In these three arithmetic problems, there are ten letters, which can be assumed to each replace a different digit from zero to nine. There are several ways to work through this puzzle, but one logical path is below:
- In the division problem, A – H = 0, and H – Y = 0. Since no two letters can represent the same digit, A must be one more than H, which is one more than Y.
- Again using the division problem, AR – HT = Y. However, since A is one more than H, and R = 0, this means that 10 – T = Y.
- Replacing R with 0 give H00 – YNE = HU. This gives the new information that 10 – E = U and 9 – N = H. Since YTE – YNE = H0, T – N = H as well. Therefore T is 9. Since 10 – T = Y, Y is 1. Based on what we know from the first step, H is 2 and A is 3. With T being 9 and H being 2, 9 – N = 2, so N is 7.
- Also in the division problem, E – E = R. Since this occurs in the ones column, R must be equal to 0. In the tens column of the addition problem, this means that C + 0 = C. Since this is true, nothing has been borrowed, and E + E = S. The only possible values for E are 1-4. Since 1, 2, and 3 are already taken, E is 4 and S is 8.
- Only C and U are left to identify, and there are many places to do so. The division problem shows that HT x U = YNE, or U = 174 / 29. U is 6, leaving C as 5.
The final translated math facts are 90654 + 5304 = 95958, 539 x 247 = 133133, and 309401 / 29 = 10669.
The question marks from one to seven are replacing the digits 3, 5, 9, 1, 3, 0, and 1. Translated to letters, these spell out the answer ACTUARY.
