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 Ternary train (Posted on 2014-03-28)
Please restore the 5 by 5 grid containing 10 ternary numbers with no leading zeroes.
The crossword-like definitions follow:

Across:
I. a cube
II. twice a permutable (in base 10) prime
III. divisible by eleven
IV. the number of trees with 10 vertices
V. a square

Down:
1. repdigit in base 12
2. a factorion in base 10
3. a semiprime
4. a square of a prime
5. its cube in base 10 uses only 3 distinct digits

Rem: I have built it bottoms-up. Hope there is only one solution...

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Solution | Comment 1 of 3

The condition that there were no leading zeroes means the solutions ranged from 81 (10000 in base 3) to 242 (22222 in base 3), and in particular that 1 across and 1 down don't contain any zeroes.

The number of trees with 10 vertices is 106, and the only valid base 10 factorion is 145, so 4 across and 2 down could be filled in immediately with (10221) and (12101), respectively.

1 across: The only cubes in the range of solutions are 125 (11122) and 216 (22000) but since this number can't contain zeroes it must be 125 (11122).

4 down: The only squares of primes in this range are 121 (11111) and 169 (20021), and we know that 4 down starts with a 2, so it must be 169 (20021).

1 down: The only solutions in the range that are both repdigit in base 12 and don't contain any zeroes in base 3 are 130 (11211) and 157 (12211).  So we don't know what the second digit is, but we can fill in the rest (1x211).

5 across: We know that it's 11x1x.  The only square that fulfills that condition is 121 (11111).

5 down: We know that it's 2xx11 and that it's second digit is either 1 or 2.  There are only six possible solutions, and only one of them matches the clue, 211 (21211).

2 across: We know that it's x2x01.  The only values in the range that are double a permutable prime are 146 (12102) and 226 (22101).  It must be 226 (22101).

3 across: We know it's 21x02.  There are only three possibilities, and only one of them is divisible by 11, it's 209 (21202).

The final solution is:

1 1 1 2 2

2 2 1 0 1

2 1 2 0 2

1 0 2 2 1

1 1 1 1 1

Edited on March 29, 2014, 8:11 am
 Posted by tomarken on 2014-03-28 14:08:28

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