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Mind sharing the explanation.

I’ll help if I can figure it out

Thank you.

This is from Magoosh.


Let’s think about, say, 40! This number, 40!, is the product of all the integers from one to 40. Let’s think about its prime factorization. It would have at least one factor of 2 for every even number from 2 to 40, and a second factor for every multiple of 4, and a third factor for every multiple of 8, etc.; a lot of factors of two. Think about the factors, say, of 7: there are five multiples of 7 from 7 to 35, so in the prime factorization of 40!, the factor 7 would have an exponent of 5. Which factors would have exponents of 1? Well, the prime numbers that are less than N, but have no other multiples less than N. For example, in 40!, the factor 37 would have an exponent of 1 since it appears once and no other multiple of it is less than 40.



Statement #1: 32 ≤ N ≤ 40

As we move through different N’s in this region, we cross the prime number 37, which will have an exponent of 1 if it appears. Some N’s include this prime number and some don’t, so the number of factors with an exponent of 1 is different for different values of N. This statement, alone and by itself, is not sufficient.



Statement #2: 27 ≤ N ≤ 35

As we move through different N’s in this region, we cross two prime numbers, 29 and 31, each of which will have an exponent of 1 if it appears. Some N’s include neither, some include 29 and not 31, and some include both, so the number of factors with an exponent of 1 is different for different values of N. This statement, alone and by itself, is not sufficient.



Combined: 32 ≤ N ≤ 35

Now, there are no prime values in the range specified. But, here’s a tricky thing. If N = 32 or 33, then either 32! or 33! contains exactly one factor of the prime numbers {17, 19, 23, 29, 31}: five prime factors with an exponent of one. BUT, if N = 34 or 35, there are now two factors of 17 (one from 17 and one from 34), either 34! or 35! contains exactly one factor of the prime numbers {19, 23, 29, 31}: four prime factors with an exponent of one. Even in this narrow range, different choices lead to different answers for the prompt question. Even together, the statements are not sufficient.

Answer = (E)

We have a good discussion on this question here: https://gmatclub.com/forum/if-f-is-the- ... 96002.html

Answer is neither both statements together can give an answer?

My guess was also E. Cause X^1=X. And neither statement was able to explain that.
Hope ya’ll didnt mind the extra explanation beyond what was posted in the Forum.

The solution I thought of is if we go with statement 1 then from 32 to 40, we are adding 1 additional prime factor 37 so we can’t get a unique answer. With statement 2 also we add two new prime factors 31 and 37 so we can’t get a unique answer. If we use both together then also we don’t get a unique answer so it must be E

Could I ask? Why 37 cannot be unique answer (statement 1) ?

Have you checked this topic:
if-f-is-the-prime-factorization-of-n-how-many-factors-in-f-have-an-e-196002.html ????
It has all the answers

The purpose of ds question is if you can get a unique answer. If N varies from 32 to 40 your answer changes from 9 prime factors (5, 7, 11, 13, 17, 19, 21, 23, 29, 31) to 10 (all those factors and 37) in statement 1. Similarly you can figure out what statement 2 gives you. In each statement you won’t get a unique answer and so E is the correct choice

The fulcrum of this question is to realise that it is asking you about the number of primes in a range

Because large primes will have an exponent of 1 -> therefore the number of primes in a range of number determines how many numbers have an exponent of 1

As long as N! is in a range that does not introduce any new primes that are not present at the bottom of the set, the number of exponents to the power of 1 is the same

try a smaller example: like 5! to 6! and then 6! to 7!

5! = 5*4*3*2 = 5*(2^2)*3*2 = 5*3*(2^3) >> There are 2 prime factors with 1 power: 5 and 3

6! = 6*5*4*3*2 = (2*3)*5*(2*2)*3*2 = 5*(3^2)*(2^4) >> There are 1 prime factor with 1 power: 5

Right?

Yes, unluckily I still did not understand that. But i think I understand more right now.

Hi All. I have my GMAT online exam scheduled for tomorrow morning.
I wanted to know if we can send out our score to schools well after the exam (let us say after couple of days) free of cost or we have to choose schools then and there?
I know, in case of in-centre exams, we can send afterwards but it costs some amount. But in online exam, for sending the scores free, do we have to submit schools then and there?

If memory serves me well, you have some days to send it for free after receiving your score.
After that time, each school costs about USD 35
Good luck

Can someone solve and explain this sum

Yea it should be D

actual value will be 1-(1/20)^20= Very very close to 1 but smaller than 1

How did you get that value

The expression is in geometric progression. Formula for getting the sum is a(1-r^n)/1-r. Check what is a, r and n online for a geometric progression. Fill the values and you’ll get this value.

A is the first term, r is common factor and n is no of terms.