1973 AHSME Problems/Problem 31

Problem

In the following equation, each of the letters represents uniquely a different digit in base ten:

$(YE) \cdot (ME) = TTT$

The sum $E+M+T+Y$ equals

$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 24$

Solution

The right side of the equation can be rewritten as $111T = 37 \cdot 3T$ . With trial and error and prime factorization as a guide, we can test different digits of $T$ to see if we can find two two-digit numbers that have the same units digit and multiply to $111T$ .

The only possibility that works is $37 \cdot 27 = 999$ . That means $E+M+T+Y = \boxed{\textbf{(C)}\ 21}$ .

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 30	Followed by Problem 32
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1973 AHSME Problems/Problem 31

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