = dict((seg[k], k) for k in normal)

# the display
display = "d45calcul1a"

# compute possible replacement (superset) digits for the symbols
r = dict((k, list(d for d in normal if seg[d].issuperset(seg[k]))) for k in set(display))

# choose possible replacement digits for the symbols
for ds in product(*(r[x] for x in display)):
  if ds[0] == '0': continue
  # two of the digits are unaffected
  if sum(x == y and x in normal for (x, y) in zip(display, ds)) < 2: continue
  # make the number
  n = int(join(ds))
  # and the result
  s = div(n, 9 * 8 * 7 * 6)
  if s is None: continue
  # remove broken segments from the result
  # x = original display, y = original digit, z = result digit
  rs = list(
    norm.get(seg[z].difference(seg[y].difference(seg[x])), '?')
      for (x, y, z) in zip(display[::-1], ds[::-1], str(s)[::-1])
  )
  # there should be 3 normal digits
  if len(rs) - rs.count('?') != 3: continue
  # output solution
  printf("{n} -> {s} -> {rs}", rs=join(rs[::-1]))
```
  Solution: dYSCALCULIA = 84588800688.
  
  The digits displaying 4 and 5 must be the undamaged ones.
  
  So the segments that are definitely broken are as shown below:
  
  There are 22 segments that are definitely broken, and a further 3 that we cannot determine if they are broken or not.
  
  The result of dividing the number by 9×8×7×6 = 3024 is 27972487, which would look something like this:
  
  (Only showing the segments that we definitely know to be broken, there are two segments shown lit that may be broken).
  
  The 2nd digit (7) displays as a 7. The 7th digit (8) displays as a 1. The 8th digit (7) displays as a 7.
  
  LikeLike

	Jim Randell on Teaser 2415: [Double shif…
	Jim Randell on Teaser 3305: Things with scale…
	Jim Randell on Teaser 3305: Things with scale…
	Jim Randell on Teaser 2462: [What NOW?]
	Frits on Teaser 2462: [What NOW?]
	ruudvanderham on Teaser 2462: [What NOW?]
	Jim Randell on Teaser 2462: [What NOW?]
	Frits on Teaser 2363: [A round num…

Search

Recent Posts

Recent Comments

Archives

Categories

Subscribe

RSS Links

Site Stats

Meta

Updates from Jim Randell Toggle Comment Threads | Keyboard Shortcuts

Jim Randell 9:01 am on 16 July 2020 Permalink | Reply Tags: by: Graham Smithers ( 82 )

Rate this:

Share:

Jim Randell 9:02 am on 16 July 2020 Permalink | Reply

Reply Cancel reply

Jim Randell 12:11 pm on 14 July 2020 Permalink | Reply Tags: by: Victor Bryant ( 140 )

Rate this:

Share:

Jim Randell 12:13 pm on 14 July 2020 Permalink | Reply

GeoffR 5:44 pm on 14 July 2020 Permalink | Reply

Jim Randell 5:12 pm on 15 July 2020 Permalink | Reply

Jim Randell 9:55 am on 12 July 2020 Permalink | Reply Tags: by: John Feather ( 3 )

Rate this:

Share:

Jim Randell 9:56 am on 12 July 2020 Permalink | Reply

Jim Randell 5:11 pm on 10 July 2020 Permalink | Reply Tags: by: Andrew Skidmore ( 85 ), flawed ( 53 )

Rate this:

Share:

Jim Randell 6:26 pm on 10 July 2020 Permalink | Reply

Jim Randell 9:22 am on 14 July 2020 Permalink | Reply

Jim Randell 10:09 am on 25 March 2023 Permalink | Reply

GeoffR 9:02 am on 14 July 2020 Permalink | Reply

Dilwyn Jones 11:02 am on 17 July 2020 Permalink | Reply

Jim Randell 2:37 pm on 18 July 2020 Permalink | Reply

Jim Randell 8:44 am on 9 July 2020 Permalink | Reply Tags: by: Ptolemaeus ( 5 )

Rate this:

Share:

Jim Randell 8:45 am on 9 July 2020 Permalink | Reply

Jim Randell 9:36 am on 7 July 2020 Permalink | Reply Tags: by: Alan Bergson ( 4 )

