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If X = the product of four distinct prime numbers, how many

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If X = the product of four distinct prime numbers, how many [#permalink] New post 06 Sep 2013, 12:32
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Question Stats:

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If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?

(a) 10
(b) 11
(c) 12
(d) 13
(e) 14


My approach:

Let's take the four primes: 2, 3, 5 and 7 --> each of the 4 primes will be a factor of the product

All the possible combinations of the four primes will be factors (10 possible combinations):
2x3
2x5
2x7
3x5
3x7
5x7
2x3x5
2x3x7
2x5x7
3x5x7

Thus, in total there are 10 + 4 = 14 factors.

Is my reasoning correct? Is there a faster way to figure out the number of possible combinations than listing them?
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Sep 2013, 13:37, edited 1 time in total.
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Re: If X = the product of four distinct prime numbers, how many [#permalink] New post 06 Sep 2013, 13:42
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salsal wrote:
If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?

(a) 10
(b) 11
(c) 12
(d) 13
(e) 14


My approach:

Let's take the four primes: 2, 3, 5 and 7 --> each of the 4 primes will be a factor of the product

All the possible combinations of the four primes will be factors (10 possible combinations):
2x3
2x5
2x7
3x5
3x7
5x7
2x3x5
2x3x7
2x5x7
3x5x7

Thus, in total there are 10 + 4 = 14 factors.

Is my reasoning correct? Is there a faster way to figure out the number of possible combinations than listing them?


No that;s not correct.

Finding the Number of Factors of an Integer

First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: 450=2^1*3^2*5^2

Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors.

BACK TO THE QUESTION:

We are told that x=abcd, where a, b, c, and d are four distinct prime numbers. According to the above the # of factors of x including 1 and x is (1+1)(1+1)(1+1)(1+1)=16. Excluding 1 and x the # of factors is 16-2=14.

Answer: E.

Hope it's clear.
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Re: If X = the product of four distinct prime numbers, how many [#permalink] New post 06 Sep 2013, 13:45
salsal wrote:
If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?

(a) 10
(b) 11
(c) 12
(d) 13
(e) 14


My approach:

Let's take the four primes: 2, 3, 5 and 7 --> each of the 4 primes will be a factor of the product

All the possible combinations of the four primes will be factors (10 possible combinations):
2x3
2x5
2x7
3x5
3x7
5x7
2x3x5
2x3x7
2x5x7
3x5x7

Thus, in total there are 10 + 4 = 14 factors.

Is my reasoning correct? Is there a faster way to figure out the number of possible combinations than listing them?


I believe the formula for the amount of factors in an number was to prime factorize it, and then add 1 to all the exponents and multiply them. Since it's only going to be 4 prime numbers all to the 1st power that are multiplied, each of the exponents is 1. So basically this number will have 16 factors, because (2^4) = 16. The question asks you to exclude 2 of the factors, so the answer is 16.
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Re: If X = the product of four distinct prime numbers, how many [#permalink] New post 29 Nov 2013, 11:25
salsal wrote:
If X = the product of four distinct prime numbers, how many factors does X have besides 1 and itself?

(a) 10
(b) 11
(c) 12
(d) 13
(e) 14


My approach:

Let's take the four primes: 2, 3, 5 and 7 --> each of the 4 primes will be a factor of the product

All the possible combinations of the four primes will be factors (10 possible combinations):
2x3
2x5
2x7
3x5
3x7
5x7
2x3x5
2x3x7
2x5x7
3x5x7

Thus, in total there are 10 + 4 = 14 factors.

Is my reasoning correct? Is there a faster way to figure out the number of possible combinations than listing them?


I took the combinations approach as well. Instead of listing them out you I picked combinations of 3, 2 and 1. Combination of all 4 numbers is out because we are told that number and itself should be excluded.

4C3 + 4C2 + 4C1
4!/(3!1!) + 4!/(2!2!) + 4!/(3!1!)
4 + 6 + 4 = 14
Re: If X = the product of four distinct prime numbers, how many   [#permalink] 29 Nov 2013, 11:25
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