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Jim Randell 12:43 pm on 9 May 2019 Permalink | Reply
Tags: by: John Walker
Brainteaser 1568: Back to the future
From The Sunday Times, 27th September 1992 [link]

Many dates, such as my birthday 27th September 1972, are palindromic when written down in the form day/month/year without breaks, my birthday being 2791972.

What about palindromic dates in the millennium beginning on 1st January 2000?

(a) How many palindromic dates will there be in this millennium?

(b) Of these, which two are closest together?

The text above is taken from the book Brainteasers (2002), and is different from the originally published puzzle, which is reproduced below:

With another palindromic date occurring on Tuesday (29.9.1992) I am reminded of a Teaser I set last year in which readers were asked to find the longest sequence of consecutive non-palindromic dates in the next 200 years. Here’s another idea:

Assuming the standard numerical format of a date as day.month.year with the year as a four-figure number and with no zeros to start the day or month, find the two palindromic dates in the future which are closest together.

[teaser1568]
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Lise Andreasen, Jim Randell, and Jim Randell are discussing. Toggle Comments
- Jim Randell 12:44 pm on 9 May 2019 Permalink | Reply
  This Python program considers dates from 1-1-2000 to 31-12-2999. It runs in 780ms.
  
  Run: [ @replit ]
```
from datetime import (date, timedelta)
from enigma import (repeat, inc, concat, tuples, Accumulator, printf)

# record palindromic dates from 2000 - 2999
ds = list()
for d in repeat(inc(timedelta(days=1)), date(2000, 1, 1)):
  if not (d.year < 3000): break
  s = concat(d.day, d.month, d.year)
  if s == s[::-1]:
    ds.append(d)
printf("[{n} palindromic dates]", n=len(ds))

# find the two dates closest together
r = Accumulator(fn=min)
r.accumulate_data_from((b - a, (a, b)) for (a, b) in tuples(ds, 2))

# output solution
(t, (a, b)) = (r.value, r.data)
printf("{a.day}-{a.month}-{a.year} -> {b.day}-{b.month}-{b.year} ({t.days} days)")
```
  Solution: (a) There are 323 palindromic dates. (b) The closest pair is: 23-2-2232 and 2-3-2232 (8 days apart).
  
  Pleasingly the number of palindromic dates is itself a palindrome.
  
  If we extend the date range to the year 9999 there are 2107 more palindromic dates (assuming the calendar remains the same). And the closest two are:
  
  29-8-8892 and 2-9-8892 (4 days apart)
  
  And if we extend the current calendar backwards to year 1 we can manage even better:
  
  31-10-113 and 3-11-113 (3 days apart)
  
  LikeLike
- Lise Andreasen 12:20 am on 8 May 2024 Permalink | Reply
  
  Does your algorithm count 1112111 (to give an example) as one or 2 dates? 1/11/2111 and 11/1/2111.
  
  LikeLike
  - Lise Andreasen 12:30 am on 8 May 2024 Permalink | Reply
    
    Oh, never mind. Figured it out.
    
    LikeLike
- Lise Andreasen 12:34 am on 8 May 2024 Permalink | Reply
  
  It’s possible to count the variations.
  There are 81 dates of the form a/b/22ba.
  There are 27 dates of the form a/bc/2cba.
  There are 192 dates of the form ab/c/2cba.
  There are 22 dates of the form ab/12/21ba.
  
  Neat algorithm though. 👍🏻
  
  LikeLike
  - Jim Randell 10:26 am on 8 May 2024 Permalink | Reply
    
    I don’t know if it is a typo, but these numbers don’t add up to 323.
    
    I found there should 193 dates of the form ab/c/2cba. (Or maybe you forgot: 29/2/2292).
    
    LikeLike
    - Lise Andreasen 12:53 pm on 8 May 2024 Permalink | Reply
      
      It was a typo. And it’s probably also a remnant of me getting that leap year date wrong the first time through.
      
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Jim Randell 11:22 am on 8 May 2019 Permalink | Reply
Tags: by: John Owen ( 33 )

Teaser 2914: Bank statement

From The Sunday Times, 29th July 2018 [link] [link]

My last bank statement was most interesting. The first line showed the opening balance in pounds and pence, then each of the following four lines showed a debit together with the resulting balance. I did not go overdrawn.

Remarkably, each of the five balances used the same five different digits once only, and the largest of the digits was less than three times the smallest. Each of the four debits also used five different digits once only, but the digits in the debits were all different from those in the balances.

What was the final balance?

[teaser2914]

Jim Randell 11:23 am on 8 May 2019 Permalink | Reply

This Python 3 program runs in 415ms.

Run: [ @repl.it ]

from itertools import (combinations, permutations)
from collections import defaultdict
from enigma import (irange, nconcat, nsplit, join, sprintf as f)

digits = irange(0, 9)

# from transactions <ts> and balance <b> find <k> transactions
def solve(ts, b, k, s=[]):
  if k == 0:
    yield s
  elif b in ts:
    for (d, a) in ts[b]:
      yield from solve(ts, a, k - 1, s + [(b, d, a)])

# choose 5 different digits for the balance
for ds in combinations(digits, 5):

  # the largest is less than 3x the smallest
  if not (ds[-1] < 3 * ds[0]): continue

  # make all possible numbers from the digits
  ns = sorted(nconcat(s) for s in permutations(ds) if s[0] != 0)

  # find transactions that make viable debits
  ts = defaultdict(list)
  for (a, b) in combinations(ns, 2):
    d = b - a
    xs = set(nsplit(d))
    if len(xs) == 5 and xs.isdisjoint(ds):
      ts[b].append((d, a))

  # find a sequence of 4 transactions
  for b in ts.keys():
    for s in solve(ts, b, 4):
      print(f("final = {f} [{s}]", s=join((f("{b} - {d} = {a}") for (b, d, a) in s), sep=", "), f=s[-1][2]))

Solution: The final balance was £ 347.58.

There are two possibilities for the initial amount and the first debit amount, which give the same value for the balance:

(1) 853.47 – 109.62 = 743.85; or:
(1) 873.45 – 129.60 = 743.85

After that the next three debits are:

(2) 743.85 – 169.02 = 574.83
(3) 574.83 – 120.96 = 453.87
(4) 453.87 – 106.29 = 347.58

The condition that “the largest of the digits was less than three times the smallest” reduces the search space, but there are no further solutions without this condition.

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GeoffR 10:28 am on 9 May 2019 Permalink | Reply

A brute force approach in MiniZinc uses 45 integer variables and sets to solve this teaser.

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

% Balance/Debit Sums - uses 5-digit integers for £ and pence parts
% ABCDE - abcde == FGHIJ;
% FGHIJ - fghij == KLMNO;
% KLMNO - klmno == PQRST;
% PQRST - pqrst == UVWXY;

% Balances integers
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; var 0..9: J;
var 0..9: K; var 0..9: L; var 0..9: M; var 0..9: N; var 0..9: O;
var 0..9: P; var 0..9: Q; var 0..9: R; var 0..9: S; var 0..9: T;
var 0..9: U; var 0..9: V; var 0..9: W; var 0..9: X; var 0..9: Y;

constraint A != 0 /\ F != 0 /\ K != 0 /\ P != 0 /\ U != 0;

% max balance digit < 3 * min digit for the five balances  
constraint max([A,B,C,D,E]) < 3 * min([A,B,C,D,E]);
constraint max([F,G,H,I,J]) < 3 * min([F,G,H,I,J]);
constraint max([K,L,M,N,O]) < 3 * min([K,L,M,N,O]);
constraint max([P,Q,R,S,T]) < 3 * min([P,Q,R,S,T]);
constraint max([U,V,W,X,Y]) < 3 * min([U,V,W,X,Y]);

% Balances - sets of integers
var set of int: s1 = { A,B,C,D,E };
var set of int: s2 = { F,G,H,I,J };
var set of int: s3 = { K,L,M,N,O };
var set of int: s4 = { P,Q,R,S,T };
var set of int: s5 = { U,V,W,X,Y };

% Sets s1 to s5 have the same integers
constraint card (s1 intersect s2 ) == 5;
constraint card (s1 intersect s3 ) == 5;
constraint card (s1 intersect s4 ) == 5;
constraint card (s1 intersect s5 ) == 5;

% Debits integers
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; var 0..9: j;
var 0..9: k; var 0..9: l; var 0..9: m; var 0..9: n; var 0..9: o;
var 0..9: p; var 0..9: q; var 0..9: r; var 0..9: s; var 0..9: t;

constraint a != 0 /\ f != 0 /\ k != 0 /\ p != 0;

% Debits - sets of integers
var set of int: s6 = { a,b,c,d,e };
var set of int: s7 = { f,g,h,i,j };
var set of int: s8 = { k,l,m,n,o };
var set of int: s9 = { p,q,r,s,t };

