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If n is the product of 3 same prime numbers and 1 different prime numb

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If n is the product of 3 same prime numbers and 1 different prime numb [#permalink]

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New post 31 Jul 2017, 00:07
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If n is the product of 3 same prime numbers and 1 different prime number, then how many different factors of n are there?

A. 6
B. 8
C. 9
D. 12
E. 18
[Reveal] Spoiler: OA

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Re: If n is the product of 3 same prime numbers and 1 different prime numb [#permalink]

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New post 31 Jul 2017, 00:23
Let p and q be the prime numbers.
n= (p^3)*q
Factor will be: n, 1, p, p^2, p^3, q, p*q, (p^2)*q
8 factors. Answer: B
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If n is the product of 3 same prime numbers and 1 different prime numb [#permalink]

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New post 31 Jul 2017, 01:42
If the number n can be represented as a product of 2 primes p and q,
where p has x occurrences and q has y occurrences(\(n = p^x * q^y\)),
the total number of factors possible is equal to (x+1)(y+1)

Coming back to the problem,
the prime numbers p and q occur 3 times and 1 time respectively,
the total number of factors is (1+1)(3+1) = 2*4 = 8(Option B)
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If n is the product of 3 same prime numbers and 1 different prime numb [#permalink]

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New post 31 Jul 2017, 17:20
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MathRevolution wrote:
If n is the product of 3 same prime numbers and 1 different prime number, then how many different factors of n are there?

A. 6
B. 8
C. 9
D. 12
E. 18

For those who haven't seen material in citation below: To find the number of factors of n, the method for the formula* mentioned above is

1. List n's prime factors raised to the correct power. Factors here are

"three same prime numbers and one different prime number..." Just assign values (this formula depends on exponents)

7*7*7*13 is

\(7^313^1\)

2. List the exponents and add 1 to each

3 (+ 1) = 4
1 (+ 1) = 2

3. Multiply the resultant numbers

4*2 = 8 factors of n, 1 and n included

Answer B

*Bunuel, Finding the Number of Factors of an Integer, at
https://gmatclub.com/forum/math-number- ... 88376.html
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Re: If n is the product of 3 same prime numbers and 1 different prime numb [#permalink]

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New post 01 Aug 2017, 23:58
==> You get \(n=(3)(3)(3)(5)\), which becomes \(n=3^35^1\). Thus, the number of factors of n becomes \((3+1)(1+1)=8\).

The answer is B.
Answer: B
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Re: If n is the product of 3 same prime numbers and 1 different prime numb   [#permalink] 01 Aug 2017, 23:58
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