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Difference between revisions of "1961 AHSME Problems/Problem 33"
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==Problem 33==
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==Problem ==
  
 
The number of solutions of <math>2^{2x}-3^{2y}=55</math>, in which <math>x</math> and <math>y</math> are integers, is:
 
The number of solutions of <math>2^{2x}-3^{2y}=55</math>, in which <math>x</math> and <math>y</math> are integers, is:
  
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<math> \textbf{(A)} \ 0 \qquad\textbf{(B)} \ 1 \qquad \textbf{(C)} \ 2 \qquad\textbf{(D)} \ 3}\qquad \textbf{(E)} \ \text{More than three, but finite} } } </math>
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<cmath>\textbf{(A)} \ 0 \qquad\textbf{(B)} \ 1 \qquad \textbf{(C)} \ 2 \qquad\textbf{(D)} \ 3\qquad \textbf{(E)} \ \text{More than three, but finite}</cmath>
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{{MAA Notice}}
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==Solution==
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Let <math>a = 2^x</math> and <math>b = 3^y</math>.  Substituting these values results in
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<cmath>a^2 - b^2 = 55</cmath>
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Factor the [[difference of squares]] to get
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<cmath>(a + b)(a - b) = 55</cmath>
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If <math>y < 0</math>, then <math>55 + 3^{2y} < 64</math>, so <math>y</math> can not be negative.  If <math>x < 0</math>, then <math>2^{2x} < 1</math>.  Since <math>3^{2y}</math> is always positive, the result would be way less than <math>55</math>, so <math>x</math> can not be negative.  Thus, <math>x</math> and <math>y</math> have to be [[nonnegative]], so <math>a</math> and <math>b</math> are [[integers]].  Thus,
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<cmath>a+b=55 \text{ and } a-b=1</cmath>
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<cmath>\text{or}</cmath>
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<cmath>a+b=11 \text{ and } a-b=5</cmath>
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From the first case, <math>a = 28</math> and <math>b = 27</math>.  Since <math>2^x = 28</math> does not have an integral solution, the first case does not work.
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From the second case, <math>a = 8</math> and <math>b = 3</math>.  Thus, <math>x = 3</math> and <math>y = 1</math>.  Thus, there is only one solution, which is answer choice <math>\boxed{\textbf{(B)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1961|num-b=32|num-a=34}}
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[[Category:Intermediate Number Theory Problems]]

Latest revision as of 14:11, 31 May 2018

Problem

The number of solutions of $2^{2x}-3^{2y}=55$, in which $x$ and $y$ are integers, is:

\[\textbf{(A)} \ 0 \qquad\textbf{(B)} \ 1 \qquad \textbf{(C)} \ 2 \qquad\textbf{(D)} \ 3\qquad \textbf{(E)} \ \text{More than three, but finite}\]

Solution

Let $a = 2^x$ and $b = 3^y$. Substituting these values results in \[a^2 - b^2 = 55\] Factor the difference of squares to get \[(a + b)(a - b) = 55\] If $y < 0$, then $55 + 3^{2y} < 64$, so $y$ can not be negative. If $x < 0$, then $2^{2x} < 1$. Since $3^{2y}$ is always positive, the result would be way less than $55$, so $x$ can not be negative. Thus, $x$ and $y$ have to be nonnegative, so $a$ and $b$ are integers. Thus, \[a+b=55 \text{ and } a-b=1\] \[\text{or}\] \[a+b=11 \text{ and } a-b=5\] From the first case, $a = 28$ and $b = 27$. Since $2^x = 28$ does not have an integral solution, the first case does not work. From the second case, $a = 8$ and $b = 3$. Thus, $x = 3$ and $y = 1$. Thus, there is only one solution, which is answer choice $\boxed{\textbf{(B)}}$.

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 32 		Followed by
Problem 34
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
All AHSME Problems and Solutions
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