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If n=2^5*3^4*5^3*7^2*11, how many factors of n are there?

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MathRevolution
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If n=2^5*3^4*5^3*7^2*11, how many factors of n are there? [#permalink] New post   Updated on: 01 Dec 2016, 02:01
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If \(n=2^5*3^4*5^3*7^2*11\), how many factors of n are there?

A. 180 B. 360 C. 540 D. 720 E. 810
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D

Originally posted by MathRevolution on 29 Nov 2016, 17:37.
Last edited by MathRevolution on 01 Dec 2016, 02:01, edited 1 time in total.
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Re: If n=2^5*3^4*5^3*7^2*11, how many factors of n are there? [#permalink] New post   29 Nov 2016, 17:49
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Number of factors of any positive integer of the form \(k=p^a*q^b*r^c\) where p,q,r are prime number is given by the equation => \((a+1)*(b+1)*(c+1)\)

Hence the number of factors => 6*5*4*3*2

For quick calculation notice the above value is 6! and the value of 6! is 720
Hence D


Side Note =>
Its good if you know the fist few factorials =>
1!->1
2!->2
3!->6
4!->24
5!->120
6!->720
Additionally 0!->1
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Re: If n=2^5*3^4*5^3*7^2*11, how many factors of n are there? [#permalink] New post   01 Dec 2016, 02:00
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==> The number of factors become (5+1)(4+1)(3+1)(2+1)(1+1)=720, and therefore the answer is D.

Answer: D
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Re: If n=2^5*3^4*5^3*7^2*11, how many factors of n are there? [#permalink] New post   13 Feb 2018, 08:27
MathRevolution wrote:
If \(n=2^5*3^4*5^3*7^2*11\), how many factors of n are there?

A. 180 B. 360 C. 540 D. 720 E. 810


(5+1)(4+1)(3+1)(2+1)(1+1) = 6*5*4*3*2 = 720

Answer will be (D) 720
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If n=2^5*3^4*5^3*7^2*11, how many factors of n are there? [#permalink] New post   13 Feb 2018, 20:43
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MathRevolution wrote:
If \(n=2^5*3^4*5^3*7^2*11\), how many factors of n are there?

A. 180 B. 360 C. 540 D. 720 E. 810

The number of factors of n, including n and 1, can be found by:
1) finding the prime factors of n (given)

2) adding 1 to every factor's exponent, thus
6 5 4 3 2

3) multiplying the results from #2.
6 * 5 * 4 * 3 * 2 = 720 factors of n, including n and 1

See Bunuel , Finding the Number of Factors of an Integer (scroll down)

Answer D
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Re: If n=2^5*3^4*5^3*7^2*11, how many factors of n are there? [#permalink] New post   14 Feb 2018, 10:55
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MathRevolution wrote:
If \(n=2^5*3^4*5^3*7^2*11\), how many factors of n are there?

A. 180 B. 360 C. 540 D. 720 E. 810


The procedure for determining the total number of factors of a number is to first prime factorize the number, then add 1 to each exponent, and then multiply those values. Thus, the number of factors of n is:

(5 + 1)(4 + 1)(3 + 1)(2 + 1)(1 + 1)

6 x 5 x 4 x 3 x 2 = 720

Answer: D
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Re: If n=2^5*3^4*5^3*7^2*11, how many factors of n are there? [#permalink]
14 Feb 2018, 10:55
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