From The Sunday Times, 8th June 1969 [link]
Fred was an exasperating person; he would never give a straight answer to a straight question. The other day I was seeing him off at a bus stop and he said, just as the bus was coming, “Give me a ring some time”.
“I will”, I said, and asked him for his telephone number.
“It has five digits”, he replied, and the first two are the same as my wife’s age while the last two are my age”.
“But I don’t know how old you are”, I said, “except that you are less than 70”.
“There’s no need to be rude”, he said. “I am older than my wile by as many years as our son’s age”.
“What has he got to do with your telephone number?”, I asked.
“Oh”, he replied, “his age is the middle digit”.
“That doesn’t help much”, I said.
“It’s divisible by 259, he shouted”, as the bus whisked him away.
“That’s a great help”, I shouted back, and it really was.
What was Fred’s telephone number?
This puzzle is included in the book Sunday Times Brain Teasers (1974).
[teaser422]
Hugo, 
GeoffR, and 1 other are discussing. 
Frits are discussing. 
Ender Aktulga are discussing. 
S2T2 
Jim Randell 9:19 am on 30 November 2023 Permalink |
If we consider the four rows that make up the square, each row sums to the magic constant k, and between all 4 rows each number is included exactly once. So:
We can now use the [[
SubstitutedExpression]] solver from the enigma.py library to fill out the remaining squares.The following run file executes in 96ms. (Internal runtime of the generated program is 853µs).
Run: [ @replit ]
Solution: Here is the completed square:
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NigelR 6:22 pm on 30 November 2023 Permalink |
Hi Jim
I agree your solution, but I’m not sure how you reached it with your short diagonal condition of
“E + J + P < 34" (which isn't satisfied by the solution). Shouldn't it be E+J+O?
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NigelR 6:25 pm on 30 November 2023 Permalink |
Apologies: Just realised that your notation didn’t include ‘O’!!
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Jim Randell 7:55 pm on 30 November 2023 Permalink |
@NigelR: I often skip O as a variable, as it is easily confused with 0 (zero).
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GeoffR 7:02 pm on 1 December 2023 Permalink |
# A B C D # E F G H # I J K L # M N P Q from itertools import product, permutations from enigma import all_different # Given grid values A, J, N, Q = 9, 13, 12, 14 # Find remaining 12 digits digits = set(range(1,17)).difference({9, 13, 12, 14 }) # 1st row of grid for B, C, D in product(digits, repeat=3): if not all_different(A, B, C, D):continue if A + B + C + D != 34:continue R1 = list(digits.difference({B, C, D})) # 2nd row of grid for E, F, G, H in product(R1, repeat=4): if not all_different(E, F, G, H):continue if E + F + G + H != 34:continue R2 = set(R1).difference({E, F, G, H}) # 3rd row of grid for I, K, L in product(R2, repeat=3): if not all_different(I, J, K, L):continue if I + J + K + L != 34:continue if B + G + L >= 34:continue if C + F + I >= 34:continue # only M and P are missing from the 4th row of grid R3 = list(set(R2).difference({I, K, L})) for M, P in permutations(R3, 2): # columns if A + E + I + M != 34:continue if B + F + J + N != 34:continue if C + G + K + P != 34:continue if D + H + L + Q != 34:continue # diagonals if E + J + P >= 34:continue if H + K + N >= 34:continue if A + F + K + Q != 34:continue if D + J + G + M != 34:continue print('A, B, C, D = ', A,B,C,D) print('E, F, G, H = ', E,F,G,H) print('I, J, K, L = ', I,J,K,L) print('M, N, P, Q = ', M,N,P,Q) exit() # only 1 solution needed # A, B, C, D = 9 6 15 4 # E, F, G, H = 16 3 10 5 # I, J, K, L = 2 13 8 11 # M, N, P, Q = 7 12 1 14LikeLike
Hugo 10:01 am on 2 December 2023 Permalink |
I found a second solution. Did I do something wrong?
9 2 15 8
6 7 10 11
16 13 4 1
3 12 5 14
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Jim Randell 11:03 am on 2 December 2023 Permalink |
Yes, the shorter diagonals have to sum less than 34.
You have 16 + 7 + 15 = 38 in your diagram.
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Hugo 11:16 am on 2 December 2023 Permalink |
Aha! I misread the condition as “not equal to that sum”.
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