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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]

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06 Sep 2013, 13:32
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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, 14:37, edited 1 time in total.
RENAMED THE TOPIC.
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Re: If X = the product of four distinct prime numbers, how many [#permalink]

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06 Sep 2013, 14:42
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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.

Hope it's clear.
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Re: If X = the product of four distinct prime numbers, how many [#permalink]

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06 Sep 2013, 14:45
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]

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29 Nov 2013, 12:25
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
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Re: If X = the product of four distinct prime numbers, how many [#permalink]

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26 Feb 2015, 07:03
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Re: If X = the product of four distinct prime numbers, how many [#permalink]

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28 Feb 2015, 04:58
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?

Let the numbers be 2,3,5,7.
Single factors are 4C1 = 4
Product of 2 numbers are also factors = 4C2 = 6
Product of 3 numbers are also factors = 4C3 = 4

Hence 14 other factors besides 1 and the number itself.

Hence option E.

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Re: If X = the product of four distinct prime numbers, how many [#permalink]

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02 Mar 2015, 11:04
If we use 2*3*5*7 = 210

210 = 2*5*3*7

Total number of factors: (1+1) *(1+1) *(1+1) *(1+1) = 16

16 - 2 = 14, without 1 and itself. ANS E
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Re: If X = the product of four distinct prime numbers, how many [#permalink]

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30 Aug 2016, 23:34
Hello from the GMAT Club BumpBot!

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Re: If X = the product of four distinct prime numbers, how many [#permalink]

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31 Aug 2016, 00:05
[quote="salsal"]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

No. of Factors =(1+1) x (1+1) x(1+1) x(1+1)
= 16
Besides 1 & itself no. of factors X will have = 16-2 =14
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Re: If X = the product of four distinct prime numbers, how many   [#permalink] 31 Aug 2016, 00:05
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