Greatest common divisor

How many unordered pairs of integers ( x , y ) (x, y) exist such that

x × y × gcd ( x , y ) = 1080 ? x \times y \times \text{gcd} (x,y) =1080?

Clarification: gcd ( x , y ) \text{gcd}(x,y) is the greatest common divisor of x x and y y .


8 52 5 12 11 10 7 9

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1 solution

Harsh Poonia
Jul 6, 2019

Let g c d ( x , y ) = k . gcd(x,y)=k. Let x = k m , y = k n x=km,y=kn where g c d ( m , n ) = 1 gcd(m,n)=1 . Now the given equation reduces to k m k n k = 1080 k 3 m n = 3 3 2 3 5. km \cdot kn \cdot k=1080 \implies k^3 mn=3^3 2^3 5.

To solve these cases, it's very important to recall that g c d ( m , n ) = 1 gcd(m,n)=1 AND that we have to search for unordered pairs.

Possible cases to consider :

1) k = 1 m n = 27 × 8 × 5 m ( , n ) = ( 1080 , 1 ) ; ( 135 , 8 ) ; ( 216 , 5 ) ; ( 27 , 40 ) k=1 \implies mn=27×8×5 \implies m(,n)={{(1080,1); (135, 8); (216 ,5); (27, 40)} }

2) k = 2 m n = 135 ( m , n ) = ( 135 , 1 ) ; ( 27 , 5 ) k=2 \implies mn=135 \implies (m,n)={(135,1); (27,5) }

3) k = 3 m n = 40 ( m , n ) = ( 40 , 1 ) ; ( 8 , 5 ) k=3 \implies mn=40 \implies (m,n)={(40,1); (8,5)} .

4) k = 6 m n = 5 ( m , n ) = ( 5 , 1 ) k=6 \implies mn=5 \implies (m,n)={(5,1)}

Thus total 9 \boxed{9} pairs.

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