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 $N$

Answer: D

Concepts:digits perfect power systematic listing

Difficulty rating: 1630

Small Hint:

If the digits are $a>b,$ the difference from the reversal is $9(a-b)$

Big Hint:

The difference is at most $81,$ so test the positive cubes no larger than $81$

Solution:

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.

1957 AMC 12 Problem 38

Answer: D

Difficulty rating: 1630

Small Hint:

Big Hint:

Solution:

Problem 38 in Other Years