If F is the prime factorization of N!

[Mo2men]

If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

(2) 27 ≤ N ≤ 35

Source: Magoosh

OA: E

[abhijit@prepitt]

If F is the prime factorization of N!, how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

(2) 27 ≤ N ≤ 35

Source: Magoosh

OA: E

Lets understand what the question is asking with a simple example:
If N=6. N! = 6! = 6*5*4*3*2*1. Prime factorization for this would look something like: F= 5^1 * 3^2 * 2^4 ==> So 1 factor (i.e 5) in F has exponent of 1

As you can see the prime factor closest to N (i.e. 5) will have the exponent of 1. However, as the value of N increases, there may be many more such prime factors with exponent of 1. For example, if N=22: 19, 17 ... etc. will have exponent as 1. If N=23: 23, 19, 17 .. will have exponent as 1. If you note, N=23 has one additional prime factor with exponent of 1 than N=22.

Once we understand this, the question actually becomes easier to solve:

how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

Take N=30. The prime factors with exponent 1 will be lets say X i.e. 29, 23, 19 ... etc.
Take N=31. The prime factors with exponent 1 will be X+1 i.e. 31, 29, 23, 19 ... etc.

So the answer could be either X or X+1 depending on value of N. So with these examples itself (1) is insufficient.

(2) 27 ≤ N ≤ 35

Same principle can be applied here, so leaving for you to practice. Hint: Take example of N=28 and N=29.
(2) is insufficient also.

(1) + (2) you get an even bigger range 30 <= N <= 35. So that will be definitely insufficient as well.

Hence answer (E)

Hope this helps!

[Mo2men]

Lets understand what the question is asking with a simple example:
If N=6. N! = 6! = 6*5*4*3*2*1. Prime factorization for this would look something like: F= 5^1 * 3^2 * 2^4 ==> So 1 factor (i.e 5) in F has exponent of 1

As you can see the prime factor closest to N (i.e. 5) will have the exponent of 1. However, as the value of N increases, there may be many more such prime factors with exponent of 1. For example, if N=22: 19, 17 ... etc. will have exponent as 1. If N=23: 23, 19, 17 .. will have exponent as 1. If you note, N=23 has one additional prime factor with exponent of 1 than N=22.

Once we understand this, the question actually becomes easier to solve:

how many factors in F have an exponent of 1?

(1) 30 ≤ N ≤ 40

Take N=30. The prime factors with exponent 1 will be lets say X i.e. 29, 23, 19 ... etc.
Take N=31. The prime factors with exponent 1 will be X+1 i.e. 31, 29, 23, 19 ... etc.

So the answer could be either X or X+1 depending on value of N. So with these examples itself (1) is insufficient.

(2) 27 ≤ N ≤ 35

Same principle can be applied here, so leaving for you to practice. Hint: Take example of N=28 and N=29.
(2) is insufficient also.

(1) + (2) you get an even bigger range 27 <= N <= 40. So that will be definitely insufficient as well.

Hence answer (E)

Although answer is E, I believe range should be 30 ≤ N ≤ 35. For example, if N= 28, it will satisfy statement 2 but not statement 1. Also, if N= 37, it will satisfy statement 1 but not statement 2.

[abhijit@prepitt]

My bad, I have edited the solution.

(1) 30 ≤ N ≤ 40

(2) 27 ≤ N ≤ 35

Combined range should be an intersection of the two : 30<=N<=35