% Debit sets s6 to s9 have the same integers
constraint card (s6 intersect s7 ) == 5;
constraint card (s6 intersect s8 ) == 5;
constraint card (s6 intersect s9 ) == 5;

% Sets s1 and s6 have no digits in common
constraint card(s1 intersect s6) == 0;

% Form all 5-digit integers
var 10000..99999: ABCDE = 10000*A + 1000*B + 100*C + 10*D + E;
var 10000..99999: FGHIJ = 10000*F + 1000*G + 100*H + 10*I + J;
var 10000..99999: KLMNO = 10000*K + 1000*L + 100*M + 10*N + O;
var 10000..99999: PQRST = 10000*P + 1000*Q + 100*R + 10*S + T;
var 10000..99999: UVWXY = 10000*U + 1000*V + 100*W + 10*X + Y;

var 10000..99999: abcde = 10000*a + 1000*b + 100*c + 10*d + e;
var 10000..99999: fghij = 10000*f + 1000*g + 100*h + 10*i + j;
var 10000..99999: klmno = 10000*k + 1000*l + 100*m + 10*n + o;
var 10000..99999: pqrst = 10000*p + 1000*q + 100*r + 10*s + t;

% Balance/Debit sums in sequence
constraint ABCDE - abcde == FGHIJ;
constraint FGHIJ - fghij == KLMNO;
constraint KLMNO - klmno == PQRST;
constraint PQRST - pqrst == UVWXY;

solve satisfy;

output ["Final Balance = " ++ "£"++ show(U),show(V),show(W) ++ "." 
++ show(X),show(Y) ] 
++ ["\n" ++ show(ABCDE) ++ " - " ++ show(abcde)  ++ " = " ++ show(FGHIJ) ]
++ ["\n" ++ show(FGHIJ) ++ " - " ++ show(fghij)  ++ " = " ++ show(KLMNO) ]
++ ["\n" ++ show(KLMNO) ++ " - " ++ show(klmno)  ++ " = " ++ show(PQRST) ]
++ ["\n" ++ show(PQRST) ++ " - " ++ show(pqrst)  ++ " = " ++ show(UVWXY) ];

% Final Balance = £347.58
% 85347 - 10962 = 74385 - five digit integer format for £/pence sums
% 74385 - 16902 = 57483
% 57483 - 12096 = 45387
% 45387 - 10629 = 34758
% % time elapsed: 0.27 s
% ----------
% Final Balance = £347.58
% 87345 - 12960 = 74385
% 74385 - 16902 = 57483
% 57483 - 12096 = 45387
% 45387 - 10629 = 34758
% % time elapsed: 0.27 s
% ----------
% ==========
% Finished in 666msec

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Jim Randell 12:58 pm on 7 May 2019 Permalink | Reply
Tags: by: P Woolfenden ( 4 )
Brain-Teaser 479: [Sixpences]
From The Sunday Times, 2nd August 1970 [link]

My nephew Anselm collects good old-fashioned sixpences in a good old-fashioned piggy bank. These he recently extracted to audit and, while experimenting by putting them into two or more equal piles of two or more coins, found that there were exactly six different ways in which this could be done.

From one of these combinations he returned one pile to the bank, and found that the balance could again be distributed into equal piles in exactly six different ways. On this second occasion he returned two equal piles from one of the combinations to the bank, and again found that the balance could be distributed into equal piles in six different ways only. He conducted this experiment a total of seven times, on the third occasion returning three piles, on the fourth four piles and so on, and after the seventh found that he had 24 sixpences remaining.

On none of these occasions did he deposit as much as 25s. in the bank. His third deposit was twice as much as his sixth, and three times as much as his second. His fifth deposit was five times as large as his first.

How many sixpences had he saved up?

I have modified the wording slightly to clarify the puzzle.

This puzzle was originally published with no title.

[teaser479]
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- Jim Randell 12:59 pm on 7 May 2019 Permalink | Reply
  I assumed that making one pile with all the coins in does not count (as there are no solutions if it does count as one of the six configurations).
  
  1s = 12d, so at no point does Anselm deposit more than 52 coins. So he cannot start with more than (24 + 7×52 = 388) coins. This gives us a bounded solution space to search in, but it is more interesting to run the process backwards starting with the 24 final coins.
  
  This Python 3 program works backwards recursively to solve the problem. It runs in 79ms.
```
from enigma import (divisors, div, printf)

# possible pile sizes for n coins
piles = lambda n: list(p for p in divisors(n) if 1 < p < n)

# work backwards from <n> coins and <i> piles deposited to 1 pile deposited
# i = number of piles deposited
# n = number of coins
# ps = possible pile sizes
# return [(<n coins>, <n piles>), ...]
def solve(n, i, ps, ss=[]):
  if i == 0:
    yield ss
  else:
    # consider the number of coins in each pile
    for k in ps:
      # i equally sized piles were deposited in the bank
      d = i * k
      # but not more than 52 coins
      if d > 52: break
      # the total number of coins before the deposit
      t = n + d
      # and t must have 6 possible piles numbers
      ds = piles(t)
      if not (len(ds) == 6): continue
      yield from solve(t, i - 1, ds, [(t, k)] + ss)

# we end up with 24 coins
N = 24
for ss in solve(N, 7, piles(N) + [N]):

  # calculate the numbers of coins returned to the bank at each stage
  ds = list(i * k for (i, (_, k)) in enumerate(ss, start=1))
  # 3rd deposit was twice the 6th and 3 times the 2nd
  if not (ds[2] == 2 * ds[5] == 3 * ds[1]): continue
  # 5th deposit was 5 times the 1st
  if not (ds[4] == 5 * ds[0]): continue

  # output solution
  for (i, ((t, k), d)) in enumerate(zip(ss, ds), start=1):
    printf("{i}: {t} coins in {p} piles of {k}; deposit {i} piles = {d} coins", p=div(t, k))
  printf("started with {n} coins; finished with {N} coins", n=ss[0][0])
  printf()
```
  Solution: Anselm has 138 sixpences.
  
  LikeLiked by 1 person

Jim Randell 4:41 pm on 6 May 2019 Permalink | Reply
Tags: by: Graham Smithers ( 84 )

Teaser 2915: £sd

From The Sunday Times, 5th August 2018 [link] [link]

50 years ago, the coins in circulation were the halfpenny (1/2d), penny (1d), threepenny bit (3d), sixpence (6d), shilling (12d), florin (2 shillings) and half crown (2 shillings and sixpence).

One day, having at least one of each of the above coins, I decided to bank them.

The cashier set out the coins in separate piles, checked them (discovering that the number of coins in each pile was a square), and gave me exactly a 10-shilling note in exchange.

If I told you how many different numbers of coins were in those piles, you should be able to work out the numbers of each coin.

In the order half crown down to halfpenny, how many coins of each type were there?

[teaser2915]

Jim Randell 4:42 pm on 6 May 2019 Permalink | Reply

We know there is at least one of each denomination of coin. So that is 1/2 + 1 + 3 + 6 + 12 + 24 + 30 = 76.5d accounted for, leaving us only 43.5d to worry about, and we need to make this amount from quantities that are 1 less than a perfect square.

It turns out there are 6 different ways to achieve this. Five of them use 3 different quantities, the sixth way uses 4 different quantities, and gives us the solution.

This Python 3 program runs in 73ms.

Run: [ @repl.it ]

from enigma import (defaultdict, isqrt, irange, arg, printf)

# express total <t> using denominations <ds>, quantities <qs>
def express(t, ds, qs, s=[]):
  if t == 0:
    if not ds:
      yield s
    elif 0 in qs:
      yield s + [0] * len(ds)
  elif ds:
    d = ds[0]
    for i in qs:
      if d * i > t: break
      yield from express(t - d * i, ds[1:], qs, s + [i])

# denominations of coins (in half-pennies)
coins = (1, 2, 6, 12, 24, 48, 60)

# total amount (of half-pennies)
T = arg(240, 0, (lambda x: int(2 * float(x))))
printf("[T={T}(1/2)d]")

# subtract one of each coin from the total
t = T - sum(coins)

# squares - 1 (up to t)
squares1 = list(i * i - 1 for i in irange(1, isqrt(t)))

# collect results by how many different quantities there are
r = defaultdict(list)

# consider ways to express t using (i^2 - 1) quantities
for qs in express(t, coins, squares1):
  # record result by the numbers of different quantities
  k = len(set(qs))
  printf("[{qs} -> {k} values]")
  r[k].append(qs)

# look for keys with unique values
for (k, vs) in r.items():
  if len(vs) == 1:
    # include the first coin of each denomination
    qs = list(n + 1 for n in vs[0])
    # output quantities and denominations (in half-pennies)
    printf("{k} values -> {T}(d/2) = {qs} x {coins}(d/2)")

Solution: The numbers of coins in the piles (from half-crowns down to half-pennies) are: 1, 1, 1, 4, 1, 9, 36.

