Question 764648: 2, 6, 12, ?, 108
Answer by josgarithmetic(39618)   (Show Source):
You can put this solution on YOUR website! Factor into each ones prime factorization.
index______________prime factorization
1__________________2
2__________________2*3
3__________________2*2*3
4__________________(_________)
5__________________2*2*3*3*3
Some kind of pattern is happening. Maybe not very neat, but something predictable. We want to find the element of index 4. We find that the other elements have the same count of prime factors as the index value. Index 5 element has five factors ( although some repetions), element at index 3 has three factors, and on like that.
As index increases, the count or multiplicity for 2 either increases or stays the same. Starting at index 2, the multiplicity for 3 either increases or stays the same.
Clearly we want element for index 4 to have FOUR factors, and we want at least 2*2, very likely NOT 2*2*2, but just 2*2...
For the factor 3, since we want to keep the same or increase the multiplicity and we want to have exactly FOUR factors, the conclusion seems 2*2 and 3*3,
so best conclusion is element at index 4 is 
-----ANSWER----
2*2*3*3=36
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