Difference between revisions of "1957 AHSME Problems/Problem 24"
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== Solution == | == Solution == | ||
− | <math>\fbox{\textbf{(B) }the product of the digits}</math>. | + | Let the first number have digits <math>\underline a</math><math>\underline b</math>. Then, it equals <math>10a+b</math>, and the second number equals <math>10b+a</math>. 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 <math>9</math>, <math>11</math>, <math>(a-b)</math>, and <math>(a+b)</math>. This fact eliminates all of our options except <math>\fbox{\textbf{(B) }the product of the digits}</math>. | ||
== See Also == | == See Also == |
Latest revision as of 14:59, 25 July 2024
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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All AHSME Problems and Solutions |
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