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Frits 3:36 pm on 5 June 2024 Permalink | Reply

“The cashier set out the coins in separate piles”.

I understand you have assumed that the cashier made 7 piles per denomination. I would also have considered fi 4four piles (with 4, 4, 36 and 196 coins).

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- Jim Randell 5:02 pm on 5 June 2024 Permalink | Reply
  
  @Frits: My natural interpretation was that each denomination is in a separate pile, so there is (exactly) one pile per denomination.
  
  LikeLike

GeoffR 9:26 pm on 6 May 2019 Permalink | Reply

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

% HP = No. HalfPennies, P3 = No. Threepence, P6 = No. Sixpence....
var 1..150:HP; var 1..100:P; var 1..40:P3; var 1..20:P6;     
var 1..10:SH; var 1..5:FL; var 1..4:HC; 

% even number of half-pennies needed to make 120p
constraint HP mod 2 == 0;  

% the number of coins in each pile was a square
set of int: sq = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100};

% all types of coins are squares
constraint HP in sq /\ P in sq /\ P3 in sq /\ P6 in sq
/\ SH in sq /\ FL in sq /\ HC in sq;

% number of coins/types to make 120p 
constraint HP * 1 div 2 + P*1 + P3*3  + P6*6 + SH*12 
+ FL*24 + HC*30 == 120;

var set of int: coins = {HP, P, P3, P6, SH, FL, HC};

solve satisfy;
output [ show([HP,P,P3,P6,SH,FL,HC]) ++ "  " ++ show(coins) ];

% [HP, P, P3,P6,SH,FL,HC] Set of Coins
% ------------------------------------ 
% [64, 4, 4, 1, 1, 1, 1]  {1,4,64}
% [4, 25, 1, 4, 1, 1, 1]  {1,4,25}
% [4, 16, 4, 4, 1, 1, 1]  {1,4,16}
% [4, 1, 9, 4, 1, 1, 1]  {1,4,9}
% [36, 9, 1, 4, 1, 1, 1]  {1,4,9,36} << only set of 4 coin types
% [16, 1, 1, 1, 4, 1, 1]  {1,4,16}
% ==========
% Answer: 1 Half Crown, 1 Florin, 1 Shilling, 4 Sixpence
% 1 Threepence, 9 Pennies and 36 HalfPennies

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Jim Randell 1:37 pm on 5 May 2019 Permalink | Reply
Tags: by: Susan Bricket ( 17 )
Teaser 2916: Pointless batting averages
From The Sunday Times, 12th August 2018 [link] [link]

When my son was chosen to play cricket for his School First XI, I kept a record of the number of runs he scored in each of his first five innings. After each innings I calculated his batting average (the total number of runs scored so far divided by the number of innings) and found an interesting pattern:

(i) Each score was under 30
(ii) They [the scores] were all different
(iii) Each of the five averages was a whole number

When he asked me how he could maintain this pattern with his sixth innings, I was able to tell him the smallest score that would achieve this.

What is the largest number this could have been?

[teaser2916]
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- Jim Randell 1:37 pm on 5 May 2019 Permalink | Reply
  If the scores in the first five innings are (a, b, c, d, e) and there are scores for the sixth innings f, (between 0 and 29), that continue the pattern. And there will be a smallest such value:
  
  (a, b, c, d, e, f) → f_min
  
  So, we can look at all possible (a, b, c, d, e, f) values and find the largest possible f_min value.
  
  This Python program runs in 569ms.
  
  Run: [ @repl.it ]
```
from enigma import (irange, Accumulator, printf)

# possible next scores
def scores(ss, avgs):
  t = sum(ss)
  n = len(ss) + 1
  for s in irange(-t % n, 29, step=n):
    if s not in ss:
      yield (ss + [s], avgs + [(t + s) // n])

# find sequences of length k
def solve(ss=[], avgs=[], k=5):
  if len(ss) == k:
    yield (ss, avgs)
  else:
    for (ss1, avgs1) in scores(ss, avgs):
      yield from solve(ss1, avgs1, k)

# find the largest of the smallest values
r = Accumulator(fn=max)

for (ss, avgs) in solve():
  # find the smallest score to maintain this
  for (ss1, avgs1) in scores(ss, avgs):
    s = ss1[-1]
    r.accumulate(s)
    if s == r.value: printf("scores={ss} (avgs={avgs}) -> score={s} (avg={avg})", avg=avgs1[-1])
    break  # we only want the smallest value

# output the solution
printf("max value = {r.value}")
```
  Solution: The largest number it could have been is 23.
  
  I found 142 sequences that give this largest possible f_min value, although only 15 of these also have the averages take on 5 different values.
  
  One possible sequence is:
  
  (5, 11, 17, 27, 25)
  
  which give the corresponding averages of:
  
  (5, 8, 11, 15, 17)
  
  A score in the sixth innings of 23, would give an average of 18 over the six innings.
  
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Jim Randell 2:13 pm on 4 May 2019 Permalink | Reply
Tags: by: S G Worthington

Brain-Teaser 478: [Alphametic family]

From The Sunday Times, 26th July 1970 [link]

Brian and Betty have a daughter, Moira, and this can be proved by addition where each letter is represented by a different digit:

BRIAN + BETTY = MOIRA

The sum of the individual digit values of MOIRA gives her age. The age of her brother ROBERT is similarly obtained. The sum of the children’s ages is a perfect square.

How old is Robert?

This puzzle was originally published with no title.

[teaser478]

Jim Randell 2:14 pm on 4 May 2019 Permalink | Reply
Once we have the solved the alphametic sum the value of ROBERT is determined.

We can use the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve this puzzle.

The following run file executes in 703ms. (Internal runtime of the generated program is 490ms).
```
#! python -m enigma -rr

SubstitutedExpression

"BRIAN + BETTY = MOIRA"

"is_square(sum([M, O, I, R, A, R, O, B, E, R, T]))"

--answer="sum([R, O, B, E, R, T])"
```
But it is faster to write a short program that uses the [[ SubstitutedSum() ]] solver.

This Python program runs in 79ms. (Internal runtime is 7.0ms).
```
from enigma import (SubstitutedSum, is_square, printf)

# make the alphametic sum
p = SubstitutedSum(['BRIAN', 'BETTY'], 'MOIRA')

# solve it
for s in p.solve():
  # compute the ages
  (m, r) = (sum(s[x] for x in w) for w in ('MOIRA', 'ROBERT'))
  # check they sum to a square
  if not is_square(m + r): continue
  # output solution
  printf("r = {r} [m = {m}] [{s}]", s=p.substitute(s, "BRIAN + BETTY = MOIRA"))
```
Solution: Robert is 23.

Moira is 26. Together the ages sum to 49 (= 7²).

There are 2 possible sums as the values of N and Y (in the units column of the sum) are interchangeable.

So the sum is one of:

14836 + 15007 = 29843
14837 + 15006 = 29843

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- Jim Randell 2:40 pm on 22 November 2024 Permalink | Reply
  Or using the [[ SubstitutedExpression.split_sum ]] solver generates a program with an internal runtime of just 1.1ms.
```
#! python3 -m enigma -rr

SubstitutedExpression.split_sum

"BRIAN + BETTY = MOIRA"

--extra="is_square(sum([M, O, I, R, A, R, O, B, E, R, T]))"

--answer="sum([R, O, B, E, R, T])"
```
  LikeLike

GeoffR 3:10 pm on 6 May 2019 Permalink | Reply

% A solution in MinZinc 
include "globals.mzn";

var 0..9: B; var 0..9: R; var 0..9:I; var 0..9: A; var 0..9: N;
var 0..9: E; var 0..9: T; var 0..9:Y; var 0..9: M; var 0..9: O;

constraint B > 0 /\ M > 0 /\ R > 0;
constraint all_different ([B, R, I, A, N, E, T, Y, M, O]);

var 10000..99999: BRIAN = 10000*B + 1000*R + 100*I + 10*A + N;
var 10000..99999: BETTY = 10000*B + 1000*E + 100*T + 10*T + Y;
var 10000..99999: MOIRA = 10000*M + 1000*O + 100*I + 10*R + A;

constraint BRIAN + BETTY == MOIRA;

% The sum of the children’s ages is a perfect square
set of int: sq = {9, 16, 25, 36, 49, 64, 81, 100};
constraint  (M + O + I + R + A + R + O + B + E + R + T) in sq;

solve satisfy;

% The sum of the individual digit values gives each child's age
output [ "ROBERT = " ++ show(R+O+B+E+R+T)] ++  
[", MOIRA = " ++ show(M+O+I+R+A) ]
++ ["\nSum is " ++ show(BRIAN) ++ " + " ++ show(BETTY) ++ 
" = " ++ show(MOIRA) ] ;

% ROBERT = 23, MOIRA = 26
% Sum is 14837 + 15006 = 29843
% % time elapsed: 0.03 s
% ----------
% ROBERT = 23, MOIRA = 26
% Sum is 14836 + 15007 = 29843
% % time elapsed: 0.03 s
% ----------
% ==========
% Finished in 254msec

The parents ages (22 and 13) or (23 and 12) don’t seem to make much sense in relation to the children’s ages! ie Robert (23) and Moira(26)

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Jim Randell 5:41 pm on 2 May 2019 Permalink | Reply
Tags: by: Stephen Hogg ( 64 )
Teaser 2954: Lovely meter, RITA made!
From The Sunday Times, 5th May 2019 [link] [link]

Revisiting the old curiosity shop, I bought an unusual moving-iron ammeter (made by Rho Iota Tau Associates). On the non-linear scale “9” was the last possible “whole amperes” scale graduation marked before the “full scale deflection” end stop (which was a half turn of the pointer from zero).

