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If n is the product of 3 different prime numbers, how many factors doe

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[GMAT math practice question]

If n is the product of 3 different prime numbers, how many factors does n have except 1 and n?

A. 6
B. 7
C. 8
D. 9
E. 10
[Reveal] Spoiler: OA

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If n is the product of 3 different prime numbers, how many factors doe [#permalink]

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New post 02 Oct 2017, 01:09
MathRevolution wrote:
[GMAT math practice question]

If n is the product of 3 different prime numbers, how many factors does n have except 1 and n?

A. 6
B. 7
C. 8
D. 9
E. 10


Let the 3 different prime numbers be a,b,and c.

If \(n = a^x*b^y*c^z\) where x,y, and z are the powers of the respective prime numbers
then the number "n" has \((x+1)(y+1)(z+1)\) factors.

In this case, x=y=z=1, the number n will have 2*2*2 or 8 factors.

Therefore, excluding 1 and n, the number n has 6 factors(Option A)
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Re: If n is the product of 3 different prime numbers, how many factors doe [#permalink]

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Or try plugging in numbers:

We need 3 different prime numbers, so try {2,3,5}
2*3*5=30
Find the factors of 30 other than 1 and 30
3*10 (that's two)
2*15 (four)
6*5 (six)

Answer is A

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New post 04 Oct 2017, 00:43
=>

n = p*q*r where p, q and r are different prime numbers.
Then the number of factors of n is (1+1)(1+1)(1+1) = 8 including 1 and n, since the exponents of p, q and r are 1.
Thus the number of factors of n except 1 and n is 8 – 2 = 6.

Answer: A
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Re: If n is the product of 3 different prime numbers, how many factors doe   [#permalink] 04 Oct 2017, 00:43
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