If the greatest common factor (GCF) of two positive integers x and y i
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19 Apr 2020, 06:38
IMO: D
Here's how I solved it:*
x can be written as GCF(x,y)*n1, where n1 is any positive integer
Similarly, y can be written as GCF(x,y)*n2
substituting the above expressions in x+y+GCF(x,y)=91, we get,
GCF(x,y)*n1 + GCF(x,y)*n2 +GCF(x,y)=91
=> GCF(x,y)*(n1 + n2 +1) = 91
The GCD(x,y) should be a factor of 91 i.e 1,7,13 or 91.
We reject "1" as It's given in the question that the GCD(x,y)>1
and we reject "91" as n1 and n2 can't be 0.
So the final 2 expressions and the ordered pairs we get for n1 and n2 are:
1. 7*(n1 + n2 + 1)=91
=> n1 + n2 =12
so (n1 ,n2) could be (1,11) , (2,10), (3,9) , (4,8) , (5,7) , (6,6), (7,5) , (8,4) , (9,3) , (10,2), (11,1)
i.e 11 values
2. 13(n1 + n2 +1)=91
=> n1 + n2=6
so (n1 ,n2) could be (1,5) , (2,4) , (3,3) , (4,2), (5,1)
i.e. 4 values
Now, since we are taking GCD(x,y) as 7 & 13 there should be no common factor between n1 and n2 i.e n1 and n2 should be coprime.
The coprime ordered pair of n1 and n2 are:
(1,11), (5,7), (7,5), (11,1), (1,5), (5,1)
6 pair. So the answer is D