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# If n is the product of the 5 different prime numbers, how many facto

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6978
GMAT 1: 760 Q51 V42
GPA: 3.82
If n is the product of the 5 different prime numbers, how many facto  [#permalink]

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28 Jun 2016, 17:40
00:00

Difficulty:

45% (medium)

Question Stats:

55% (00:56) correct 45% (01:32) wrong based on 110 sessions

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If n is the product of the 5 different prime numbers, how many factors n have except 1 and n?
A. 20
B. 22
C. 24
D. 30
E. 32

*An answer will be posted in 2 days.

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Current Student Joined: 22 Jun 2016 Posts: 242 Re: If n is the product of the 5 different prime numbers, how many facto [#permalink] ### Show Tags 28 Jun 2016, 20:48 1 1 Number of all factors of a number: if N=(a^m)*(b^n)*(c^o)*(d^p)... (where a,b,c,d.. are prime factors of N) so, number of all factors of a N = (m+1)*(n+1)*(o+1)*(p+1).... Applying the above in the question: We have a no. N which has 5 prime factors. So, we can say N = (a^m)*(b^n)*(c^o)*(d^p)*(e^q) (where a,b,c,d and e are 5 prime factor of N) Lowest possible N will have m=n=o=p=q=1. No. of factors of such N = (m+1)*(n+1)*(o+1)*(p+1)*(q+1) = 2*2*2*2*2 = 32 Now, just remove 1 and N from 32 and we are left with 30 factors. Hence, answer is D. _________________ P.S. Don't forget to give Kudos Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6978 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If n is the product of the 5 different prime numbers, how many facto [#permalink] ### Show Tags 01 Jul 2016, 01:17 (1+1)(1+1)(1+1)(1+1)(1+1)=32. Hence, from 32-2=30, we can see that D is the correct answer. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Joined: 20 Feb 2015
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If n is the product of the 5 different prime numbers, how many facto  [#permalink]

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01 Jul 2016, 02:13
1
n=2*3*5*7*11
no of factors of n = 2*2*2*2*2=32 (take any other (set of) distinct prime number for n and the result will always be the same)
no of factors except 1 and n = 32 -2 =30
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Joined: 22 May 2016
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If n is the product of the 5 different prime numbers, how many facto  [#permalink]

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17 Dec 2017, 14:24
MathRevolution wrote:
If n is the product of the 5 different prime numbers, how many factors n have except 1 and n?
A. 20
B. 22
C. 24
D. 30
E. 32

*An answer will be posted in 2 days.

If the method does not look familiar*:
To find the number of factors in a number, n, including n and 1:

1) Find n's prime factors raised to the proper power. Here they are given.
This n could be ANY five distinct primes

$$2^15^111^113^131^1$$

2) Take each prime factor's power and add 1 to the exponent.
E.g., factor 2 has a power of 1. 1+1 = 2. Do the same for all factors.

2 2 2 2 2

3) Multiply the resultant numbers

$$2^5 = 32$$ factors of n, including n and 1

The question excludes n and 1, so without those two factors, there are
(32 - 2) = 30 factors

*See Bunuel , Finding the Number of Factors of an Integer
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If n is the product of the 5 different prime numbers, how many facto   [#permalink] 17 Dec 2017, 14:24
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