Bunuel wrote:

In the expression \(\frac{13!}{x^3}\), for which of the following values of x will the expression will NOT be an integer?

A. 3

B. 4

C. 5

D. 6

E. 8

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 \(n!\)
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