Dividers

Call $d_1<d_2<d_3<d_4<d_5, d_6$ the 6 distinct divisors of $N$ and total divisors $d(N) =6$ . Then we have the product of divisors $N$ is given by $N^{\frac{d(N)}{2}}= d_1\times d_2\times \cdots \times d_6\\ N^3=648 \times d_6\implies N = \sqrt[3]{648\times d_6}$ it's shows $N$ can be reduced to an integer if $648\times d_6$ is the equal cube of $N$ . Now by prime factorization of $648$ we can claim as follows $N = \sqrt[3]{2^3\times 3^4\times {\color{#3D99F6}d_6}} = 2\sqrt[3]{3^4\times{\color{#3D99F6}d_6}}$ $N$ is an integer iff ${\color{#3D99F6}d_6}= 3^2 =9$ . Hence the answer is $9$ .

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Note :

$N$ can also be the integer $d_6 = 3^5 , 3^{11}, 3^{14}$ etc. If so , then the divisors of $N$ will be differ not equal to 6.

We can consider the factors of N as { 1 , N, a , b, A, B }

where a * b = N and A * B = N

Looking at the possible answers, 1 is not an option. That means it's part of the 5. So the question reduces to finding 4 out of 5 numbers from { N, a, b, A, B } that multiply to 648.

Let's factorize 648 before we go any further: 648 = 2 * 324 = 2 * 2 * 162 = 4 * 2 * 81 = 8 * 81 = 2^3 * 3^4

There are two options here: either N is in the 4, or it isn't.

Let us assume N is not in the 4. Then the four must consist of a, b, A, and B. By our definitions above, multiplying them all together yields N^2, so 648 must be a perfect square. But 648 is not a perfect square (see the 2^3 term), so we cannot find such an N value. This means our assumption is incorrect; N must be in the 4. (via proof by contradiction)

Now that we know N is in the four, we need to select 3 out of { a, b, A, B } to multiply with. You see how necessarily we have to select two items that multiply together to N and a third (via pigeonhole principle). Therefore, our multiplication must be in the form N * a * b * A (without loss of generality, since these are all just constants.). Since a*b = N, we need to find a factorization of 648 into the form N^2 * A, where A is a factor of N.

Our factorization is 8 * 81 = 9^2 * 8. However, 8 is not a factor of 9. We can rewrite this as 9^2 * 2^2 *2 = (9 * 2)^2 * 2 = 18^2 * 2

Since 18^2 * 2 = 648, and 2 is a factor of 18, we've found our answer:

N = 1

a = 3

b = 6 (3 and 6 come from the factorization of 18 that doesn't contain 2)

A = 2

The only number we haven't used is N/A = 18/2 = 9.

So the answer is 9