Dilemma is a purely numerical puzzle, very neatly constructed
and capable of solution without Mathematica or other online programming aids (though AI is useful
in narrowing down searches). Oyler is to be congratulated for his concise clues!
As ever, making headway and reaching the goal of 2 full grids, each with 2 instances
of every digit from 0-9, relies on finding the unique path through the clues.
I won’t bore the reader with the numerous wrong turns, dead ends and careless
mistakes that abounded in my handcrafted workings. Instead I will set out the path
which led me to the solution.
NB: some squares can act as both last digits and intermediate digits, such as the last
digit of 5 across which is also an intermediate digit of 3 down. After shading the
squares containing the last digits (whether or not they were also intermediate digits),
I observed that in each grid there are 10 grey squares and 10 white squares, and
made the assumption that a square containing a last digit always counts as a last
digit, irrespective of any other role that digit might play.
It’s always gratifying when an endgame reinforces and confirms the gridfill. In
addition to checking that I had a white and grey version of each digit, I sought to
derive words by converting numbers to letters; however, I could only find BADE at
4dn in the left-hand grid, and CEDI at 3dn in the right-hand grid. Had there been
anything significant I’m sure Oyler would have incorporated it.
Finally, Dilemma threw an interesting fact my way: in squares, penultimate odd digits
are always followed by the digit 6. I must tuck that away for future use somewhere.
My methodology is set out in Appendix A.
Appendix A to Listener 4908 solution: the gridfill
Left-hand grid
1. Shade all squares containing last digits grey or a colour of choice to help avoid
duplication of digits.
2. Consider in the left grid 3dn, 3ac and 6ac. 6ac must begin with 1; the smallest
value for 3ac is 232 (palindrome) or 233 (prime); 6ac is either triangular or even:
only 10 or 12 give a small enough 3dn to fit. This is true for other first digits of
3ac/3dn. 10 does not work because both 6ac and 3dn would end in 0. Therefore
6ac is 12, even.
3. 3ac must begin with 2 or 3 or 4 to satisfy 3dn. Possible values, once we take into
consideration that we’ve used up a white 1 and a grey 2, are
3ac 233 323 ) either way, both 2s and 3s are used
3dn 2796 3876 )
4. Thus 9ac ends in 6.
5. 4dn is either prime: 2179 3109 3167 3169 3187
or triangular 2145
6. 8ac is either square: 784 900 961 (so 4dn is not 2179)
or triangular 741
7. 10ac is either Fib: no solutions
or prime: 59 71 79 97
8. 1ac is either reversed square: 94 61
or reversed triangular: 51 54 55 87
9. If 3ac = 233, then 3dn = 2796, 8ac = 900 or 961 and 4dn = 3109 or 3167 or 3169.
10ac = 71 or 79 or 97. All of these solutions put 2 7s on grey squares.
10. Therefore 3ac = 323, 3dn = 3876, 4dn = 2145, 8ac = 741 and 10ac = 59. Both 1s
are used.
11. 1ac is 94 or 54 or 87.
12. The 0 on the white square must be the middle digit of 5ac. There is no triangular
number x08 so 5ac is a palindrome, 808.
13. 2dn ends in 0 1 3 4 5 6 7 or 9. Therefore options for 9ac are: 576 676
14. 2dn ends in 7 therefore prime. The 0 on the grey square must be the last digit of
7ac
15. 7ac is twice a square, ie 50, 1ac = 94 and 9ac = 676.
Right-hand grid
1. The clues relating to the right-hand grid are
Across
1 Reverse of a triangular number
3 Prime
5 Triangular
6 Triangular
7 Odd
8 Square
9 Square or reverse of a square
10 Fibonacci number
Down
2 Square
3 7ac x 9ac
4 Prime
2. 4dn ends in 1 3 7 or 9
3. 10ac is 13 or 34; either way the 3 on the grey square is used up
4. Looking at 3dn, 7ac and 9ac: 7ac is odd, so either
5. 7ac ends in 1, or
6. 9ac ends in 5
7. 2dn is a square so ends in 0 1 4 5 6 or 9
8. 9ac is a square or reverse of a square, but there is no solution with 0 1 2 4 5 6 or
9 as the middle digit.
9. So 7ac ends in 1, using 1 on a grey square. Thus 10ac must be 34.
10. Options for 2dn, square are 2116 2916 9216 (NB for squares, penultimate
odd digits are always followed by 6s)
11. 9ac ends in 2 7 8 9 so must be 169
12. White square 0 must be middle digit of 8ac 100 400 900
13. Only option for 7ac is 21; 3dn is 3549; 2dn is 2916; 1ac is 82
14. 5ac is 595; 6ac is 78; 3ac is 367 and 4dn is 6703
