Problem 30 · 2019 Math Kangaroo
Stretch
Number Theory
factorizationperfect-square
What is the minimum number of elements of the set \(\{10, 20, 30, 40, 50, 60, 70, 80, 90\}\) that have to be removed so that the product of the remaining elements is a square number?
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Answer: B — 2
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Hint 1 of 3
A product is a square exactly when every prime appears an even number of times.
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Hint 2 of 3
Only one element, 70, carries the prime 7, so 70 must go — then check what its removal does to the other primes.
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Hint 3 of 3
Removing 70 throws the counts of 2 and 5 off, so a second element is needed to fix them.
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Approach: make every prime exponent even
- In the full product \(2^{16}\cdot 3^{4}\cdot 5^{10}\cdot 7^{1}\) the only odd exponent is the lone 7, which comes solely from 70.
- So 70 must be removed; but 70 = \(2\cdot 5\cdot 7\), and dropping it turns the exponents of 2 and 5 odd.
- Removing one more element that supplies an odd 2 and an odd 5 (for example 40 = \(2^{3}\cdot 5\)) makes everything even, so the minimum is 2.
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