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The greatest common factor of two positive integers is 12 and their pr

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Math Expert
Joined: 02 Sep 2009
Posts: 64990
The greatest common factor of two positive integers is 12 and their pr  [#permalink]

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24 Feb 2020, 05:10
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85% (hard)

Question Stats:

47% (02:31) correct 53% (03:05) wrong based on 47 sessions

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The greatest common factor of two positive integers is 12 and their product is 31104. How many pairs of such numbers are possible?

A. 6
B. 5
C. 3
D. 2
E. 1

Are You Up For the Challenge: 700 Level Questions

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Joined: 14 Dec 2019
Posts: 673
Location: Poland
GMAT 1: 570 Q41 V27
WE: Engineering (Consumer Electronics)
Re: The greatest common factor of two positive integers is 12 and their pr  [#permalink]

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24 Feb 2020, 06:00
Bunuel wrote:
The greatest common factor of two positive integers is 12 and their product is 31104. How many pairs of such numbers are possible?

A. 6
B. 5
C. 3
D. 2
E. 1

Are You Up For the Challenge: 700 Level Questions

The product of 2 integers be a*b = $$31104 = 12^2*216 = 12^2*m*n =$$ where m and n are integers not having a 2 or 3 as a prime factor in it since having any of those will change the GCF

Therefore $$m*n=216 = 2^3*3^3$$

As we can see there is no pair other than 216 and 1 that will satisfy the condition hence only 1 such pair. And that pair would be $$12*2^3*3^3$$ and 12

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Joined: 26 Sep 2017
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Re: The greatest common factor of two positive integers is 12 and their pr  [#permalink]

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24 Feb 2020, 12:20
1

If (2^3 *12 and 3^3*12) 12 and ( 12 and 12*2^3*3^3)
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Re: The greatest common factor of two positive integers is 12 and their pr  [#permalink]

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24 Feb 2020, 12:30
1
3
a= 12*x
b=12*y

where x and y are co-primes

$$12^2*x*y = 31104= 12^2 * 6^3$$

$$x*y = 2^3 *3^3$$

Since x and y are co-prime, there are only 2 unordered pair possible.

(x,y) = ($$1, 2^3*3^3$$) and ($$2^3, 3^3$$)

Bunuel wrote:
The greatest common factor of two positive integers is 12 and their product is 31104. How many pairs of such numbers are possible?

A. 6
B. 5
C. 3
D. 2
E. 1

Are You Up For the Challenge: 700 Level Questions
CEO
Joined: 03 Jun 2019
Posts: 3182
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)
Re: The greatest common factor of two positive integers is 12 and their pr  [#permalink]

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29 Mar 2020, 05:47
Bunuel wrote:
The greatest common factor of two positive integers is 12 and their product is 31104. How many pairs of such numbers are possible?

A. 6
B. 5
C. 3
D. 2
E. 1

Are You Up For the Challenge: 700 Level Questions

Given: The greatest common factor of two positive integers is 12 and their product is 31104.

Asked: How many pairs of such numbers are possible?

Let the pair of positive integers be (x,y) and gcd & lcm be g & l respectively

xy = 12 * l = 31104 = 2^2*3 * l = 2^7*3^5
l = 2^5*3^4

Let x be 2^a*3^b & y be 2^c*3^d
min (a, c) = 2; min (b, d) = 1
max (a, c) = 7; max(b, d) = 5

(a, c) = {(2,7),(7,2)}
(b, d) = {(1,5),(5,1)}

(x,y)= {(2^2*3^1,2^7,*3^5),(2^2*3^5,2^7,*3^1),(2^7*3^1,2^2,*3^5),(2^7,3^5,2^2*3^1)}

There are 4 such ordered pairs possible.

If order is not important, as may be the case here, (2^2*3^1 & 2^7*3^5) & (2^2*3^5 & 2^7*3^1) are possible

IMO D
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Kinshook Chaturvedi
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Joined: 08 Sep 2019
Posts: 62
Location: India
Schools: LBS '22, IESE
Re: The greatest common factor of two positive integers is 12 and their pr  [#permalink]

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09 Apr 2020, 17:52
Kinshook wrote:
Bunuel wrote:
The greatest common factor of two positive integers is 12 and their product is 31104. How many pairs of such numbers are possible?

A. 6
B. 5
C. 3
D. 2
E. 1

Are You Up For the Challenge: 700 Level Questions

Given: The greatest common factor of two positive integers is 12 and their product is 31104.

Asked: How many pairs of such numbers are possible?

Let the pair of positive integers be (x,y) and gcd & lcm be g & l respectively

xy = 12 * l = 31104 = 2^2*3 * l = 2^7*3^5
l = 2^5*3^4

Let x be 2^a*3^b & y be 2^c*3^d
min (a, c) = 2; min (b, d) = 1
max (a, c) = 7; max(b, d) = 5

(a, c) = {(2,7),(7,2)}
(b, d) = {(1,5),(5,1)}

(x,y)= {(2^2*3^1,2^7,*3^5),(2^2*3^5,2^7,*3^1),(2^7*3^1,2^2,*3^5),(2^7,3^5,2^2*3^1)}

There are 4 such ordered pairs possible.

If order is not important, as may be the case here, (2^2*3^1 & 2^7*3^5) & (2^2*3^5 & 2^7*3^1) are possible

IMO D

Shouldn't the max(a,c) be 5 and max(b,d) be 4 as the LCM is 2^5*3^4, and thus is the answer not only 1 pair. Experts kindly help.
Re: The greatest common factor of two positive integers is 12 and their pr   [#permalink] 09 Apr 2020, 17:52