Difference between revisions of "1987 AHSME Problems/Problem 16"

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\textbf{(C)}\ 82 \qquad
 
\textbf{(C)}\ 82 \qquad
 
\textbf{(D)}\ 108 \qquad
 
\textbf{(D)}\ 108 \qquad
\textbf{(E)}\ 113  </math>
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\textbf{(E)}\ 113  </math>
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== Solution ==
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Since <math>VYX + 1 = VVW</math>, i.e. adding <math>1</math> causes the "fives" digit to change, we must have <math>X = 4</math> and <math>W = 0</math>. Now since <math>VYZ + 1 = VYX</math>, we have <math>X = Z + 1 \implies Z = 4 - 1 = 3</math>. Finally, note that in <math>VYX + 1 = VVW</math>, adding <math>1</math> will cause the "fives" digit to change by <math>1</math> if it changes at all, so <math>V = Y + 1</math>, and thus since <math>1</math> and <math>2</math> are the only digits left (we already know which letters are assigned to <math>0</math>, <math>3</math>, and <math>4</math>), we must have <math>V = 2</math> and <math>Y = 1</math>. Thus <math>XYZ = 413_{5} = 4 \cdot 5^{2} + 1 \cdot 5 + 3 = 100 + 5 + 3 = 108</math>, which is answer <math>\boxed{\text{D}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 14:32, 1 March 2018

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
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