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

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

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New post 28 Jun 2016, 18:40
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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.
[Reveal] Spoiler: OA

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Re: If n is the product of the 5 different prime numbers, how many facto [#permalink]

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New post 28 Jun 2016, 21:48
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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.
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Re: If n is the product of the 5 different prime numbers, how many facto [#permalink]

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New post 01 Jul 2016, 02:17
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(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.
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If n is the product of the 5 different prime numbers, how many facto [#permalink]

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New post 01 Jul 2016, 03:13
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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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If n is the product of the 5 different prime numbers, how many facto [#permalink]

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New post 17 Dec 2017, 15:24
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

Answer D

*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, 15:24
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