## Brain Teaser: Noughts and crosses

**From The Sunday Times, 23rd December 1956** [link]

Each square is to be occupied by either a nought or a cross.

The cross (or

x) stands for one particular digit which you may discover for yourself.No light begins with nought.

The following clues give the prime factors of the various lights, each letter representing a different prime:

Across:

(I) abcd

(II) a²b²ef

(III) ab²gh

(IV) abij²k

(V) a²bejl

(VI) ab²ikm

Down:

(i) ab²ijkm

(ii) a²beij²k

(iii) ab²mnp

(iv) a²bcde

(v) abijkl

This is one of the occasional *Holiday Brain Teasers* published in *The Sunday Times* prior to the start of numbered *Teasers* in 1961. Prizes of £5, £3 and £1 were offered for the first 3 correct solutions.

[teaser-1956-12-23] [teaser-unnumbered]

## Jim Randell 11:04 am

on28 November 2021 Permalink |Once you get going I think it is easier to solve this puzzle by hand. But here is a fully programmed solution:

The numbers (I) and (i) must be repdigits of the appropriate length with the appropriate prime factorisation patterns.

This immediately gives us the required digit

x, then we just have to fill out the remaining numbers, matching the factorisation patterns.This Python program runs in 49ms.

Run:[ @replit ]Solution:The completed grid looks like this:We can determine the primes as:

Note that although it may look like the prime factorisation may have been presented in numerical order, this is not the case for at least 2 of the clues.

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## Jim Randell 9:57 am

on29 November 2021 Permalink |Here is my manual solution:

(I) 11111 factorises as (41, 271), so that gives us two of

a, b, c, d, and the remaining two come from the single digitx. Sox= 6, and(a, b, c, d)= (2, 3, 41, 271) (in some order).(i) 666666 factorises as (2, 3², 7, 11, 13, 37), so

b= 3,a = 2, and(i, j, k, m)= (7, 11, 13, 37),(c, d)= (41, 271).(iv) = (a², b, c, d, e) = 2 × (I) ×

e= 133332 ×e, and consists of 6 digits chosen from (0, 6). The only possible prime value foreis 5, giving (iv) = 666660.(VI) = (a, b², i, k, m) = 18 ×

ikm. Choosing one of the values from (7, 11, 13, 37) forjwe get:So

j= 11, and (VI) = 60606.(IV) = (a, b, i, j², k) = 726 ×

ik. Choosing one of the values from (7, 13, 37) formwe get:So

m= 37,ik= 7 × 13, and (IV) = 66066.(ii) = (a², b, e, i, j², k) = 660660.

(II) = (a², b², e, f) = 180 × f, which matches 66?6?. The possible numbers are:

So,

f= 367, and (II) = 66060.(V) = (a², b, e, j, l) = 660 ×

l, and matches 66?6?. The possibilities are:So

l= 101 and (V) = 66660.(v) = (a, b, i, j, k, l) = 606606.

(III) = (a, b², g, h) = 18 ×

gh, and matches 60?66. The possibilities are:So (III) = 60066, and the grid is complete.

The solution was published in the 6th January 1957 issue of

The Sunday Timesalong with the following note:LikeLike

## Frits 1:32 pm

on28 November 2021 Permalink |LikeLike

## Jim Randell 4:41 pm

on28 November 2021 Permalink |A neat use of the [[

`SubstitutedExpression()`

]] solver.But do you think [[

`mfac()`

]] is strong enough? A number like 9240 would give an answer of [4, 0], but not match the patterna⋅b⋅c⋅d.Although checking the numbers have the right number of single and double prime factors turns out to be enough to narrow down the solution space to a single candidate, even without matching up the primes in the patterns.

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## Frits 9:37 pm

on28 November 2021 Permalink |With a better “find multiplicities” function (thanks Jim).

Most of the work is now done by [[ SubstitutedExpression() ]].

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## Sunday Times Brainteaser – Noughts and Crosses | PuzzlingInPython 5:01 pm

on29 November 2021 Permalink |[…] Published 23rd December 1956 (link) […]

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