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100!=1*2*3*...*100, so 100! has all prime factors which are less than 100 and only them. It happens to be that there are 25 primes less than 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Answer: B.

I really doubt that this is a realistic GMAT question: either you should remember the fact that there are 25 primes less than 100 or you should jut use brute force to count them. _________________

you're right B. i was going through the topic " finding the Number of powers of a prime number P in the n!" in Number Theory article and thought that we could find Number of prime factors in any factorial given through the formula n/p + n/p^2 ........till p^x < n _________________

100!=1*2*3*...*100, so 100! has all prime factors which are less than 100 and only them. It happens to be that there are 25 primes less than 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Answer: B.

I really doubt that this is a realistic GMAT question: either you should remember the fact that there are 25 primes less than 100 or you should jut use brute force to count them.

Brunel, can we do this for every number?

for example, can we say that 55! would have as many prime number as the count of total prime numbers between 1 and 55? _________________

Appreciation in KUDOS please! Knewton Free Test 10/03 - 710 (49/37) Princeton Free Test 10/08 - 610 (44/31) Kaplan Test 1- 10/10 - 630 Veritas Prep- 10/11 - 630 (42/37) MGMAT 1 - 10/12 - 680 (45/34)

100!=1*2*3*...*100, so 100! has all prime factors which are less than 100 and only them. It happens to be that there are 25 primes less than 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Answer: B.

I really doubt that this is a realistic GMAT question: either you should remember the fact that there are 25 primes less than 100 or you should jut use brute force to count them.

Brunel, can we do this for every number?

for example, can we say that 55! would have as many prime number as the count of total prime numbers between 1 and 55?

Sure. 55!=1*2*3*...*55.

Now, as prime number cannot be written as a product of two factors, both of which are greater than 1 then prime more than 55 can not be a factor of 55! (prime more than 55, for example 71, just no way can occur in 55!).

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