Given that $392040 = 2^3 \times 3^4 \times 5 \times 11^2$ when expressed as a product of its prime

Question

Given that $392040 = 2^3 \times 3^4 \times 5 \times 11^2$ when expressed as a product of its prime factors, find the smallest positive integer $n$ such that $392040\sqrt{n}$ is a perfect square, and[1] the largest positive integer $m$ such that $\frac{392040}{m}$ is a perfect cube.[1]

Accepted Answer

(i) $n=100$ (ii) $m=392040$

IP1 Class Test (May 2025) Q5

IP1 Class Test (May 2025) Q5

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