Prime Factorization of 288

To find the smallest natural number N such that $288 \times N$ is a perfect cube, we first find the prime factorization of 288.

• $288 = 2 \times 144$
• $144 = 12 \times 12 = (2^2 \times 3) \times (2^2 \times 3) = 2^4 \times 3^2$
• Therefore, $288 = 2 \times (2^4 \times 3^2) = 2^5 \times 3^2$.

Perfect Cube Condition

A number is a perfect cube if all exponents in its prime factorization are multiples of 3.

• The prime factorization of 288 is $2^5 \times 3^2$.
• The exponent for the prime factor 2 is 5. The next multiple of 3 is 6. We need $6 - 5 = 1$ more factor of 2.
• The exponent for the prime factor 3 is 2. The next multiple of 3 is 3. We need $3 - 2 = 1$ more factor of 3.

Determining the Smallest N

To make $288 \times N$ a perfect cube, N must supply the missing factors.

• N must contain $2^1$.
• N must contain $3^1$.
• The smallest natural number N is the product of these required factors: $N = 2^1 \times 3^1 = 6$.

Checking the result:

$288 \times N = 288 \times 6 = (2^5 \times 3^2) \times (2^1 \times 3^1) = 2^{5+1} \times 3^{2+1} = 2^6 \times 3^3$.

Both exponents (6 and 3) are multiples of 3, confirming $2^6 \times 3^3$ is a perfect cube.