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Jim Randell 8:57 am on 3 March 2022 Permalink | Reply
Tags: by: Angela Newing
Teaser 2874: An age-old progression
From The Sunday Times, 22nd October 2017 [link] [link]

It is my birthday today — the same day as two of my younger relatives, Betty and Charles. I commented that the digits involved in our three ages are all different. Betty noted that the square of her age is equal to my age multiplied by Charles’s age. Then Charles added that on one of our birthdays in the next ten years the sum of our ages will be one hundred.

What are our three ages today?

[teaser2874]
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Jim Randell is discussing. Toggle Comments
- Jim Randell 8:58 am on 3 March 2022 Permalink | Reply
  We have:
  
  A > B, C
  B² = A⋅C (i.e. B is the geometric mean of A and C)
  A, B, C digits are all different
  A + B + C + 3n = 100, for some n = 1 .. 10
  
  We can get an upper bound on B by setting A = B = C:
  
  A + B + C ≤ 97
  3B ≤ 97
  B ≤ 32
  
  This Python program finds possible values for A, B, C. It runs in 47ms.
  
  Run: [ @replit ]
```
from enigma import (irange, divisors_pairs, is_duplicate, div, printf)

# consider values of B
for B in irange(1, 32):

  # B^2 = A.C; C < A
  for (C, A) in divisors_pairs(B, 2):
    if not (C < B < A): continue

    # calculate n
    n = div(100 - (A + B + C), 3)
    if n is None or n < 1 or n > 10: continue

    # check for repeated digits
    if is_duplicate(A, B, C): continue

    # output solution
    printf("A={A} B={B} C={C}; n={n}")
```
  The program finds 2 possible solutions:
  
  B=8; C=1 A=64; n=9
  B=21; C=7 A=63; n=3
  
  The second of these seems the most likely as Charles joins in the conversation. (Although it would have been nice if the first case had been ruled out explicitly by the puzzle text).
  
  Solution: Angela is 63. Betty is 21. Charles is 7.
  
  We have 21² = 441 = 7 × 63 as required.
  
  And in 3 years time: (63 + 3) + (21 + 3) + (7 + 3) = 100.
  
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Jim Randell 4:25 pm on 28 January 2022 Permalink | Reply
Tags: by: Angela Newing
Teaser 3097: Crazy golf
From The Sunday Times, 30th January 2022 [link] [link]

Ian was trying to entertain John and Ken, his 2 young nephews, in the local park, which had a 9-hole crazy golf course. He decided to play a round with each of them separately, and gave them money for each hole he lost on a rising scale. He would pay £1 if he lost the first hole (or the first hole after winning one), then £2 if he lost the next consecutive hole, and £3 for the third, and so on.

In the event, Ian won only 5 holes in total between the two rounds, including the first hole against John and the last hole against Ken. There were no ties. At the reckoning after both rounds, both boys received equal amounts of money.

How much did it cost Uncle Ian?

[teaser3097]
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- Jim Randell 4:47 pm on 28 January 2022 Permalink | Reply
  Each boy loses a hole at one end of the course, so we can just consider course with 8 holes, as we are only interested in groups of consecutive wins.
  
  This Python program runs in 47ms.
  
  Run: [ @replit ]
```
from enigma import (irange, subsets, intersect, printf)

# consider a course with 8 holes
holes = irange(1, 8)

# calculate the winnings for a nephew who wins holes <ws>
def winnings(ws):
  (t, h) = (0, 0)
  for n in holes:
    if n in ws:
      h += 1
      t += h
    else:
      h = 0
  return t

# choose 5 to 8 extra holes for J and K
X = dict()
for n in irange(5, 8):
  X[n] = set(winnings(ws) for ws in subsets(holes, size=n))

# consider number of holes won by J
for j in irange(5, 8):
  k = 13 - j
  # look for equal amounts
  for w in intersect([X[j], X[k]]):
    printf("I pays {t} [J wins {j}; K wins {k}]", t=w + w)
```
  Solution: Ian paid £32 to the boys.
  
  Ian lost to one of the boys on 5 consecutive holes and 1 other, paying (1 + 2 + 3 + 4 + 5) + 1 = £16.
  
  And to the other boy on 4 consecutive holes and another 3 consecutive holes, playing (1 + 2 + 3 + 4) + (1 + 2 + 3) = £16.
  
  So, in total Ian pays £32 for the 13 holes he lost.
  
  From 8 holes there are 6 ways that runs of 5 and 1 can be made:
  
  (1 2 3 4 5) (7)
  (1 2 3 4 5) (8)
  (2 3 4 5 6) (8)
  (1) (3 4 5 6 7)
  (1) (4 5 6 7 8)
  (2) (2 5 6 7 8)
  
  And there are 2 ways that runs of 4 and 3 can be made:
  
  (1 2 3 4) (6 7 8)
  (1 2 3) (5 6 7 8)
  
  So if John wins 5+1 and Ken wins 4+3 there are 12 possibilities.
  
