2002 AMC 10P Problems/Problem 8

Problem

How many ordered triples of positive integers $(x,y,z)$ satisfy $(x^y)^z=64?$

$\text{(A) }5 \qquad \text{(B) }6 \qquad \text{(C) }7 \qquad \text{(D) }8 \qquad \text{(E) }9$

Solution 1

The given expression, $(x^y)^z=64$, is equivalent to $x^{yz}=2^6$. Next, notice how $x$ must be a power of $2$ less than $64$ and the exponent of its prime factorization must be a factor of $6,$ otherwise, $yz$ won't be an integer. We can use a bit of case work to solve this problem.

Case 1: $x=2^1$

$yz=6.$ Clearly, our only four solutions are $y=1, z=6$ and $y=6, z=1,$ along with $y=2, z=3$ and $y=3, z=2.$

Case 2: $x=2^2$

$yz=3.$ Clearly, our only two solutions are $y=1, z=3$ and $y=3, z=1.$

Case 3: $x=2^3$

$yz=2.$ Clearly, our only two solutions are $y=1, z=2$ and $y=2, z=1.$

Case 4: $x=2^6$

$yz=1.$ Clearly, our only solution is $y=z=1.$

Adding up all our cases gives $4 + 2 + 2 + 1 =9.$

Thus, our answer is $\boxed{\textbf{(E) }9}$.

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
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

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