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For any integer P greater than 1, P! denotes the product of all the in

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

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New post 05 Sep 2018, 13:15
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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

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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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New post 05 Sep 2018, 13:36
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GmatDaddy 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!=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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New post 22 Mar 2019, 11:52
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  [#permalink]

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New post 22 Mar 2019, 12:32
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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New post 22 Mar 2019, 12:57
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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


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

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)
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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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New post 23 Mar 2019, 01:33
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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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