9
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This is just a straightforward crossword puzzle. Every answer is some positive integer. To make it a bit more interesting I have withheld the clue for H5. Use the fact that with the clue for H5 the puzzle has a unique solution. For a bonus you can make up your own clue for H5.


enter image description here

Horizontal

1 Divisor of H8.
3 Halve of V2.
4 The digits of this number are ascending.
5 ???
6 This triangle number consists of three of the same digits.
8 Fourth power.

Vertical

2 Divisible by 17.
3 A square.
5 V10 times 111.
7 Sum of the digits is 10.
9 This palindromic number is also palindromic in binary.
10 One of the digits is twice the other.


EDIT: Gareth McCaughan mentioned quite rightly that there is a slight ambiguity in the way the question is phrased now. To make the solution unique you must assume that H5 does not reference any other clues.

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  • 2
    $\begingroup$ This is a very cool concept! $\endgroup$ – GoldenGremlin Mar 10 '17 at 22:36
  • $\begingroup$ Very nice puzzle, well done :) $\endgroup$ – Rubio Mar 10 '17 at 23:09
  • $\begingroup$ It is cool, but it isn't new. A magazine called "Tough Puzzles" used to do these (but much larger) something like 20-30 years ago. $\endgroup$ – Paul Sinclair Mar 11 '17 at 0:02
  • $\begingroup$ Thank you guys. Indeed, I didn't think of the concept, I've seen these puzzles here and there and I always like them. So I decided to try making my own. $\endgroup$ – Pjotr5 Mar 11 '17 at 8:46
7
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I believe this is the intended solution. Explanation to follow shortly.

enter image description here

OK, so here's one way to get there.

First of all, 9d is a palindrome in both decimal and binary. That means it's either 33 or 99. If it's 33 then 8a can only be 923521, and then neither possibility for 10d yields a consistent result for 5d. Therefore, 9d is 99, and then 8a has to be 279841. And now exactly one of the two possibilities for 10d does work: it's 48 and 5d is 5328.

Now

1a divides 8a, but the latter is 23^4 so its only 2-digit divisor is 23. So 2d is one of {306,323,340,357,374,391}. It can't be 306 because that would make 4a start with 0. And it must be even because 3a is half of it. So it's 340 or 374. So 3a is either 170 or 187. In either case it starts with 1.

Next

consider 6a; the very strong condition given in the clue forces it to equal 666. Now 3d is a square of form 1x6 so must be 196. And 7d's digits 6x1 add up to 10, so we can fill in the 3 in the middle.

What's left?

Only 5a (whose clue is missing), 2d (either 340 or 374), and 4a (ends in 9, clue says digits are in ascending order).

Now, if

5a allows its answer to be 50, then we could have 2d = 340 and then 4a would be 4x9 and its middle digit could be any of {5,6,7,8}. That won't do (but see below). So the missing clue must require 5a to be 54, so that 2d = 374 and then the ascending-digit condition on 4a allows only 789.

The clue for 5a might be, say,

Twice a cube.

But there's a little loophole making the solution ambiguous (which is why I was careful to say "I believe this is the intended solution" above):

the clue for 5a could also reference 4a and thus pin down its middle digit. I don't see any nice simple clue that would do that, but e.g. it could be "1 less than a multiple of the largest prime factor of 4a". Then we'd have 5a=50 and 4a=459 (whose largest factor happens to be 17).

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  • 2
    $\begingroup$ You are too fast. *mutter* :) $\endgroup$ – Rubio Mar 10 '17 at 22:14
  • 1
    $\begingroup$ People keep telling me that. I can't imagine why. :-) $\endgroup$ – Gareth McCaughan Mar 10 '17 at 22:26
  • 1
    $\begingroup$ Very fast and very smart. That was indeed the intended solution and I didn't even think of that loophole, but you are absolutely right that it does make the solution ambiguous. Next time I'll try to make it more difficult, while checking even harder for unambiguity. Thank you :) $\endgroup$ – Pjotr5 Mar 10 '17 at 22:41

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