Problem 6 · 2020 Math Kangaroo
Medium
Number Theory
factorizationfactor-triples
Let a, b, c be integers with \(1 \le a = b \le c\) and \(abc = 2020^2\). What is the highest possible value of a?
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Answer: D — 101
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Hint 1 of 2
With a = b you need a²·c = 2020², so a must divide 2020.
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Hint 2 of 2
Push a as high as you can while keeping c ≥ a.
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Approach: a divides 2020; test the largest such a with c at least a
- Since a = b, the condition is a²c = 2020² with a ≤ c, so a divides 2020.
- 2020 = 2²·5·101, and taking a = 101 gives c = (2020/101)² = 400 ≥ 101.
- No larger divisor keeps c ≥ a, so the greatest possible a is 101.
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