After attempting this problem, I got E.



1) x is sum of two distinct single prime numbers

Single prime number 2,3,5,7

12! = 1 x 2 x 3 x 2 x 2x 5x 2x3x7x2x2x2x9x2x5x11x2x2x3

Since x can be 6(a factor) or 12(not a factor), this is insufficient

2.) 0<x>1 <= Is this written correctly? If it is it is insufficient.

Using both 1.) is 2.) together is still insufficient since x can still be 6 or 12.



What's the official answer?

The thing to add is that 2^x cannot be bigger than 2^10
given 12!=2^10*m with m composed of facctors different from 2

For the 2)
it cannot be 0<x<1 because x would no longer be an integer...
so it is 0<x>1... that is strange could you check because it means x>1

In that case I go for E

hi ,

Regret for mis-reporting, Clue 2 is 0<x<10.

I thought that 2^x has to be 2^10 to evenly divide the 12!. 2^10 would be the max value which can divide. Because thru below breakdown , maximum number of 2's are 10.
12! =

12= 2*2*2*3
11
10=2*5
9
8=2*2*2
7
6=2*3
5
4=2*2
3
2=2*1
1

By that toke , 10 would be max value for 2.

And since clue# 1 tells us that 2-signle digits add up to that value . clue 1 can be sufficent . Isnt it ?

And second option tells same clue in different phrase. SO cant ans be D