1987 AHSME Problems/Problem 16

Problem

A cryptographer devises the following method for encoding positive integers. First, the integer is expressed in base $5$ . Second, a 1-to-1 correspondence is established between the digits that appear in the expressions in base $5$ and the elements of the set $\{V, W, X, Y, Z\}$ . Using this correspondence, the cryptographer finds that three consecutive integers in increasing order are coded as $VYZ, VYX, VVW$ , respectively. What is the base- $10$ expression for the integer coded as $XYZ$ ?

$\textbf{(A)}\ 48 \qquad \textbf{(B)}\ 71 \qquad \textbf{(C)}\ 82 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 113$

Solution

Since $VYX + 1 = VVW$ , i.e. adding $1$ causes the "fives" digit to change, we must have $X = 4$ and $W = 0$ . Now since $VYZ + 1 = VYX$ , we have $X = Z + 1 \implies Z = 4 - 1 = 3$ . Finally, note that in $VYX + 1 = VVW$ , adding $1$ will cause the "fives" digit to change by $1$ if it changes at all, so $V = Y + 1$ , and thus since $1$ and $2$ are the only digits left (we already know which letters are assigned to $0$ , $3$ , and $4$ ), we must have $V = 2$ and $Y = 1$ . Thus $XYZ = 413_{5} = 4 \cdot 5^{2} + 1 \cdot 5 + 3 = 100 + 5 + 3 = 108$ , which is answer $\boxed{\text{D}}$ .

1987 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1987 AHSME Problems/Problem 16

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