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

Question

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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nick1816 · Accepted Answer

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 )

shameekv1989 · Answer

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

jrk23 · Answer

D is the answer...

If (2^3 *12 and 3^3*12) 12 and ( 12 and 12*2^3*3^3)

Kinshook · Answer

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

SanjaySrini · Answer

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

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

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