1957 AHSME Problems/Problem 38

Solution

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
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