Problem 25 · 2013 Math Kangaroo
Stretch
Number Theory
perfect-squarefactorization
Let Q be the number of square numbers among the natural numbers from 1 to \(2013^{6}\), and K the number of cube numbers among the natural numbers from 1 to \(2013^{6}\). Which of the following holds true?
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Answer: A — \(Q = 2013\times K\)
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Hint 1 of 2
The count of perfect squares up to a number is the square root of that number (rounded down); the count of cubes is the cube root.
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Hint 2 of 2
Take the square root and cube root of 2013⁶ exactly.
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Approach: count squares and cubes by taking roots
- Squares up to 2013⁶: Q = √(2013⁶) = 2013³.
- Cubes up to 2013⁶: K = (2013⁶)^(1/3) = 2013².
- So Q = 2013³ = 2013 × 2013² = 2013 × K.
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