Difference between revisions of "1961 AHSME Problems/Problem 33"
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| − | ==Problem | + | ==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: | ||
| − | < | + | <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> |
| + | |||
| + | ==Solution== | ||
| + | Let <math>a = 2^x</math> and <math>b = 3^y</math>. Substituting these values results in | ||
| + | <cmath>a^2 - b^2 = 55</cmath> | ||
| + | Factor the [[difference of squares]] to get | ||
| + | <cmath>(a + b)(a - b) = 55</cmath> | ||
| + | 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, | ||
| + | <cmath>a+b=55 \text{ and } a-b=1</cmath> | ||
| + | <cmath>\text{or}</cmath> | ||
| + | <cmath>a+b=11 \text{ and } a-b=5</cmath> | ||
| + | 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. | ||
| + | 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>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1961|num-b=32|num-a=34}} | ||
| + | |||
| + | [[Category:Intermediate Number Theory Problems]] | ||
Latest revision as of 14:11, 31 May 2018
Problem
The number of solutions of 
, in which 
and 
are integers, is:
![\[\textbf{(A)} \ 0 \qquad\textbf{(B)} \ 1 \qquad \textbf{(C)} \ 2 \qquad\textbf{(D)} \ 3\qquad \textbf{(E)} \ \text{More than three, but finite}\]](http://latex.artofproblemsolving.com/a/6/b/a6bc1b96cd570e295625859fd6c23a66e6ab7b6b.png)
Solution
Let 
and 
. Substituting these values results in
![\[a^2 - b^2 = 55\]](http://latex.artofproblemsolving.com/2/4/3/2435d232019168f1aec7aad9c2b56ce16372f5b8.png)
Factor the difference of squares to get
![\[(a + b)(a - b) = 55\]](http://latex.artofproblemsolving.com/d/e/f/def4058df3ce9a984bab0e4649e65633bbd73931.png)
If 
, then 
, so can not be negative. If 
, then 
. Since 
is always positive, the result would be way less than 
, so can not be negative. Thus, and have to be nonnegative, so 
and 
are integers. Thus,
![\[a+b=55 \text{ and } a-b=1\]](http://latex.artofproblemsolving.com/0/f/c/0fc81447e1ef64e9fc387f003f3bb5a4a9bc03ea.png)
![\[\text{or}\]](http://latex.artofproblemsolving.com/e/d/7/ed748d15ea9de34b1afaf8ddf810cbedece4f122.png)
![\[a+b=11 \text{ and } a-b=5\]](http://latex.artofproblemsolving.com/c/0/f/c0f1551ef82af54a0c8da1c18c66a211dad4c512.png)
From the first case, 
and 
. Since 
does not have an integral solution, the first case does not work.
From the second case, 
and 
. Thus, 
and 
. Thus, there is only one solution, which is answer choice 
.
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 | ||
