1987 AHSME Problems/Problem 16

Revision as of 13:32, 1 March 2018 by Hapaxoromenon (talk | contribs) (Fixed the layout)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $\boxed{\text{D}}$.

See also

1987 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS