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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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New post 06 Sep 2013, 12:32
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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, 13:37, 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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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]

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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]

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

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

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New post 28 Feb 2015, 03:58
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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New post 02 Mar 2015, 10:04
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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Re: If X = the product of four distinct prime numbers, how many [#permalink]

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New post 30 Aug 2016, 23:05
[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] 30 Aug 2016, 23:05
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