1957 AMC 12 Problem 38

Question

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:

N

cannot end in

5

N

can end in any digit other than

5

N

does not exist

there are exactly

7

values for

N

there are exactly

10

values for

N

If the digits are

a>b,

the difference from the reversal is

9(a-b)

The difference is at most

81,

so test the positive cubes no larger than

81

Writing

N=10a+b,

the positive difference is

\begin{aligned} &(10a+b)-(10b+a)\\ &\quad=9(a-b). \end{aligned}

It is at most

81.

Among

1,

8,

27,

and

64,

only

27

is divisible by

9,

so

a-b=3.

The digit pairs are

\begin{gathered} (3,0),(4,1),(5,2),(6,3),\\ (7,4),(8,5),(9,6), \end{gathered}

giving exactly seven values of

N.

Thus, the correct answer is D.

Problem 38 in other years