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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 ==
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{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category:AHSME]][[Category:AHSME Problems]]
 
[[Category:AHSME]][[Category:AHSME Problems]]
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[[Category:Introductory Algebra Problems]]

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:

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

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
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

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