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The greatest common factor of two positive integers is 12 and their pr 24 Feb 2020, 06:10
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36% (02:35) correct 64% (03:04) wrong based on 147 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

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D
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The greatest common factor of two positive integers is 12 and their pr 24 Feb 2020, 13:30
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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
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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The greatest common factor of two positive integers is 12 and their pr 24 Feb 2020, 07:00
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Bunuel
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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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

Answer - E
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The greatest common factor of two positive integers is 12 and their pr 24 Feb 2020, 13:20
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D is the answer...

If (2^3 *12 and 3^3*12) 12 and ( 12 and 12*2^3*3^3)
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The greatest common factor of two positive integers is 12 and their pr 29 Mar 2020, 06:47
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Bunuel
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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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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The greatest common factor of two positive integers is 12 and their pr 09 Apr 2020, 18:52
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Kinshook
Bunuel
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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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.
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The greatest common factor of two positive integers is 12 and their pr 24 Jul 2025, 08:33
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