Difference between revisions of "2002 AMC 10P Problems/Problem 8"
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| − | == | + | == Problem == |
| + | How many ordered triples of positive integers <math>(x,y,z)</math> satisfy <math>(x^y)^z=64?</math> | ||
| + | <math> | ||
| + | \text{(A) }5 | ||
| + | \qquad | ||
| + | \text{(B) }6 | ||
| + | \qquad | ||
| + | \text{(C) }7 | ||
| + | \qquad | ||
| + | \text{(D) }8 | ||
| + | \qquad | ||
| + | \text{(E) }9 | ||
| + | </math> | ||
| + | |||
| + | == Solution 1 == | ||
| + | The given expression, <math>(x^y)^z=64</math>, is equivalent to <math>x^{yz}=2^6</math>. Next, notice how <math>x</math> must be a power of <math>2</math> less than <math>64</math> and the exponent of its prime factorization must be a factor of <math>6,</math> otherwise, <math>yz</math> won't be an integer. We can use a bit of case work to solve this problem. | ||
| + | |||
| + | Case 1: <math>x=2^1</math> | ||
| + | |||
| + | <math>yz=6.</math> Clearly, our only four solutions are <math>y=1, z=6</math> and <math>y=6, z=1,</math> along with <math>y=2, z=3</math> and <math>y=3, z=2.</math> | ||
| + | |||
| + | Case 2: <math>x=2^2</math> | ||
| + | |||
| + | <math>yz=3.</math> Clearly, our only two solutions are <math>y=1, z=3</math> and <math>y=3, z=1.</math> | ||
| + | |||
| + | Case 3: <math>x=2^3</math> | ||
| + | |||
| + | <math>yz=2.</math> Clearly, our only two solutions are <math>y=1, z=2</math> and <math>y=2, z=1.</math> | ||
| + | |||
| + | Case 4: <math>x=2^6</math> | ||
| + | |||
| + | <math>yz=1.</math> Clearly, our only solution is <math>y=z=1.</math> | ||
| + | |||
| + | Adding up all our cases gives <math>4 + 2 + 2 + 1 =9.</math> | ||
| + | |||
| + | Thus, our answer is <math>\boxed{\textbf{(E) }9}</math>. | ||
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=P|num-b=7|num-a=9}} | {{AMC10 box|year=2002|ab=P|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:17, 14 July 2024
Problem
How many ordered triples of positive integers 
satisfy 
Solution 1
The given expression, 
, is equivalent to 
. Next, notice how 
must be a power of 
less than 
and the exponent of its prime factorization must be a factor of 
otherwise, 
won't be an integer. We can use a bit of case work to solve this problem.
Case 1: 
Clearly, our only four solutions are 
and 
along with 
and 
Case 2: 
Clearly, our only two solutions are 
and 
Case 3: 
Clearly, our only two solutions are 
and 
Case 4: 
Clearly, our only solution is 
Adding up all our cases gives 
Thus, our answer is 
.
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