The booklet gave the pointer swing from zero (in degrees) equal to the current (in amperes) raised to a fixed single-figure positive whole number power, then divided by a positive whole number constant. Curiously, the angle between “exact” pointer swings, calculated using this formula, for two single-figure “whole amperes” values was exactly a right angle.

What, in amperes, were these two currents (lower first) and to what power were they raised?

[teaser2954]
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- Jim Randell 5:44 pm on 2 May 2019 Permalink | Reply
  The title of the puzzle, of course, is a reference the song Lovely Rita by The Beatles.
  
  This Python program runs in 82ms.
```
from enigma import (irange, subsets, div, fdiv, printf)

# the deflection in degrees is d(x) = (x^n)/k

# consider possible single figure powers n
for n in irange(2, 9):

  # choose a and b the two single figure values
  for (a, b) in subsets(irange(0, 9), size=2):

    # determine the constant
    k = div(b**n - a**n, 90)
    if k is None: continue

    # 9 amps -> deflection =< 180 degrees
    # 10 amps -> deflection > 180 degrees
    if not (9**n <= 180 * k < 10**n): continue
    
    d = lambda x: fdiv(x**n, k)
    printf("n={n}, a={a} b={b}, k={k} [a->{da:.2f} b->{db:.2f} 9->{d9:.2f} 10->{d10:.2f}]", da=d(a), db=d(b), d9=d(9), d10=d(10))
```
  Solution: The two currents are 3 A and 9 A. They are raised to the power of 8.
  
  So the formula for the deflection (in degrees) as a function of the current (in amps) is:
  
  d(x) = (x^8) / 478224
  
  Giving the corresponding readings for currents from 0 to 10 amps:
```
 0 A →   0.000°
 1 A →   0.000°
 2 A →   0.001°
 3 A →   0.014° (d = 9/656)
 4 A →   0.137°
 5 A →   0.817°
 6 A →   3.512°
 7 A →  12.055°
 8 A →  35.082°
 9 A →  90.014° (d = 90 + 9/656)
10 A → 209.107° (d > 180)
```
  So if you want to read small currents you had better be good at measuring very small angles.
  
  There are two further solutions for single digit powers and single digit currents that give a difference of exactly 90°.
  
  These are:
  
  d(x) = (x^4) / 72; for 3A and 9A
  d(x) = (x^6) / 2912; for 2A and 8A
  
  But these are disallowed by the condition that 9A should be on the scale and 10A off the scale. In the first case both 9A and 10A are on the scale, in the second case they are both off the scale. But they would be solutions to similar puzzles where the scale goes to 10A (but not 11A), or 8A (but not 9A), respectively.
  
  LikeLike
Jim Randell 3:04 pm on 2 May 2019 Permalink | Reply
Tags: by: G S Green
Brain-Teaser 477: [Tribute of goats]
From The Sunday Times, 19th July 1970 [link]

His four wives have to make tribute of goats to the Khan of Bakristan. My job is to collect the goats.

“Kahn Sahib says each gift to be divisible by donor’s age which consists of first and third digits of her number of goats”, said Munshi. “I don’t know ages, but know the goats required from each”.

I knew their ages. To celebrate each wife’s 31st birthday the Khan had married a teenager, so I quickly jotted down the four numbers of goats — no names because Munshi was looking over my shoulder and reading each of my four numbers from right to left, as they do in Bakristan. “Numbers correct”, he said.

Having done things the opposite way my numbers agreed with his.

What are the ages of the 3rd and 4th wife?

This puzzle was originally published with no title.

[teaser477]
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- Jim Randell 3:05 pm on 2 May 2019 Permalink | Reply
  If we suppose the current ages of the four wives are:
  
  AB > CD > EF > GH
  
  Then the difference in the ages of consecutive wives is a number between 12 and 18.
  
  And the numbers of goats must be:
  
  APB, CQD, ERF, GSH
  
  (assuming each is a three digit number), where each number is divisible by the corresponding age.
  
  And this set of numbers can also be read as:
  
  BPA, DQC, FRE, HSG
  
  This is enough to allow us to use the general alphametic solver [[ SubstitutedExpression() ]] from the enigma.py library to find an answer.
  
  The following run file executes in 188ms.
  
  Run: [ @replit ]
```
#! python -m enigma -rr

SubstitutedExpression

--distinct=""

# ages in order
"AB > CD > EF > GH > 12"

# number of goats is divisible by the age
"APB % AB = 0"
"CQD % CD = 0"
"ERF % EF = 0"
"GSH % GH = 0"

# age difference of consecutive wives
"11 < AB - CD < 19"
"11 < CD - EF < 19"
"11 < EF - GH < 19"

# the set of numbers is the same in reverse
"ordered(APB, CQD, ERF, GSH) == ordered(BPA, DQC, FRE, HSG)"

# answer is the current ages of the wives
--answer="(AB, CD, EF, GH)"
```
  Solution: The third wife is 35. The fourth wife is 17.
  
  When the first wife is 31, the new (second) wife was 13.
  
  18 years later: The first wife is 49, the second wife is now 31, and the new (third) wife is 13.
  
  18 years later: The first wife is 67, the second wife is 49, the third wife is now 31, and the new (fourth) wife is 13.
  
  4 years later (brings us to now): The first wife is 71, the second wife is 53, the third wife is 35, and the fourth wife is 17.
  
  The corresponding numbers of goats are: 781, 583, 385, 187 (1936 goats in total).
  
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Jim Randell 3:40 pm on 1 May 2019 Permalink | Reply
Tags: by: Danny Roth ( 88 )
Teaser 2917: Polygoland
From The Sunday Times, 19th August 2018 [link] [link]

We’ve all heard of Heligoland but George and Martha are on holiday in Polygoland. This is an island and when it was first inhabited, the authorities created as many national parks as possible subject to the restriction that each such park will be a regular polygon with the angle between each side being an integral number of degrees, and all the parks have different numbers of sides. Walking along the border between two such parks, Martha commented that the angle (A) of one of them was an exact multiple of the number of sides (S) in the other. George mentioned that the multiple was equal to the number of parks on the island.

What are A and S?

[teaser2917]
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- Jim Randell 3:40 pm on 1 May 2019 Permalink | Reply
  For a regular n-gon, the internal angles are (180 − 360 / n) degrees.
  
  So we can determine the number of possible n-gons with integer angles t from the number of divisors of 360 (greater than 2).
  
  There are 22 possible polygons, so t = 22
  
  So we need to find a solution to the equation:
  
  180 − 360 / n = 22m
  n = 360 / (180 − 22m)
  
  where n is a divisor of 360 (greater than 2).
  
  There are only a few values of m to consider and only one of them gives a viable value for n.
  
  Run: [ @repl.it ]
```
from enigma import (divisors, irange, div, printf)

# find divisors of 360 where n > 2
ns = list(n for n in divisors(360) if n > 2)
t = len(ns)
printf("[{t} different polygons]")

# now consider possible values of m
for m in irange(1, 180 // t):
  n = div(360, 180 - t * m)
  if n is None or n == m or n not in ns: continue
  printf("A={A} degrees, S={m} [m={m} n={n}]", A=(180 - 360 // n))
```
  Solution: A = 176°, S = 8.
  
  The solution to the equation is m = 8, which gives a value of n = 90.
  
  The angle 176° is 4° off a straight line, so if we walk along the perimeter each side turns us by 4°. After 90 sides we have turned through 360°.
  
  176 = 22 × 8, and 8 is a divisor of 360, so there will be an octagonal park (with interior angles of 180° − 45° = 135°).
  
  LikeLike
Jim Randell 8:10 am on 30 April 2019 Permalink | Reply
Tags: by: Victor Bryant ( 144 )
Brainteaser 1567: How many teams?
From The Sunday Times, 20th September 1992 [link]

The teams in our local football league each play each of the others once in a season, earning three points for a win and one point for a draw. After the last Saturday of the season (when all the teams had played one game that day) I worked out the league table (with the teams in alphabetical order). I then consistently replaced digits in the table by letters, using different letters for different digits.

