Numbers having 12 factors

$96=2^5*3^1$ 6x2 factors

$90=2^1*3^2*5^1$ 2x3x2 factors

$84=2^2*3^1*7^1$ 3x2x2 factors

$72=2^3*3^2$ 4x3 factors

$60=2^2*3^1*5^1$ 3x2x2 factors

$96+90+84+72+60=402$

In order to find the 5 numbers, I first decided on a set of the first 6 factors, then I tried to find a number that has those factors. Each of those 6 factors would need to multiply by another unique factor to equal the number. So a number $n$ with factors

$a, b, c, d, e, f, u, v, w, x, y, z$

would be equal to

$a \times z = b \times y = c \times x = d \times w = e \times v = f \times u = n$

In order for this to work, the square root of the number must be greater than the first 6 factors and less than the last 6 factors.

So, I first started with $1, 2, 3, 4, 5, 6$ as the first 6 factors. The LCM of those numbers happens to be $60$ . So does $60$ work?

Factors of $60 = (1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60)$

Yes it does. However this didn't always work. For example, the LCM of $1, 2, 3, 4, 6, 8$ is only $24$ .

Factors of $24 = (1, 2, 3, 4, 6, 8, 12, 24)$

Only 8 factors. However, if you multiply $24$ by an existing factor, like $3$ or $4$ , you can get a number that works.

$24 \times 3 = 72$

Factors of $72 = (1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72)$

$24 \times 4 = 96$

Factors of $96 = (1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96)$

Both $72$ and $96$ have 12 factors.

So you can continue this process to find the other 2 numbers. One thing that is useful to note is that $1, 2, 3, 4$ will always be the first 4 factors, because if they aren't, you will not be able to find 6 factors less than 10, which means the resulting number will be greater than 100, or will not have 12 factors.

$LCM(1, 2, 3, 4, 6, 7) = 84$

Factors of $84 = (1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84)$

$LCM(1, 2, 3, 5, 6, 9) = 90$

Factors of $90 = (1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90)$

Both of those numbers also have 12 factors. So since we have found the 5 factors, all we need to do is add them.

$60 + 72 + 84 + 90 + 96 = 402$

12=2 $\times$ 3 = 4 $\times$ 3 =2 $\times$ 2 $\times$ 3

let N= $2^{p}3^{q}5^{r}7^{s}\cdots$

so from factors of 12 we can get p,q r,s and also can get five required integers