Problem 28 · 2011 Math Kangaroo
Stretch
Number Theory
factorizationprime-test
Let a, b and c be positive whole numbers for which \(a^{2}=2b^{3}=3c^{5}\). What is the smallest possible number of divisors of \(abc\), counting 1 and \(abc\) themselves?
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Answer: D — 77
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Hint 1 of 2
Write the common value as 2x3y and impose the square, cube and fifth-power conditions on the exponents.
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Hint 2 of 2
Each exponent must satisfy three modular conditions at once.
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Approach: force the exponents to meet all three power conditions
- Let a² = 2b³ = 3c⁵ = 2x3y; then x,y even, x≡1 and y≡0 (mod 3), x≡0 and y≡1 (mod 5).
- Smallest solution: x = 10, y = 6, giving abc = 2103⁶.
- Number of factors = (10+1)(6+1) = 77.
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