Question 1129602
The product of all the positive divisors of a positive integer N is 6∙16∙36∙96. What is N?
<pre>
We prime factor those factors:
6 = 2&#8729;3
16 = 2&#8729;2&#8729;2&#8729;2
36 = 2&#8729;2&#8729;3&#8729;3
96 = 2&#8729;2&#8729;2&#8729;2&#8729;2&#8729;3

Therefore 6&#8729;16&#8729;36&#8729;96 = 2<sup>12</sup>3<sup>4</sup> 

From above we see that N contains prime factors 2 and 3 and ONLY those two
prime factors. Now we must determine how many times N contains each as a
factor. 

N cannot contain the factor 3 more than once. Because:  If so, N would
contain 2&#8729;3&#8729;3 as a factor.  Then the factors of N would contain factors 
1, 2, 3, 2&#8729;3, 3&#8729;3, 2&#8729;3&#8729;3, which would produce a product of factors 2<sup>3</sup>&#8729;3<sup>6</sup>.
 
Then the product of factors of N would contain 3 as a factor 6 times.
However we know that the product of factors of N contains 3 as a factor only
4 times. 

Therefore N contains 3 as a factor exactly once, no more and no less!

So we only need determine how many times N contains 2 as a factor.

It must contain 2 as a factor more than once, for if N contained 2 as a
factor only once, then N would be 2&#8729;3=6, with product of factors
(1)(2)(3)(2&#8729;3) = 2<sup>2</sup>&#8729;3<sup>3</sup>, not 2<sup>12</sup>3<sup>4</sup>.

N also must contain 2 as a factor more than twice, for if it contained 2 as
a factor only twice, then N would be 2&#8729;2&#8729;3=12, with product of factors
(1)(2)(3)(2&#8729;2)(2&#8729;3)(2&#8729;2&#8729;3) = 2<sup>6</sup>&#8729;3<sup>3</sup>, not 2<sup>12</sup>3<sup>4</sup>.
 
So we try N as containing 2 as a factor 3 times.  That would make N = 2&#8729;2&#8729;3
= 24. Then the product of factors of N would be
(2)(3)(2&#8729;2)(2&#8729;3)(2&#8729;2&#8729;2)(2&#8729;2&#8729;3)(2&#8729;2&#8729;2&#8729;3) = 2<sup>12</sup>3<sup>4</sup>, which
is exactly what we're looking for!

So the answer is N = 24.

Edwin</pre>