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 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 30	Followed by Problem 32
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
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1973 AHSME Problems/Problem 31

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