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If set G is composed of all the prime numbers n such that

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kevincan
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If set G is composed of all the prime numbers n such that [#permalink] New post   18 Sep 2006, 01:08
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If set G is composed of all the prime numbers n such that n! has less than 1000 odd factors, then how many elements are there in G?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11



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Re: If set G is composed of all the prime numbers n such that [#permalink] New post   Updated on: 25 Sep 2008, 01:55
Hey Kevin..........
The answer must be 8 ie option B

Originally posted by cicerone on 19 Sep 2006, 10:17.
Last edited by cicerone on 25 Sep 2008, 01:55, edited 1 time in total.
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Re: If set G is composed of all the prime numbers n such that [#permalink] New post   19 Sep 2006, 20:16
kevincan wrote:
If set G is composed of all the prime numbers n such that n! has less than 1000 odd factors, then how many elements are there in G?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11


= nc1 + nc2 +......... + ncn<1000

any choice that fits into the above expression should be OA.
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Re: If set G is composed of all the prime numbers n such that [#permalink] New post   14 Aug 2007, 09:49
kevincan wrote:
If set G is composed of all the prime numbers n such that n! has less than 1000 odd factors, then how many elements are there in G?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11


G = {3,7,9,...,n}
Knowing this, we know that:
n! = 3*7*9*...*n and so on.
Since they all odd, we don't have to worry about "odd factors" because odd*odd = odd.
So, we are just trying to find the number of factors.
If set G has one number, then the number of factor is 2
If set G has two numbers, then the number of factor is (2*2) = 4
We know that
2^10 = 1024
The number of factor must be less than that; thus, 2^9 fits.
Therefore, C is the answer.
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Re: If set G is composed of all the prime numbers n such that [#permalink] New post   14 Aug 2007, 13:14
bkk145 wrote:
kevincan wrote:
If set G is composed of all the prime numbers n such that n! has less than 1000 odd factors, then how many elements are there in G?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11


G = {3,7,9,...,n}
Knowing this, we know that:
n! = 3*7*9*...*n and so on.
Since they all odd, we don't have to worry about "odd factors" because odd*odd = odd.
So, we are just trying to find the number of factors.
If set G has one number, then the number of factor is 2
If set G has two numbers, then the number of factor is (2*2) = 4
We know that
2^10 = 1024
The number of factor must be less than that; thus, 2^9 fits.
Therefore, C is the answer.


I got 9 too. But a small correction though. G will not contain 9 as it is not a prime no. I got this problem done in 1.5 mins
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Re: If set G is composed of all the prime numbers n such that [#permalink] New post   14 Aug 2007, 13:20
dahcrap wrote:
bkk145 wrote:
kevincan wrote:
If set G is composed of all the prime numbers n such that n! has less than 1000 odd factors, then how many elements are there in G?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11


G = {3,7,9,...,n}
Knowing this, we know that:
n! = 3*7*9*...*n and so on.
Since they all odd, we don't have to worry about "odd factors" because odd*odd = odd.
So, we are just trying to find the number of factors.
If set G has one number, then the number of factor is 2
If set G has two numbers, then the number of factor is (2*2) = 4
We know that
2^10 = 1024
The number of factor must be less than that; thus, 2^9 fits.
Therefore, C is the answer.


I got 9 too. But a small correction though. G will not contain 9 as it is not a prime no. I got this problem done in 1.5 mins


Good call, I must have blurred out thinking 9 is prime...zzzz
Anyways, one problem I have with this problem is that the question asked for "less than 1000". To me, 7,8,9 are all true, but I picked 9 anyways. It would be clearer if the question ask for "closest to 1000"...don't know.
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Re: If set G is composed of all the prime numbers n such that [#permalink] New post   14 Aug 2007, 14:02
kevincan wrote:
If set G is composed of all the prime numbers n such that n! has less than 1000 odd factors, then how many elements are there in G?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11


The question was not clear to me. I though

n! = 1*2*3* ........*n

Any how now the concept is clear to me.

This is a good question.
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Re: If set G is composed of all the prime numbers n such that [#permalink] New post   14 Aug 2007, 16:06
kevincan wrote:
If set G is composed of all the prime numbers n such that n! has less than 1000 odd factors, then how many elements are there in G?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11


The 7th largest prime number is 17 (2,3,5,7,11,13,15,17)
Note that 17! =3^6 x 5^3 x 7^2 x 11x13x17x 2^n, which has 7x4x3x2x2x2=96x 7 odd factors

The next largest prime number is 19, and 19! surely has more than 1000 odd factors:

19! =3^7x5^3x7^2x11x13x17x19x 2^m, which has 8x4x3x2^4 > 1000 odd factors

Answer A
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Re: If set G is composed of all the prime numbers n such that [#permalink] New post   14 Aug 2007, 19:01
I get A too..though it took some time ...

here is how..

list all the numbers in Set G..

{3,5,7,11,13,17,19,23}

ok now its ask the odd factors of this should be less than 1000!

so the factors could be
3*5, 3*7, 3*11, 3*13...
5*7, 5*11, 5*13...etc
7*11, 7*13...
and so on..you begin to realize that the number of factors will be (n-1)! where n is the number of terms in set G...

the only number that gives (n-1)! <1000 is when n=7...then 6! =720..

Anyone agree with my approach???



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