2000 AMC 8 Problems/Problem 9

Problem

Three-digit powers of $2$ and $5$ are used in this cross-number puzzle. What is the only possible digit for the outlined square? \[\begin{array}{lcl} \textbf{ACROSS} & & \textbf{DOWN} \\ \textbf{2}. 2^m & & \textbf{1}. 5^n \end{array}\]

[asy] draw((0,-1)--(1,-1)--(1,2)--(0,2)--cycle); draw((0,1)--(3,1)--(3,0)--(0,0)); draw((3,0)--(2,0)--(2,1)--(3,1)--cycle,linewidth(1));  label("$1$",(0,2),SE); label("$2$",(0,1),SE); [/asy]

$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution

The $3$-digit powers of $5$ are $125$ and $625$, so space $2$ is filled with a $2$. The only $3$-digit power of $2$ beginning with $2$ is $256$, so the outlined block is filled with a $\boxed{\text{(D) 6}}$.

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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