If we expand 15!, we have:
$$1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15$$
If we pull out every possible 3, we get:
$$1\cdot2\cdot3\cdot4\cdot5\cdot\left(2\cdot3\right)\cdot7\cdot8\cdot\left(3\cdot3\right)\cdot10\cdot11\cdot\left(3\cdot4\right)\cdot13\cdot14\cdot\left(3\cdot5\right)$$
$$3^6\cdot1\cdot2\cdot4\cdot5\cdot2\cdot7\cdot8\cdot10\cdot11\cdot4\cdot13\cdot14\cdot5$$
This means that 3^6 is the greatest multiple of 3 that is still a factor of 15!. This also means that 3^0, 3^1, 3^2, 3^3, 3^4, and 3^5 are also possible factors of 15!. So if 0 ≤ x ≤ 6, 3^x is a factor of 15!. If x is an integer less than 0 or greater than 6, it will not be a factor of 15!.
Statement 1
To test, we can simply try numbers to see if we can find an x inside our range as well as an x outside of our range.
We could have x = 2 + 3 = 5, which would make 3^x a factor of 15!. However, we could also have x = 3 + 5 = 8, which would make 3^x NOT a factor of 15!. Not sufficient.
Statement 2
This tells us that x must be 1, 2, 3, 4, or 5. All of these numbers are within our range, and thus would make 3^x a factor of 15!. Sufficient.
