Question

If the letters used in the following expression denote uniquely a different digit in base $10$ and if $\left(YE\right)\left(ME\right)=TTT,$ find $E+M+Y+T$.

Answer 1

Since $TTT=T\times 111=T\times 37\times 3$, therefore one of the numbers on the left hand side must be multiple of 37 and the other must be a multiple of 3. Since the only two digit numbers which are multiplies of 37 are and 74, therefore one of the numbers must be either 37 and 74.
Case 1: One of the numbers on LHS is 37. The other number must be a two-digit number ending in 7, and must be a multiple of 3. the only possible such numbers are 27, 57, 87. Out of these 57, and 87 are too big in he sense that when either of these is multiplied by 37, we get a four digit number. Thus we find that the only choice for the other number is 27, and in this case we find that $E=7,T=9andY+M=5$ (because either $YE=37,ME=27orME=37,YE=27$), so that $E+M+Y+T=21$
Case 2: One of the numbers on the LHS is 74. Arguing as in case I we find that the other number can be one of the numbers 24, 54, 84. Since all of them are too big, therefore we find that it is not possible for either number on LHS to be 74.Thus we find that $E+M+Y+T=21$