In the expression 13!/x^3, for which of the following values of x will

Method I - Raise each answer choice to the power of 3 because we're looking for \(x^3.\). Prime factorize each answer choice if not prime.

Then find the number of powers of the prime(s) in 13! The answer will be an integer if and only if there are enough powers of the prime factor(s) in the factorial. Using formula:

\(\frac{13}{n} + \frac{13}{n^2} + \frac{13}{n^3}\) . . . only to the point where n raised to whatever power is ≤ 13.**

A. \(3^3\). We need three powers of 3.

\(\frac{13}{3^1}= 4\)

\(\frac{13}{3^2} = 1\)

13 cannot be divided by \(3^3\) to get an integer, so stop there.

There are 4 + 1 = 5 powers of 3 in 13!. We only need 3. \(\frac{13!}{3^3}\) will be an integer.

B. \(4^3 = (2^2)^3 = 2^6\). We need six powers of 2.

\(\frac{13}{2}\) = 6

\(\frac{13}{2^2}\) = 3

\(\frac{13}{2^3 =\\
1}\), so

There are 6 + 3 + 1 = 10 powers of 2 in 13! We need six powers of 2 to make \(4^3\). We have ten. \(\frac{13!}{4^3}\) will be an integer.

C. \(5^3\)

\(\frac{13}{5} = 2\)

13 cannot be divided by \(5^2\) to get an integer.

We only have two powers of 5 and we need three. \(\frac{13!}{5^3}\) will not be an integer.

D. \(6^3 = 2^3*3^3\). Check to see that there are enough powers of both 2 and 3.

From answer B, there are ten powers of 2, and we need only three.

From answer A, there are five powers of 3, also enough. \(\frac{13!}{6^3}\) will be an integer.

E.\(8^3 = (2^3)^3\), or \(2^9\)

From Answer B: there are ten powers of 2, and we need only nine. \(\frac{13!}{8^3}\)will be an integer.

Method II - Look at the factors of 13! and see whether or not for each answer choice there are three powers of the number in 13!

13! = 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.

A.\(3^3\). Number of powers of 3 in 13!
3 has one three
6 has one three
9 has two threes
12 has one three
Total of five powers of 3. We only need three to get even division and thereby an integer. \(\frac{13!}{3^3}\) will be an integer.

B. \(4^3\). Either prime factorize 4, or notice:
4 has one four
8 has two fours
12 has one four. We have the necessary three powers of 4. \(\frac{13!}{4^3}\) will be an integer.

C. \(5^3\)
5 has one five
10 has one five
There are no other powers of 5. We have two, we need three. \(fraction]13!/5^3\)[/fraction] will not be an integer.

D. \(6^3\) = \(2^3*3^3\). There aren't enough obvious powers of 6 in 13! There is one in 6, and one in 12. We need three powers of 6.

But by prime factorizing \(6^3\) to \(2^33^3\)

From answer A, there are five powers of 3.
We can get one power of 2 from 2, and two powers of 2 from 4, total three as needed.

There are enough powers of 2 and 3 that \(\frac{13!}{6^3}\) will be an integer.

E.\(8^3 = (2^3)^3 = 2^9\)
2 has one two
4 has two twos
6 has one two
8 has three twos
10 has one two
12 has two twos. That's 10 total. We need 9.\(\frac{13!}{8^3}\) will be an integer.

Answer:
** See Bunuel 's excellent Math Number Theory, here https://gmatclub.com/forum/math-number-theory-88376.html, and scroll down to Factorials, Finding the number of powers of a prime number \(p\), in the \(n!\)