Can you solve this

[vccjunkjunk]

13! / 2^x

If is an integer, which of the following represents all possible values of x ?
A 0 ≤ x ≤ 10
B 0 < x < 9
C 0 ≤ x < 10
D 1 ≤ x ≤ 10
E 1 < x < 10

Please include explanation

[n00bpron00bpron00b]

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

Since we are looking out for the number of factors of "2" ; let's re-write the above expression in prime factorization form

13! = 13 x (2x2x3) x 11 x (2x5) x 9 x (2x2x2) x 7 x (2x3) x 5 x (2x2) x 3 x 2 x 1

So total factors of 2 in the expression 13! = 2^10

In order for 13!/2^x to be a integer value, we can have the value of "x" from 0 to 10 (inclusive) => 0 <= x <=10

Note: any non zero value raised to a exponent of "0" is 1, so 2^0 = 1

And "1" is the factor of all positive integers.

"A"

[tommywallach]

Exactly right. (And just to be clear for other readers, the question stem should have read: y = 13! / 2^x, if y is an integer, which of the following represents the range of values for x).

This is a primes and divisibility question. We should recognize that because it discusses division resulting in an integer. Once you recognize that category, you know you're going to be looking at primes. Another way to think of this question is "How many 2s are in the factor tree of 13!" At that point, you could just look at the even numbers in 13!, because we know the odds won't have any 2s.

12 --> this has two 2s in its factor tree
10 --> this has one 2 in its factor tree
8 --> this has three 2s in its factor tree
6 --> this has one 2 in its factor tree
4 --> this has two 2s in its factor tree
2 --> this has one 2 in its factor tree

That makes 10 altogether, so x could be anything between 0 and 10.

-t