1957 AMC 12 Problem 24

Question

If the square of a number of two digits is decreased by the square of the number formed by reversing the digits, then the result is not always divisible by:

9

the product of the digits

the sum of the digits

the difference of the digits

11

Answer: B

Concepts:digits difference of squares divisibility

Difficulty rating: 1590

Small hint:

Represent the number as

10a+b

and its reversal as

10b+a

Big hint:

Factor the difference of their squares completely

Solution:

If the digits are

a,b,

then

\begin{aligned} &(10a+b)^2-(10b+a)^2\\ &\quad=[9(a-b)][11(a+b)]\\ &\quad=99(a-b)(a+b). \end{aligned}

This is always divisible by

9,

11,

the digit sum, and the digit difference. It need not be divisible by the digit product; for example,

21^2-12^2=297

is not divisible by

2.

Thus, the correct answer is B.