# Enigma 1772

Spoiler for New Scientist Enigma 1772 (Follow the link to see the puzzle.)

Surprisingly easy for a Susan Denham puzzle. 5 across must be divisible by 5 so must end with either 0 or 5. If our even digit is 0 then one of 4 across and 5 across must be 00, which is presumably disallowed implicitly — otherwise there would be multiple trivial solutions of the puzzle, e.g. 50 at 5 across and zeros everywhere else.

Assuming all the numbers have to be nonzero, 5 across must end in 5 and the odd digit is 5. For 6 down, the only 2-digit multiple of 6 containing a 5 is 54, so our even digit is 4 and 5 across is 45. Excluding any more 5s from the third and fifth columns and third and fourth rows, 2 down is 44, as is 4 across, and 3 down is 44544.

7 across can only be 54444 or 45444 or 44454, but only the second of these is a multiple of 7. Likewise 1 down can only be 45444. The second digit of 1 across must be 4, and for divisibility by 3 the fourth digit must be 5.

So the (no-zeros) solution is unique:

 4 4 4 5 4 5 – 4 – 4 4 4 – 4 5 4 – 5 – 4 4 5 4 4 4

## 5 thoughts on “Enigma 1772”

1. Andrew Brown says:

Nice work Mr Holmes;

Pedantry:
6 down could also be “66”, but this is eliminated by considering 4 across (which would then be “66” also – not divisible by 4)
7 across could also be 44444 (but cannot be 54444 if 1 down is divisible by 4 and so is even). Doesn’t change soln as 44444 is not a multiple of 7

it is not clear “zero” solutions are invalid by the rules. All zeros is one valid solution and others include:

0 3 0 0 0
0 0 3
0 0 3 0
0 3 0
0 0 0 0 0

and

0 0 0 0 0
7 0 0
0 0 7 0
0 0 0
0 7 0 0 0

1. Good points. I did mention the multiple trivial solutions involving zeros. You could argue they’re valid, but they’re clearly not what was intended. They didn’t exclude leading zeros in the puzzle text but they also didn’t specify e.g. you must use base 10, either. I assume leading zeros are disallowed in puzzles like these unless they are explicitly allowed, or needed for a solution to exist. I’d also argue the all zeros solution is further disallowed on the grounds that it doesn’t use two different digits — though again, a lawyer could argue that requirement wasn’t made absolutely explicitly.

2. David Peterson says:

Your solution is the only one I could find, BUT I don’t think it’s valid. This is a cross number puzzle, albeit a bit unusual, and like crossword puzzles, it’s bad form to repeat an answer.

1. Well, no such restriction was stated, but that doesn’t mean you’re wrong. However the line of reasoning above says there isn’t a solution that doesn’t repeat an answer, and a computer program I wrote verifies that, so unless both reasoning and program are wrong…

1. David Peterson says:

I thought at first that perhaps the wording was poor and was meant to say that each number had at most one odd digit and the rest all the same even digit. But that interpretation leads you nowhere. The fact that the “3” number needs a 5 in the third position is so controlling. I’m just going to wait for the published answer on the 20th.