1957 AHSME Problems/Problem 38

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From a two-digit number $N$ we subtract the number with the digits reversed and find that the result is a positive perfect cube. Then:

$\textbf{(A)}\ {N}\text{ cannot end in 5}\qquad\\  \textbf{(B)}\ {N}\text{ can end in any digit other than 5}\qquad \\  \textbf{(C)}\ {N}\text{ does not exist}\qquad\\  \textbf{(D)}\ \text{there are exactly 7 values for }{N}\qquad\\  \textbf{(E)}\ \text{there are exactly 10 values for }{N}$


The number $N$ can be written as $10a+b$ with $a$ and $b$ representing the digits. The number $N$ with its digits reversed is $10b+a$. Since the problem asks for a positive number as the difference of these two numbers, than $a>b$. Writing this out, we get $10a+b-(10b+a)=9a-9b=9(a-b)$. Therefore, the difference must be a multiple of $9$, and the only perfect cube with less than $3$ digits and is multiple of $9$ is $3^3=27$. Also, that means $a-b=3$, and there are $7$ possibilities of that, so our answer is

$\boxed{\textbf{(D)}}$ There are exactly $7$ values of $N$

See Also

1957 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 37
Followed by
Problem 39
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