Stmt II
j=5 k=5 (j and k can be equal no where its given it cant be)
gcf = 5
j=2 k=5
gcf=1
INSUFF
Stmt I
k = j+1
Either k is odd j is even or j is odd k is even which means only 1 can be the greatest common factor
SUFF
Choose A
Greatest common factor
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It's not that one of the numbers is even and one odd that makes the GCD equal to 1 here -- after all, 12 is even, and 15 is odd, but their GCD is not 1 -- it's the fact that the numbers are consecutive. The GCD of consecutive integers x and x+1 is always equal to 1, since if d is a divisor of x, then x+d will be the closest multiple of d which is larger than x, so d couldn't be a divisor of x+1 unless d=1.cramya wrote: Stmt I
k = j+1
Either k is odd j is even or j is odd k is even which means only 1 can be the greatest common factor
SUFF
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