A problem from the 1973 American High School Mathematics Examination:
In this equation, each of the letters represents uniquely a different digit in base 10:
YE × ME = TTT.
What is E + M + T + Y?

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TTT = T × 111 = T × 3 × 37, so either YE or ME is 37. Either way, E is 7.
T is a digit, and T × 3 is a twodigit number ending in 7 (it’s either YE or ME, whichever of these is not 37).
So T must be 9, which means that TTT = 999 = 27 × 37, and E + M + T + Y = 2 + 3 + 7 + 9 = 21.
UPDATE: Strictly speaking we can’t assume that one of the factors must be 37 — it could be 74, a twodigit multiple of 37. But then when we consider the other factor, the only candidate that yields a product with the pattern TTT is 12, which doesn’t fit. Still, it’s necessary to check this line to be sure the solution is unique. (Thanks, Jeff and Steven.)