Here is part of that table, showing some of the entries in three of the rows:

What is the number of teams in the league? And what is the number TEAMS?

This puzzle is included in the book Brainteasers (2002). The wording here is taken from the book and is only slightly changed from the original puzzle.

[teaser1567]
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- Jim Randell 8:11 am on 30 April 2019 Permalink | Reply
  If there are n teams in the league, then at the end of the season each team has played all the other (n − 1) teams, and this total number of matches must be represented in the sum of the “won”, “drawn”, “lost” columns in the table.
  
  H + O + W = n − 1
  M + A + N = n − 1
  T + E + A = n − 1
  
  If each team played one match on the final Saturday, then there must be an even number of teams. And if n is even, then (n − 1) is odd.
  
  We are given the points in the third row:
  
  3T + E = S
  
  And we are given the “goals against” for two teams. Each “lost” game involves conceding at least one goal, so:
  
  N ≤ Y, A ≤ M
  
  (and the values cannot be equal).
  
  Together these give us enough information to assign values to the letters.
  
  We can solve these alphametic expressions using the [[ SubstitutedExpression() ]] solver from the enigma.py library.
  
  The following run file executes in 216ms.
  
  Run: [ @repl.it ]
```
#! python3 -m enigma -rr

SubstitutedExpression

# sums of "won", "drawn", "lost" are the same
"H + O + W == M + A + N == T + E + A"

# sum is odd
"(H + O + W) % 2 = 1"

# points for third row
"3 * T + E = S"

# "lost" <= "goals against"
"N < Y"
"A < M"

# answer: (number of teams, value of TEAMS)
--answer="(H + O + W + 1, TEAMS)"
```
  Solution: There are 12 teams in the league. TEAMS = 16459.
  
  The first row has “won”, “drawn”, “lost” values of 0, 3, 8, but we don’t know which order.
  
  The second row has 5 “won”, 4 “drawn”, 2 “lost”, and 7 “goals against”.
  
  The third row has 1 “won”, 6 “drawn”, 4 “lost” and 5 “goals against”, giving 9 points.
  
  There are 12 teams, so there are 66 matches played in total.
  
  LikeLike

Jim Randell 10:47 am on 29 April 2019 Permalink | Reply
Tags: by: Angela Newing ( 39 )

Teaser 2918: Prime multiplication

From The Sunday Times, 26th August 2018 [link] [link]

Almost everyone knows that the single digit prime numbers are 2, 3, 5, and 7, the number 1 having been excluded quite a long time ago.

Here is a multiplication involving prime digits:

What is the answer?

[teaser2918]

Jim Randell 10:47 am on 29 April 2019 Permalink | Reply
We can easily use the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve this puzzle.

The following run file executes in 113ms.

Run: [ @replit ]
```
#! python -m enigma -rr

#        A B C
#    *     D E
#    ---------
#      F G H J
#    K L M N
#    ---------
#    P Q R S T
#    =========

SubstitutedExpression

# use only prime digits
--digits="2,3,5,7"
--distinct=""

"ABC * DE = PQRST"
"ABC * E = FGHJ"
"ABC * D = KLMN"

# required answer
--answer="PQRST"
```
Solution: The result of the multiplication is 25575.

If the digit 1 were allowed, there would be a further 9 solutions (including one that does not have the digit 1 in the result).

LikeLike

GeoffR 1:58 pm on 29 April 2019 Permalink | Reply

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

set of int: S = {2,3,5,7};    % single digit primes

var S: a;  var S: b;  var S: c;  var S: d;
var S: e;  var S: f;  var S: g;  var S: h;
var S: i;  var S: j;  var S: k;  var S: m;
var S: n;  var S: p;  var S: q;  var S: r;
var S: s;  var S: t;  

%      a b c
%   x    d e
%    -------
%    f g h i
%  j k m n 0
%  ---------
%  p q r s t
%  ---------

var 100..999: abc = 100*a + 10*b + c;
var 10..99: de = 10*d + e;
var 10000..99999: pqrst = 10000*p + 1000*q + 100*r + 10*s + t;
var 1000..9999: fghi = 1000*f + 100*g + 10*h + i;
var 10000..99999: jkmn0 = 10000*j + 1000*k + 100*m + 10*n;

constraint abc * de == pqrst;
constraint abc * d * 10 == jkmn0;
constraint abc * e == fghi;

solve satisfy;

% a = 7; b = 7; c = 5; d = 3; e = 3;
% f = 2; g = 3; h = 2; i = 5; 
% j = 2; k = 3; m = 2; n = 5; 
% p = 2; q = 5; r = 5; s = 7; t = 5;
% time elapsed: 0.02 s
%----------
%==========
% Sum is:
%     775
%   x  33
%   ----
%    2325
%   23250
%   -----
%   25575
%   -----

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Jim Randell 8:50 am on 28 April 2019 Permalink | Reply
Tags: by: Andrew Skidmore ( 87 )

Teaser 2919: Letters puzzle

From The Sunday Times, 2nd September 2018 [link] [link]

My best friend has a special birthday later this month, and I have composed a Teaser to celebrate the occasion.

First of all, I wrote that date as a single number (with the day at the start, the month in the middle and the year at the end).

Then, replacing different letters consistently with different non-zero digits, I found that the sum of LETTERS and PUZZLE gives that number.

Find the prime value of TEASER.

[teaser2919]

Jim Randell 8:51 am on 28 April 2019 Permalink | Reply
The sum is of the form:

LETTERS + PUZZLE = abc92018

Where a, b, c allow the result to read as a valid date in September 2018 (after the date of the puzzle, which is 2nd September 2018).

If the date is read as: ab.09.2018, we have: 3 ≤ ab ≤ 30; c = 0.

If the date is read as: bc.9.2018, we have: a = 0; 3 ≤ bc ≤ 30.

Note that if the result of the sum is 3092018, the birthday could be on 03.09.2018 or 30.9.2018.

The following run file executes in 133ms. (Internal runtime of the generated code is 1.1ms).

Run: [ @replit ]
```
#! python3 -m enigma -rr

SubstitutedExpression.split_sum

# upper case letters are distinct
--distinct="AELPRSTUZ"

# but non-zero
--invalid="0,AELPRSTUZ"

# digits stand for themselves
--literal="01289"

# expressions to solve
"{LETTERS} + {PUZZLE} = {abc92018}"
--extra="is_prime({TEASER})"

# the result of the sum should read as a valid September date
--code="valid = lambda d, f, v=0: f == v and 2 < d < 31"
--extra="valid({ab}, {c}) or valid({bc}, {a})"

# required answer
--answer="{TEASER}"

# [optional] faster prime checking
--code="is_prime = is_prime_mr"
```
Solution: TEASER = 314719.

The sum is: 2133197 + 658821 = 2792018.

And so the date is 27.9.2018 (27th September 2018).

LikeLike

Frits 12:11 pm on 13 August 2025 Permalink | Reply

Two calls to is_prime().

from functools import reduce
from enigma import is_prime

# LETTERS + PUZZLE = ab92018  (as L must be even) ab < 31 or b = 0
 
# convert digits to a number
d2n = lambda *s: reduce(lambda x, y: 10 * x + y, s)

nums = set(range(1, 10))

# e < 6 as u = 6 - e 
for e in range(1, 6):
  s = 8 - e 
  for r in nums - {e, s}:
    # TEASER is a prime number
    if r not in {1, 3, 7, 9}: continue
    # (R + L) % 10 = 1
    if not (0 < (l := 11 - r) < 10): continue
    # (E + Z + 1) % 10 = 0
    if not (0 < (z := 9 - e) < 10): continue
    # (T + Z + 1) % 10 = 2,  T = 11 - Z = 2 + E
    t = e + 2
    if not (0 < (t := 11 - z) < 10): continue
    # (T + U + 1) % 10 = 9, U = 8 - T, U = 6 - E
    u = 6 - e
    # different digits
    if len(unused := nums - {e, s, r, l, z, t, u}) != 2: continue
   
    letters = d2n(l, e, t, t, e, r, s)
    # L must be even, if L >= 3: P + E = 10, if L = 2: P + E <= 10
    for p in [u for u in unused if (u + e < 11 if l == 2 else u + e == 10)]:
      puzzle = d2n(p, u, z, z, l, e)
      for a in unused - {p}:
        if is_prime(teaser := d2n(t, e, a, s, e, r)):
          print("answer:", teaser)

LikeLike

Jim Randell 8:17 am on 27 April 2019 Permalink | Reply
Tags: by: G F Anderson
Brain-Teaser 476: [Football pools]
From The Sunday Times, 12th July 1970 [link]

A football pools fan suggests a novel points scoring system which, he says, is a lot of fun and so easy to do.

