Problem 18 · 2013 Math Kangaroo
Medium
Number Theory
factorizationcasework
How many pairs of positive integers \((x, y)\) solve the equation \(x^{2} \times y^{3} = 6^{12}\)?
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Answer: E — A different number.
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Hint 1 of 2
6^12 = 2^12 · 3^12, so x and y use only the primes 2 and 3.
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Hint 2 of 2
Split the exponent equations 2p+3r = 12 and count solutions for each prime separately.
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Approach: match prime exponents
- Need 2(exp in x) + 3(exp in y) = 12 for each of primes 2 and 3.
- 2p + 3r = 12 has 3 non-negative solutions; same for the prime 3.
- Total pairs = 3×3 = 9, which is none of A–D, so E.
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