Rate this:

Share:

Jim Randell 9:37 am on 7 July 2020 Permalink | Reply

GeoffR 10:54 am on 8 July 2020 Permalink | Reply

Jim Randell 8:31 am on 5 July 2020 Permalink | Reply Tags: by: Harold Reiter ( 5 )

Rate this:

Share:

Jim Randell 8:32 am on 5 July 2020 Permalink | Reply

Jim Randell 5:14 pm on 3 July 2020 Permalink | Reply Tags: by: Victor Bryant ( 140 )

Rate this:

Share:

Jim Randell 5:32 pm on 3 July 2020 Permalink | Reply

Jim Randell 8:37 am on 2 July 2020 Permalink | Reply Tags: by: Victor Bryant ( 140 )

Rate this:

Share:

Jim Randell 8:38 am on 2 July 2020 Permalink | Reply

GeoffR 8:07 pm on 21 October 2021 Permalink | Reply

Jim Randell 9:25 am on 30 June 2020 Permalink | Reply Tags: by: W Blackman ( 3 )

Rate this:

Share:

Jim Randell 9:26 am on 30 June 2020 Permalink | Reply

Jim Randell 8:10 am on 28 June 2020 Permalink | Reply Tags: by: Colin

Rate this:

Share:

Jim Randell 8:11 am on 28 June 2020 Permalink | Reply

Jim Randell 12:38 pm on 30 June 2020 Permalink | Reply

GeoffR 4:09 pm on 29 June 2020 Permalink | Reply

GeoffR 1:46 pm on 8 July 2020 Permalink | Reply

Jim Randell 5:08 pm on 26 June 2020 Permalink | Reply Tags: by: Danny Roth ( 84 )

Rate this:

Share:

Jim Randell 5:14 pm on 26 June 2020 Permalink | Reply

GeoffR 9:19 am on 27 June 2020 Permalink | Reply

GeoffR 2:41 pm on 27 June 2020 Permalink | Reply

Jim Randell 7:16 am on 25 June 2020 Permalink | Reply Tags: by: H. Wright ( 2 )

Rate this:

Share:

Jim Randell 7:17 am on 25 June 2020 Permalink | Reply

GeoffR 4:10 pm on 25 June 2020 Permalink | Reply

Jim Randell 4:15 pm on 25 June 2020 Permalink | Reply

GeoffR 4:38 pm on 25 June 2020 Permalink | Reply

Jim Randell 9:01 am on 16 July 2020 Permalink | Reply
Tags: by: Graham Smithers ( 82 )

Jim Randell 12:11 pm on 14 July 2020 Permalink | Reply
Tags: by: Victor Bryant ( 140 )

Jim Randell 9:55 am on 12 July 2020 Permalink | Reply
Tags: by: John Feather ( 3 )

Jim Randell 5:11 pm on 10 July 2020 Permalink | Reply
Tags: by: Andrew Skidmore ( 85 ), flawed ( 53 )

Jim Randell 8:44 am on 9 July 2020 Permalink | Reply
Tags: by: Ptolemaeus ( 5 )

Jim Randell 9:36 am on 7 July 2020 Permalink | Reply
Tags: by: Alan Bergson ( 4 )

Jim Randell 8:31 am on 5 July 2020 Permalink | Reply
Tags: by: Harold Reiter ( 5 )

Jim Randell 5:14 pm on 3 July 2020 Permalink | Reply
Tags: by: Victor Bryant ( 140 )

Jim Randell 8:37 am on 2 July 2020 Permalink | Reply
Tags: by: Victor Bryant ( 140 )

Jim Randell 9:25 am on 30 June 2020 Permalink | Reply
Tags: by: W Blackman ( 3 )

Jim Randell 8:10 am on 28 June 2020 Permalink | Reply
Tags: by: Colin

Jim Randell 5:08 pm on 26 June 2020 Permalink | Reply
Tags: by: Danny Roth ( 84 )

Jim Randell 7:16 am on 25 June 2020 Permalink | Reply
Tags: by: H. Wright ( 2 )

Jim Randell 8:52 am on 23 June 2020 Permalink | Reply
Tags: by: DJT Hogg ( 3 )

Jim Randell 9:48 am on 21 June 2020 Permalink | Reply
Tags: by: R Postill ( 20 )

Jim Randell 7:06 pm on 19 June 2020 Permalink | Reply
Tags: by: Howard Williams ( 43 )

Jim Randell 7:12 am on 18 June 2020 Permalink | Reply
Tags: by: Nick MacKinnon ( 52 )

Jim Randell 8:42 am on 16 June 2020 Permalink | Reply
Tags: by: Angela Newing ( 37 )

Jim Randell 12:18 pm on 14 June 2020 Permalink | Reply
Tags: by: Peter G Chamberlain ( 11 )

Jim Randell 4:48 pm on 12 June 2020 Permalink | Reply
Tags: by: Stephen Hogg ( 62 )