1. The problem asks for the smallest integer \(n\) such that \(2250n\) is a perfect cube. 2. First, factorize 2250 into its prime factors. \[ 2250 = 2 \times 3^2 \times 5^3 \] 3. For \(2250n\) to be a perfect cube, the powers of all prime factors in the product must be multiples of 3. 4. Currently, the powers are: - For 2: 1 - For 3: 2 - For 5: 3 5. Let \(n = 2^a \times 3^b \times 5^c\). Together, \(2250n = 2^{1+a} \times 3^{2+b} \times 5^{3+c}\). Each exponent must be divisible by 3. 6. Solve for each exponent: - \(1 + a \equiv 0 \pmod 3\) implies \(a = 2\) because \(1+2=3\) - \(2 + b \equiv 0 \pmod 3\) implies \(b = 1\) because \(2+1=3\) - \(3 + c \equiv 0 \pmod 3\) implies \(c = 0\) because \(3+0=3\) 7. So \(n = 2^2 \times 3^1 \times 5^0 = 4 \times 3 = 12\). 8. Therefore, the smallest integer \(n\) such that \(2250n\) is a perfect cube is \(\boxed{12}\).