  And if John wins 4+3 and Ken wins 5+1 there are also 12 possibilities.
  
  Giving 24 possible ways to achieve the required games.
  
  Manually, we see that the boys won 13 holes between them, so the possible splits are (8, 5) or (7, 6).
  
  Winning 8 holes would give a prize of T(8) which must be more than the winnings from 5 holes. So the splits are (7, 6).
  
  Possible consecutive runs for 7 (and corresponding prizes) are:
  
  (7) → T(7) = 28
  (1, 6) → T(1) + T(6) = 22
  (2, 5) → T(2) + T(5) = 18
  (3, 4) → T(3) + T(4) = 16
  
  Possible runs for 6 (and corresponding prizes) are:
  
  (6) → T(6) = 21
  (1, 5) → T(1) + T(5) = 16
  (2, 4) → T(2) + T(4) = 13
  (3, 3) → T(3) + T(3) = 12
  (1, 1, 4) → T(1) + T(1) + T(4) = 12
  (1, 2, 3) → T(1) + T(2) + T(3) = 10
  (2, 2, 2) → T(2) + T(2) + T(2) = 9
  
  The only overlap is for a prize of £16, with consecutive runs of (3, 4) and (1, 5).
  
  LikeLike
  - Jim Randell 11:04 am on 29 January 2022 Permalink | Reply
    Here’s an alternative approach, which instead of generating the numbers of the holes won, finds possible runs of wins among the 8 holes.
    
    It is slightly faster than my original solution (above).
    
    Run: [ @replit ]
```
from enigma import (irange, decompose, tri, intersect, printf)

# generate possible runs of <n> wins for <k> holes
def generate(n, k):
  # split n into j parts
  for j in irange(1, k + 1 - n):
    yield from decompose(n, j, sep=0)

# consider winning runs totalling 5-8 holes on an 8 hole course
X = dict()
for n in irange(5, 8):
  X[n] = set(sum(tri(x) for x in ws) for ws in generate(n, 8))

# consider number of holes won by J
for j in irange(5, 8):
  k = 13 - j
  # look for equal winnings
  for w in intersect([X[j], X[k]]):
    printf("I pays {t} [J wins {j}; K wins {k}]", t=w + w)
```
    LikeLike
Jim Randell 11:18 am on 16 December 2021 Permalink | Reply
Tags: by: Angela Newing, flawed ( 54 )
Teaser 2844: Children’s children
From The Sunday Times, 26th March 2017 [link] [link]

The head teacher’s retirement celebration was attended by many former pupils. These included Adam, Brian, Colin and David, whose sons and grandsons had also attended the school – actually different numbers of grandsons for each of them, with Adam having the most. Their sons were Eric, Fred, George, Harry and Ivan, and the sons of the sons were John, Keith, Lawrence, Michael, Norman and Oliver.

Altogether Adam and Brian had sons Eric, Fred and one other; altogether Adam and Colin had sons George and one other; altogether Eric, George and Harry had sons Keith, Michael, Norman and one other; and altogether Harry and Ivan had sons Lawrence, Norman and one other.

The retirement gift was presented by the fathers of John’s and Oliver’s fathers.

Who were they?

As stated there are multiple solutions to this puzzle.

This puzzle was not included in the published collection of puzzles The Sunday Times Brainteasers Book 1 (2019).

There are now 600 puzzles available on S2T2.

[teaser2844]
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Hugh+Casement and Jim Randell are discussing. Toggle Comments
- Jim Randell 11:19 am on 16 December 2021 Permalink | Reply
  As stated the puzzle has multiple solutions.
  
  The following Python program builds the 8 possible family trees that fit the described situation.
  
  It runs in 85ms.
  