For Home wins and Draws the goals for teach match are written down (Home teams first, of course), and the figures placed next to each other. This gives the points. For example, a Home team winning by 10 goals to 3 scores 103 points, a 5-5 Draw 55 points, and so on. You have to select a given number of matches, for example, eight, from the list, which the object of making the highest points score.

I need not tell you about points for Away wins, because I did not include any among my selections when I tried out the system the other day. All my forecasts were correct. They comprised 2 Home Wins, and the remainder were all Draws. One of my Home wins was worth 2 more points than the other Home win. No match had a goals aggregate greater than 16, and the points obtained from my 2 Home wins accounted for exactly 60 per cent of the points obtained from all the matches I selected.

How many points did I score?

This puzzle was originally published with no title.

[teaser476]
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- Jim Randell 8:18 am on 27 April 2019 Permalink | Reply
  This Python 3 program runs in 93ms.
  
  Run: [ @repl.it ]
```
from enigma import (nsplit, nconcat, irange, subsets, div, join, printf)

# points for a match with a score of <x>-<y>
def points(x, y):
  return nconcat(nsplit(x) + nsplit(y))

# possible wins
wins = list()
for x in irange(1, 16):
  wins.extend((x, y, points(x, y)) for y in irange(0, min(x - 1, 16 - x)))

# possible draws
draws = list((x, x, points(x, x)) for x in irange(0, 8))

# find k draws that give t points
def solve(k, t, ds=draws, s=[]):
  if k == 0:
    if t == 0:
      yield s
  else:
    for (i, (x, y, p)) in enumerate(ds):
      if not (p > t):
        yield from solve(k - 1, t - p, ds[i:], s + [(x, y)])

# choose the two home wins (different)
for ((x1, y1, p1), (x2, y2, p2)) in subsets(wins, size=2):
  if not (abs(p1 - p2) == 2): continue

  # p1 + p2 accounts for exactly 60% of the points
  # so, calculate the total number of points
  w = p1 + p2
  t = div(100 * w, 60)
  if t is None: continue

  # find 5 draws (not necessarily all different) worth the remaining 40%
  for (i, ds) in enumerate(solve(5, t - w)):
    if i == 0: printf("total {t} pts; wins = {x1}-{y1}; {x2}-{y2}")
    printf("  draws = {ds}", ds=join((join(d, sep="-") for d in ds), sep="; "))
    break # only need one example
```
  Solution: Your selection scored 440 points.
  
  Possible draws are 0-0, 1-1, …, 8-8, scoring 0, 11, …, 88 points respectively. So draws always score some multiple of 11 points.
  
  There are 16 ways to have two wins that have scores differing by 2, but only one of these gives the remaining 40% of points being a multiple of 11.
  
  That is: wins of 13-1 and 13-3, which have scores of 131 points and 133 points respectively.
  
  Together these give 264 points, which is 60% of a total of 440 points, leaving 176 points to be made up from the draws, and 176 = 16 × 11.
  
  There are 63 ways in which we can choose 5 draws to give a total of 176 points. For example: two 8-8 matches, and three 0-0 matches.
  
  LikeLike
Jim Randell 8:01 am on 26 April 2019 Permalink | Reply
Tags: by: Andrew Skidmore ( 87 )
Teaser 2953: Marble tower
From The Sunday Times, 28th April 2019 [link] [link]

Liam has a number of bags of marbles; each bag contains the same number (more than 1) of equal-size marbles.

He is building a tetrahedron with the marbles, starting with a layer which fits snugly in a snooker triangle. Each subsequent triangular layer has one fewer marble along each edge. With just one bag left he had completed a whole number of layers; the number of marbles along the edge of the triangle in the last completed layer was equal to the number of completed layers. The last bag had enough marbles to just complete the next layer.

How many bags of marbles did Liam have?

[teaser2953]
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- Jim Randell 8:24 am on 26 April 2019 Permalink | Reply
  This Python program runs in 85ms.
```
from enigma import (irange, inf, T, div, printf)

# P(n) = the nth tetrahedral number
P = lambda n: (n * (n + 1) * (n + 2)) // 6

def solve():
  # suppose there are n bags, each containing k marbles
  # there are enough marbles in the final bag to complete a layer
  # so the number of marbles in a bag is a triangular number
  for x in irange(2, inf):
    k = T(x)

    # and there must be (x + 1) completed layers
    # and layers T(x + 1) ... T(2x + 1) have used (n - 1) k marbles
    # so: P(2x + 1) - P(x) = (n - 1) k
    # solve for (n - 1)
    n = div(P(2 * x + 1) - P(x), k)
    if n:
      yield (k, n + 1, x)

for (k, n, x) in solve():
  printf("{n} bags, each containing {k} marbles, layers 1 - {x} (of {t}) remaining", t=2 * x + 1)
  break
```
  Solution: There were 20 bags of marbles.
  
  We can simplify the equation:
  
  P(2x + 1) − P(x) = (n − 1)T(x)
  
  to:
  
  n = 7x/3 + 17/3 + 2/x
  
  For integer x the first two terms give us a whole number of thirds, so the 2/x term must also provide a whole number of thirds, i.e. x = 1, 2, 3, or 6.
  
  And the only values to give an integer value for n are:
  
  x = 1, n = 10
  x = 6, n = 20
  
  There are T(x) marbles in each of the n bags, and this is more than 1, so this eliminates the first solution leaving, T(6) = 21 marbles in each of the 20 bags.
  
  Here’s a Python program that uses the analysis:
```
from enigma import (div, printf)

for x in [2, 3, 6]:
  n = div((7 * x + 17) * x + 6, 3 * x)
  if n:
    printf("n={n} [x={x}]")
```
  LikeLike

Jim Randell 8:16 am on 25 April 2019 Permalink | Reply
Tags: by: Pete Bradley ( 7 )

Brainteaser 1563: Family sum

From The Sunday Times, 23rd August 1992 [link]

In this puzzle digits have been consistently replaced by letters, with different letters being used for different digits. This addition sum then makes sense:

also AND is divisible by three.

What is the number REAGAN?

This puzzle is included in the book Brainteasers (2002). The wording is only slightly changed from the original puzzle.

[teaser1563]

Jim Randell 8:17 am on 25 April 2019 Permalink | Reply
This puzzle can easily be solved using the [[ SubstitutedExpression() ]] solver from the enigma.py library.

The following run-file executes in 480ms.

Run: [ @replit ]
```
#! python3 -m enigma -rr

SubstitutedExpression

"RONALD + AND + NANCY = REAGAN"

"AND % 3 = 0"

--answer="REAGAN"
```
Solution: REAGAN = 396168.

The symbols L and C appear in the tens column, but not elsewhere, so their values can be interchanged. This gives rise to the two possible sums:

308627 + 687 + 86854 = 396168
308657 + 687 + 86824 = 396168

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- Jim Randell 8:41 am on 9 June 2023 Permalink | Reply
  For a faster solution we can use the [[ SubstitutedExpression.split_sum ]] solver.
  
  The following run file executes in 73ms. (Internal runtime of the generated program is 1.5ms).
  
  Run: [ @replit ]
```
#! python3 -m enigma -rr

SubstitutedExpression.split_sum

"RONALD + AND + NANCY = REAGAN"

--extra="AND % 3 = 0"

--answer="REAGAN"
```
  LikeLike

GeoffR 12:20 pm on 25 April 2019 Permalink | Reply

% A Solution in MiniZinc
include "globals.mzn";
 
var 0..9: R; var 0..9: O; var 0..9: N; var 0..9: A;  var 0..9: L; 
var 0..9: D; var 0..9: C; var 0..9: Y; var 0..9: E;  var 0..9: G; 

constraint R != 0 /\ A != 0 /\ N != 0;
constraint all_different( [R,O,N,A,L,D,C,Y,E,G]);

var 100..999: AND = 100*A + 10*N + D;
var 10000..99999: NANCY = 10000*N + 1000*A + 100*N + 10*C + Y;

var 100000..999999: RONALD = 100000*R + 10000*O + 1000*N 
+ 100*A + 10*L + D;

var 100000..999999: REAGAN = 100000*R + 10000*E + 1000*A 
+ 100*G + 10*A + N;

constraint AND mod 3 == 0;

constraint RONALD + AND + NANCY == REAGAN;

solve satisfy;

output [show(RONALD) ++ " + " ++ show(AND) ++ " + " ++ 
show(NANCY) ++ " = " ++ show(REAGAN)];

% 308657 + 687 + 86824 = 396168
% time elapsed: 0.02 s
% ----------
% 308627 + 687 + 86854 = 396168
% time elapsed: 0.02 s
%----------
%==========

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Frits 1:44 pm on 9 June 2023 Permalink | Reply

  
column 3: x + N + A = .A with x = 1,2,3 so N = 8,9,0

column 2: y + O + N = E so either:

