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Difference between revisions of "1957 AHSME Problems/Problem 24"

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== Solution ==
 
== Solution ==
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<math>\fbox{\textbf{(B) }the product of the digits}</math>.
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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:
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\begin{align*}
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(10a+b)^2-(10b+a)^2 &= ((10a+b)-(10b+a))((10a+b)+(10b+a)) \\
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&= (9a-9b)(11a+11b) \\
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&= 9 \cdot 11(a-b)(a+b)
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\end{align*}
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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 13:59, 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:

$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ \text{the product of the digits}\qquad \textbf{(C)}\ \text{the sum of the digits}\qquad \textbf{(D)}\ \text{the difference of the digits}\qquad \textbf{(E)}\ 11$

Solution

Let the first number have digits $\underline a$$\underline b$. Then, it equals $10a+b$, and the second number equals $10b+a$. 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 $9$, $11$, $(a-b)$, and $(a+b)$. This fact eliminates all of our options except $\fbox{\textbf{(B) }the product of the digits}$.

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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