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In this cryptarithm: PQR + STU + VWX = YYY, each letter represents a different base-11 digit from 0 to A, and none of the numbers can start with zero.

Many solutions exist for the above equation, but there are precisely two distinct values that Y can assume. Determine these values for Y and prove that there are no others.

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

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Y can be either 7 or A

I hope that I'm doing this correctly so that I don't spoil it for other contributors, so don't read on at this point

Solution to the problem begins with recognizing that Y >= 7 based on the restrictions that P, S, and V cannot be zero, resulting in minimum values of 1, 2, 3 and that using minimum values of 4, 5, and 6 for Q, T, and W will result in the carry of a 1.

Check the validity of values of 7, 8, 9, and A individually. As an example, for 7, the only values of P, S, and V are 1, 2, and 3. Of the remaining numbers, the middle column letters must add to 17 and the last colum to 18. Using simple substitution, shows Q, T, and W can be solved with 8, 9, and 0; R, U, and X can be 5, 6, and 7.

The process can be repeated to disprove 8 and 9 as potential answers for Y and to prove that A works P, S, and V are 1, 2, and 7; Q, T, and W are 5, 6, and 0; R, U, and X are 9, 8, and 4.

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The only issue (unless my base 11 adding is way off) is that 150 + 269 + 47A = 888 is a good solution so it can also be 8.

Y can be either 7 or A

I hope that I'm doing this correctly so that I don't spoil it for other contributors, so don't read on at this point

Solution to the problem begins with recognizing that Y >= 7 based on the restrictions that P, S, and V cannot be zero, resulting in minimum values of 1, 2, 3 and that using minimum values of 4, 5, and 6 for Q, T, and W will result in the carry of a 1.

Check the validity of values of 7, 8, 9, and A individually. As an example, for 7, the only values of P, S, and V are 1, 2, and 3. Of the remaining numbers, the middle column letters must add to 17 and the last colum to 18. Using simple substitution, shows Q, T, and W can be solved with 8, 9, and 0; R, U, and X can be 5, 6, and 7.

The process can be repeated to disprove 8 and 9 as potential answers for Y and to prove that A works P, S, and V are 1, 2, and 7; Q, T, and W are 5, 6, and 0; R, U, and X are 9, 8, and 4.

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