N = 9, is not possible
N = 8 and O = 0 and x = 2  
N = 0 causes y to be zero but then O = E

--> N = 8, O = 0, E = 9 and x = 2

column 4:
z + A + A + N = 2G so z + A + A >= 21 - 8 so A = 6,7

using AND is divisible by three:
(A, D, Y) is either (6, 7, 4) or (7, 3, 2)
(A, D, Y, L+C) is either (6, 7, 4, 7) or (7, 3, 2, 9)
(A, D, Y, {L, C}) is either (6, 7, 4, {2, 5}) or (7, 3, 2, {4, 5})

(A, G) = (6, 1) as A = 7 causes G to be 3 (which is same as D)

so (N, O, E, A, D, Y, G) = (8, 0, 9, 6, 7, 4, 1) with {L, C} = {2, 5}
This leaves R = 3

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Jim Randell 10:47 am on 24 April 2019 Permalink | Reply
Tags: by: Ian Gilchrist ( 2 )
Teaser 2920: Barcode
From The Sunday Times, 9th September 2018 [link] [link]

When Fred opened his “10 to 99 Store” (the numbers reflecting the prices of the goods in pounds) there was trouble from the start when the barcode reader scanned the prices backwards. First at the self-service checkout was Mrs Adams, who kept quiet when she was charged £59.49 for a £94.95 item. The next two customers, Mr Baker and Mr Coles, paid by card, without noticing they had in fact been overcharged. All three had bought a differently-priced item, and by coincidence, in the case of the two men, the price each had been charged was a whole multiple of the actual price.

How much had the shop lost or gained overall on the three transactions?

[teaser2920]
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- Jim Randell 10:48 am on 24 April 2019 Permalink | Reply
  This Python program runs in 157ms.
  
  (The following is for Python 2.7, see the code on repl.it for the Python 3 version).
  
  Run: [ @repl.it ]
```
from enigma import (Accumulator, irange, nreverse, div, printf)

# r accumulates the total gain for the shop
def transaction(t, (p, q)):
  d = q - p
  printf("[{n}: price = {p} -> charged = {q}, diff = {d}]", n=r.count)
  return t + d

r = Accumulator(fn=transaction, value=0)

# transaction A
p = 9495
q = nreverse(p)
assert q < p
r.accumulate((p, q))

# for transactions B and C, consider 4-digit prices
for p in irange(1000, 9999):
  # reverse the price
  q = nreverse(p)
  if not (q > p): continue
  # reversed price is an exact multiple
  m = div(q, p)
  if m is None or m < 2: continue
  r.accumulate((p, q))

# there should be three transactions in total
assert r.count == 3

# output total
printf("total shop gain = {g:.2f} pounds", g=r.value * 0.01)
```
  Solution: In total the shop gained £ 117.00 over the three transactions.
  
  The transactions for B and C are:
  
  (1) paying £ 98.01 for an item priced at £ 10.89 (9 times)
  (2) paying £ 87.12 for an item priced at £ 21.78 (4 times)
  
  LikeLike
Jim Randell 10:48 am on 23 April 2019 Permalink | Reply
Tags: by: J L Bowles ( 13 )
Brain-Teaser 475: Club lockers
From The Sunday Times, 5th July 1970 [link]

The lockers in one section of the new club-house are numbered 1 to 12 consecutively. Last Saturday, Pitcher arrived without his key. He was surprised to find that his locker could be opened by the key to Putter’s locker which is next to his.

Pitcher at once saw Wedge, the secretary, and Bunker, the captain. From them he learned that there are only 3 different key patterns for the 12 lockers (patterns A, B and C) and that the total of the numbers on the 4 A lockers is the same as the corresponding total for the 4 B lockers, which is the same as that for the 4 C lockers.

Wedge also told Pitcher that no two lockers with pattern A keys are adjoining. He added that his own key is of the same pattern as Wood’s and that the total of the numbers on Wood’s and Wedge’s lockers is 16. He further mentioned that of the 3 lockers, other than the captain’s own locker, which can be opened by the captain’s key, the nearest to the captain’s locker is further from the captain’s locker than are 5 (but not more) lockers which cannot be opened by the captain’s key.

By this time, Pitcher was making for the wall but if the number on Putter’s locker is higher than the number on Wedge’s locker — what is the number on Bunker’s locker?

This puzzle is included in the book Sunday Times Brain Teasers (1974).

[teaser475]
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- Jim Randell 10:49 am on 23 April 2019 Permalink | Reply
  This Python 3 program runs in 84ms.
  
  Run: [ @repl.it ]
```
from enigma import (irange, subsets, tuples, diff, printf)

# possible locker numbers
lockers = irange(1, 12)

# find subsets of size 4 that sum to 26 (= sum(lockers) / 3)
s4s = list(s for s in subsets(lockers, size=4) if sum(s) == 26)

# choose pairs from the given sets
def pairs(*ss):
  for s in ss:
    yield from subsets(s, size=2)

# choose a set of lockers for key pattern A
for A in s4s:
  # there are no consecutive numbers in A
  if any(y - x == 1 for (x, y) in tuples(A, 2)): continue

  # choose a (disjoint) set that includes Bunker's key
  for X in s4s:
    if set(A).intersection(X): continue

    # and the remaining set
    Y = diff(lockers, A + X)

    # choose a locker for Bunker
    for b in X:

      # find the distance of the nearest other locker to b in X
      d = min(abs(b - x) for x in X if x != b)

      # there must be exactly 5 lockers not in X closer to b
      n = sum(abs(b - x) < d for x in lockers if x not in X)
      if not (n == 5): continue

      # Wedge's and Wood's lockers (in the same set) sum to 16
      for (w1, w2) in pairs(A, X, Y):
        if not (w1 + w2 == 16): continue

        # Pitcher's and Putter's lockers (in the same set) are adjacent
        for (p1, p2) in pairs(A, X, Y):
          if not (p2 - p1 == 1): continue

          # all people are different
          if not (len(set((b, w1, w2, p1, p2))) == 5): continue

          # Putter's locker has a higher number than Wedge's
          if not (max(p1, p2) > min(w1, w2)): continue

          printf("A={A} X={X} Y={Y}, b={b} d={d}, ws=({w1}, {w2}) ps=({p1}, {p2})")
```
  Solution: Bunker has locker number 2.
  
  The A pattern keys are for lockers (1, 3, 10, 12). The other pattern keys are lockers (2, 7, 8, 9) and (4, 5, 6, 11).
  
  Wedge and Wood have lockers 5 and 11 (respectively).
  
  Pitcher and Putter have lockers 7, 8 or 8, 9 (in some order).
  
  LikeLike
Jim Randell 9:21 am on 22 April 2019 Permalink | Reply
Tags: by: Stephen Hogg ( 64 )
Teaser 2921: Octavia the octagonarian
From The Sunday Times, 16th September 2018 [link] [link]

Queen Octavia’s emblem was two concentric regular octagons (the outermost’s side length equal to the innermost’s distance between opposite sides). Seen from above her eightieth-birthday, level, cheesecake torte, with octagonal cavity all the way through and vertical side-faces, matched the emblem. Its outer side-faces were square and a two-figure whole number of inches wide. A four-feet diameter cylindrical cover protected it. At the party the whole torte was used, without wastage, to levelly fill a number of identical cuboid dishes (internally having length, width and depth, in inches, each a different single-figure whole number greater than one). Curiously, the number of dishes was a three-figure value with different numerals in descending order.

What was the torte’s volume in cubic inches?

[teaser2921]
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- Jim Randell 9:22 am on 22 April 2019 Permalink | Reply
  This Python program runs in 78ms.
```
from enigma import (irange, subsets, nsplit, tuples, sqrt, divf, div, printf)

# diameter of a regular octagon with side 1 unit
d = sqrt(4 + 2 * sqrt(2))

# possible volumes of the dishes
vs = set(a * b * c for (a, b, c) in subsets(irange(2, 9), size=3))

# is <n> a k-digit descending number?
def check(n, k=3):
  ds = nsplit(n)
  return (len(ds) == k and all(a > b for (a, b) in tuples(ds, 2)))

# consider <x>, the side length of the outer octagon = the distance
# between opposite faces of the inner octagon
# the outer octagon can fit in a 48 inch tube
for x in irange(10, min(99, divf(48, d))):

  # volume of cake
  V = 4 * x**3

  # consider the volume of each dish
  for v in vs:

    # the number of dishes is:
    n = div(V, v)
    if n is None or not check(n): continue

    printf("x={x} V={V}, v={v} n={n}")
```
  Solution: The volume of the torte was 23,328 cubic inches.
  
  The outside faces of the cake are 18 inch squares, and this is the largest cake that will fit inside a 48 inch diameter tube.
  
