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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