  Run: [ @replit ]
```
from enigma import (subsets, peek, product, map2str, join, printf)

# map ks to vs
def make_map(ks, vs):
  for ss in subsets(ks, size=len(vs), select='M'):
    d = dict((k, list()) for k in ks)
    for (k, v) in zip(ss, vs):
      d[k].append(v)
    yield d

# find viable father -> sons mappings
def check_fs(fs):

  # "A, B have sons E, F and one other"
  s = fs['A'] + fs['B']
  if not (len(s) == 3 and 'E' in s and 'F' in s): return False

  # "A, C have sons G and one other"
  s = fs['A'] + fs['C']
  if not (len(s) == 2 and 'G' in s): return False

  # looks OK
  return True

# collect viable maps
ffs = list(fs for fs in make_map('ABCD', 'EFGHI') if check_fs(fs))

# find viable son -> grandsons maps
def check_sg(sg):
    
  # "E, G, H have sons K, M, N and one other"
  s = sg['E'] + sg['G'] + sg['H']
  if not (len(s) == 4 and 'K' in s and 'M' in s and 'N' in s): return False

  # "H, I have sons L, N and one other"
  s = sg['H'] + sg['I']
  if not (len(s) == 3 and 'L' in s and 'N' in s): return False

  # looks OK
  return True

# collect viable maps
sgs = list(sg for sg in make_map('EFGHI', 'JKLMNO') if check_sg(sg))

# find the father of x in map d
def father(x, d):
  return peek(k for (k, v) in d.items() if x in v)

# choose the maps
for (fs, sg) in product(ffs, sgs):

  # A, B, C, D have different numbers of grandsons
  s = list(sum(len(sg[s]) for s in fs[f]) for f in 'ABCD')
  if len(set(s)) != 4: continue
  # A has the most
  if s[0] != max(s): continue

  # grandfather of J and O
  gJ = father(father('J', sg), fs)
  gO = father(father('O', sg), fs)
  # they must be different people
  if gJ == gO: continue

  # output solution
  f = lambda s: join(s, sep="+")
  fmt = lambda m: map2str((k, f(v)) for (k, v) in m.items())
  printf("{gs} [grandfather(J) = {gJ}; grandfather(O) = {gO}]", gs=f(sorted([gJ, gO])))
  printf("  father -> sons: {fs}", fs=fmt(fs))
  printf("  son -> grandsons: {sg}", sg=fmt(sg))
  printf()
```
  Solution: One of the presenters is Adam, but the other can be any of: Brian, Colin, or David.
  
  The originally published solution is: “Adam and David”, but a later correction was made (with Teaser 2846) noting there are three valid solutions to the puzzle (given above).
  
  However, there is a unique solution if the end of the puzzle is changed to:
  
  The retirement gift was presented by the paternal grandfather of John and Oliver.
  
  Who was he?
  
  We can change the sense of the test at line 68 to ensure that the grandfathers of John and Oliver are the same person, and we find there are 2 possible family trees that lead to this situation, but in both cases the grandfather is Brian.
  
  LikeLike
- Hugh+Casement 8:34 am on 17 December 2021 Permalink | Reply
  
  Without trying Jim’s program, I think it’s like this:
  A’s son is H whose sons are L, N, and either K or M. D’s son is I who has no son.
  C’s son is G whose son is M or K (whichever is not H’s son).
  Brian has sons E (who has no sons) and F who is the father of J and O.
  
  LikeLike

Jim Randell 5:23 pm on 24 September 2021 Permalink | Reply
Tags: by: Angela Newing

Teaser 3079: Hall of residence

From The Sunday Times, 26th September 2021 [link] [link]

Oak Hall at Woodville University has groups of five study bedrooms per flat and they share a kitchen/diner. In one flat live language students Andy, Bill, Chris, Dave and Ed. Bill, whose home town is Dunstable is reading French. The person in room 5 comes from Colchester and Dave comes from Brighton. The chap reading German has the room with a number one greater than the man from Gloucester. Chris occupies room 3, and Ed is reading Italian. The man in room 2 is reading Spanish, and the man reading English has a room whose number is two different from the student from Reigate.

What is Andy’s subject and where is his home?

[teaser3079]

Jim Randell 9:42 pm on 24 September 2021 Permalink | Reply

We can use the [[ SubstitutedExpression ]] solver from the enigma.py library to assign room numbers to the names, subjects and towns according to the given conditions.

This run-file executes in 68ms.

Run: [ @replit ]

#! python3 -m enigma -rr

# assign room numbers 1-5 to each of the groups:
#
#   Andy(A) Bill(B) Chris(C) Dave(D) Ed(E)
#   English(F) French(G) German(H) Italian(I) Spanish(J)
#   Brighton(K) Colchester(L) Dunstable(M) Gloucester(N) Reigate(P)

SubstitutedExpression

--digits="1-5"
--distinct="ABCDE,FGHIJ,KLMNP"

# "Bill, whose home town is Dunstable is reading French"
"B = M"
"B = G"

# "The person in room 5 comes from Colchester"
"L = 5"

# "Dave comes from Brighton"
"D = K"

# "The chap reading German has the room with a number one greater than the man from Gloucester"
"N + 1 = H"

# "Chris occupies room 3"
"C = 3"

# "Ed is reading Italian"
"E = I"

# "The man in room 2 is reading Spanish"
"J = 2"

# "The man reading English has a room whose number is two different from the student from Reigate"
"abs(F - P) = 2"

# mention A (so it gets assigned a value)
"A > 0"

--template="(A B C D E) (F G H I J) (K L M N P)"
--solution=""

Solution: Andy is reading Spanish, and is from Gloucester.

The complete allocations are:

Room 1: Dave, English, Brighton
Room 2: Andy, Spanish, Gloucester
Room 3: Chris, German, Reigate
Room 4: Bill, French, Dunstable
Room 5: Ed, Italian, Colchester

Manually we find that there are only two possible room values for the (Bill/French/Dunstable) group. Trying both possible values quickly leads to a solution in one case and a contradiction in the other.

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