  There are 2 possibilities for the size of the serving dishes, either 972 dishes measuring 2×3×4 inches, or 432 dishes measuring 2×3×9 inches.
  
  Although the larger dish size has a number of dishes composed of consecutive descending digits (which is possibly what the setter had in mind), I think the smaller sized dish makes the most sense. Which means there were almost 1,000 guests catered for.
  
  LikeLike

Jim Randell 12:51 pm on 21 April 2019 Permalink | Reply
Tags: by: J H Perryman ( 5 )

Brainteaser 1557: An alphabetical jigsaw

From The Sunday Times, 12th July 1992 [link]

I have some jigsaw pieces which fit together to form a 5-by-5 square with the letters all the right way up. Here is an attempt to make the jigsaw:

The attempt has failed because the last piece does not fit the right way up. In fact it is possible to make the 5-by-5 jigsaw so that the 25 different letters spell six words in reading order.

What is this arrangement?

This puzzle appears in the book Brainteasers (2002). The wording above is taken from the book, but is only slightly changed from the original puzzle.

[teaser1557]

Jim Randell 12:52 pm on 21 April 2019 Permalink | Reply

There are two parts to the program. The first is concerned with fitting the pieces into the grid. The second part is then reading the letters in the grid as a collection of words.

This Python program runs in 6.6s, but most of the time is taken up with collecting the word list. I used a collection of allowable Scrabble words, which gives 49,595 viable words.

Run: [ @replit ] (using just the selected words)

from itertools import product
from collections import (Counter, defaultdict)
from enigma import (irange, find, chunk, flatten, join, printf)

(X, Y) = (1, 7)

pieces = [
  { 0: 'H', Y: 'Z', X + Y: 'F', X + Y + Y: 'X' },
  { 0: 'P', X: 'S', X + X: 'C' },
  { 0: 'I', Y: 'J', Y + Y: 'O' },
  { 0: 'G', X: 'Y', X + Y: 'M' },
  { 0: 'T', Y: 'V', Y + Y: 'N' },
  { 0: 'E', Y: 'K', X + Y: 'W' },
  { 0: 'U', Y: 'R', X + Y: 'D' },
  { 0: 'A', X: 'L', X + Y: 'B' },
]

# create the empty grid
grid = [None] * (Y * Y)
for (x, y) in product(irange(1, 5), irange(1, 5)):
  grid[x + y * Y] = '.'

# fit piece p into grid g at index i
# return new grid or None
def fit(g, p, i):
  g = g.copy()
  for (k, v) in p.items():
    x = g[i + k]
    # non-usable square?
    if x != '.': return None
    g[i + k] = v
  return g

def solve(ps=pieces, g=grid):
  # are we done?
  if not ps:
    # return the completed (unpadded) grid
    yield list(r[1:-1] for (i, r) in enumerate(chunk(g, Y), start=1) if 1 < i < Y)
  else:
    # find an empty square
    i = find(g, '.')
    if i != -1:
      for (j, p) in enumerate(ps):
        g2 = fit(g, p, i)
        if g2 is not None:
          yield from solve(ps[:j] + ps[j + 1:], g2)

# list of words to examine
file = "words.txt"

# collect letters from the pieces
letters = Counter()
for p in pieces: letters.update(p.values())

# collect words that can be formed from the letters
words = defaultdict(list)
for w in open(file).readlines():
  w = w.strip().upper()
  if not (Counter(w) - letters):
    words[w[0]].append(w)

printf("[collected {n} words from {file}]", n=sum(len(v) for v in words.values()))

# form text t into a sequence of words
def check(t, s=[]):
  if not t:
    yield s
  else:
    for w in words[t[0]]:
      if t.startswith(w):
        yield from check(t[len(w):], s + [w])


# fit the pieces into the grid
for g in solve():
  # attempt to form the letters into words
  wss = list(check(join(flatten(g))))
  if wss:
    # output the grid
    for r in g:
      printf("{r}", r=join(r, sep=" "))
    # output the words
    for ws in wss:
      if len(ws) == 6:
        printf("words: {ws}", ws=join(ws, sep=", "))
    printf()

Solution: The completed jigsaw looks like this:

So the 6 words are: GYP, SCHMALTZ, FIB, VEX, JUNK, WORD.

There are 64 ways to fit the pieces into the grid, but only one of them makes a set of viable words.

There are 2 fifteen letter words in the list that can be made from the collection of letters in the puzzle: DERMATOGLYPHICS, UNCOPYRIGHTABLE.

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Jim Randell 9:04 am on 20 April 2019 Permalink | Reply
Tags: by: E C Parker
Brain-Teaser 474: [Triangular park]
From The Sunday Times, 28th June 1970 [link]

Three friends live at the side entrances to a park bordered by three straight roads. Alan’s entrance is on the shortest road, Colin’s on the longest East to West road, and Brian’s on the road of mid length.

Brian and Colin are directly North and South apart across the park, for half the distance they each are by road from Alan. The distance Brian and Colin are apart, plus the length of the three roads, is three miles.

Each entrance is a number of exact yards by road, from any corner of the park, and Alan is as near to one corner as possible.

How far from that corner is he?

This puzzle was originally published with no title.

[teaser474]
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- Jim Randell 9:06 am on 20 April 2019 Permalink | Reply
  If B and C’s gates are at a distance of d apart N-S across the park, and distances b and c from a corner X, then (b, c, d) form a Pythagorean triple.
  
  Once we have a suitable Pythagorean triple we can extend the lines XB and XC by lengths of 2d and bend them (at integer values) so that the bent sides form the third side of the triangle, which meet at A’s gate.
  
  If this is possible we have a viable solution. We then need to find which solution places A’s gate closest to a corner of the park.
  
  This Python program finds candidates solutions and choose the one with the smallest distance from A’s gate to a corner. It runs in 108ms.
```
from enigma import (pythagorean_triples, irange, div, Accumulator, printf)

# find the minimum distance of A from a corner
r = Accumulator(fn=min)

# BCX is right angled triangle with integer sides
for (x, y, z) in pythagorean_triples(5280):
  for (b, c, d) in [(z, x, y), (z, y, x)]:

    # check 5d + b + c = 5280 yards
    if not (5 * d + b + c == 5280): continue

    # now consider possible integer values for y
    for y in irange(1, 2 * d - 1):

      # compute viable z (using the cosine rule)
      p = (b - c) * (b * b + b * (4 * d + c) + 4 * d * (c + 2 * d - y))
      q = 2 * (b * b + b * (2 * d + y) + c * (y - c - 2 * d))
      z = div(p, q)
      if z is None or not (0 < z < 2 * d): continue

      # compute the sides of the park
      (A, B, C) = (y + z, b + 2 * d - z, c + 2 * d - y)

      # A is on the shortest side and C is on the longest side
      if not (A < B < C): continue

      # additional: the park is a right angled triangle
      #if not (A * A + B * B == C * C): continue

      printf("[A={A} B={B} C={C}, d={d} b={b} c={c}, y={y} z={z}]")

      r.accumulate_data(min(y, z), (A, B, C, d, b, c, y, z))

# output the solution
printf("min distance for A from a corner = {r.value} yards [{r.count} solutions]")
```
  Solution: The shortest distance A’s gate can be from a corner is 135 yards.
  
  The program finds 8 possible solutions:
```
A=1218 B=1400 C=2002, d=660 b=1100 c=880, y=198 z=1020
A=1155 B=1540 C=1925, d=660 b=1100 c=880, y=275 z=880
A=1122 B=1650 C=1848, d=660 b=1100 c=880, y=352 z=770
A=1110 B=1750 C=1760, d=660 b=1100 c=880, y=440 z=670
A=780 B=1800 C=2220, d=480 b=1480 c=1400, y=140 z=640
A=750 B=1850 C=2200, d=480 b=1480 c=1400, y=160 z=590
A=678 B=2050 C=2072, d=480 b=1480 c=1400, y=288 z=390
A=429 B=2196 C=2325, d=330 b=1830 c=1800, y=135 z=294
```
  The last of these has the smallest value for y or z.
  
  So the park looks like this:
  
  However, apparently this was not the intended answer to this puzzle. The puzzle text should have said that the park was a right angled triangle, but this extra condition was omitted when the puzzle was published.
  
  We can remove solution candidates that are not right-angled triangles by uncommenting the check at line 29.
  
  If we do this we find that only one of the solutions remains:
```
A=1155 B=1540 C=1925, d=660 b=1100 c=880, y=275 z=880
```
  The shortest distance being 275 yards.
  
  In this case the park looks like this:
  
  However, given there is only a single solution it seems odd to have the statement: “Alan is as near to one corner as possible”. It would make more sense to just ask: “How far is Alan from the nearest corner?”.
  
  LikeLike

	Jim Randell on Teaser 2444: [Pat’s…
	Jim Randell on Teaser 2447: [Number grid…
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