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 
, 
, (a-b), and (a+b). This fact eliminates all of our options except 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
