1957 AHSME Problems/Problem 24
Problem
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:
Solution
Let the first number have digits 
. Then, it equals 
, and the second number equals 
. Taking the difference of the squares of these two numbers and simplifying, we see that:
\begin{align*}
(10a+b)^2-(10b+a)^2 &= ((10a+b)-(10b+a))((10a+b)+(10b+a)) \\
&= (9a-9b)(11a+11b) \\
&= 9 \cdot 11(a-b)(a+b)
\end{align*}
Thus, the result in question is always divisible by 
, 
, 
, and 
. This fact eliminates all of our options except 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
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