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Retired Moderator G
Joined: 11 Aug 2016
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For any integer P greater than 1, P! denotes the product of all the in  [#permalink]

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5 00:00

Difficulty:   85% (hard)

Question Stats: 40% (01:48) correct 60% (02:02) wrong based on 45 sessions

### HideShow timer Statistics For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. In how many ways can 9! be expressed as a product of 2 positive integers?

A. 36
B. 72
C. 79
D. 80
E. 160

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Re: For any integer P greater than 1, P! denotes the product of all the in  [#permalink]

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1
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. In how many ways can 9! be expressed as a product of 2 positive integers?

A. 36
B. 72
C. 79
D. 80
E. 160

9!=1*2*3*4*5*6*7*8*9
Writing in prime factorization form,
$$9!=2^7*3^4*5*7$$
No of distinct factors of 9! is (7+1)(4+1)(1+1)(1+1)=8*5*2*2=160
We need to find out the pair of positive integers whose product is 9!.
So, No of ways we can express 9! as a product of 2 +ve integers=$$\frac{160}{2}$$=80

Ans. (D)
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For any integer P greater than 1, P! denotes the product of all the  [#permalink]

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For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. In how many ways can 9! be expressed as a product of 2 positive integers?

A. 36
B. 72
C. 79
D. 80
E. 160
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1728
Re: For any integer P greater than 1, P! denotes the product of all the  [#permalink]

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I don't like the question, because it's not clear if order should matter. If we write 9! as (1)(9!), is that different or the same as writing it as (9!)(1)?

In any case, we can prime factorize 9!

9! = (9)(8)(7)(6)(5)(4)(3)(2) = (3^2)(2^3)(7)(2*3)(5)(2^2)(3)(2) = 2^7 * 3^4 * 5^1 * 7^1

Once you have a prime factorization, you can count all of a number's divisors by adding one to each exponent, and multiplying the resulting numbers. Here our exponents are 7, 4, 1 and 1. If we add one to each of these, we get 8, 5, 2 and 2, and multiplying these numbers, we get 160. With 160 different divisors, we can make 80 pairs of divisors that will multiply together to 9!, assuming order does not matter.
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Re: For any integer P greater than 1, P! denotes the product of all the  [#permalink]

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1
kiran120680 wrote:
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. In how many ways can 9! be expressed as a product of 2 positive integers?

A. 36
B. 72
C. 79
D. 80
E. 160

$$9!$$
=$$9 * 8 * 7 * 6 * 5 *4 * 3 * 2$$
=$$3^2 * 2^3 * 7 * 2 * 3 * 5 * 2^2 * 3 * 2$$
=$$3^4 * 3^7 * 7 * 5$$

So the number of factors of 9! = (4+1)(7+1)(1+1)(1+1)
= 5 * 8 * 2 * 2
= 160
So the number of ways by which 9! can be expressed as a product of 2 integer = 160/2 = 80. (D)
Math Expert V
Joined: 02 Sep 2009
Posts: 56304
Re: For any integer P greater than 1, P! denotes the product of all the in  [#permalink]

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kiran120680 wrote:
For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. In how many ways can 9! be expressed as a product of 2 positive integers?

A. 36
B. 72
C. 79
D. 80
E. 160

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_________________ Re: For any integer P greater than 1, P! denotes the product of all the in   [#permalink] 23 Mar 2019, 01:33
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