1957 AHSME Problems/Problem 38

Revision as of 21:00, 18 June 2019 by Someonenumber011 (talk | contribs) (Solution)

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 $\textbf{(D)}$

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