Find the smallest number \(y\) such that \(y×162\) is a perfect cube.

Answer: 36

• Prime-factorize the given number: 162 = 2^1 × 3^4.

• For a product to be a perfect cube, each prime's exponent must be a multiple of 3. Current exponents are: 2 → 1, 3 → 4.

• Make each exponent a multiple of 3 by multiplying by the minimal additional powers: increase exponent of 2 from 1 to 3 (need 2 more factors of 2) and increase exponent of 3 from 4 to 6 (need 2 more factors of 3).

• Thus y = 2^2 × 3^2 = 4 × 9 = 36.

• Check: 162 × 36 = 2^3 × 3^6 = (2 × 3^2)^3 = 18^3, so the product is indeed a perfect cube.