Let the two numbers be A and B with A > B
Now K(N) = A2 - B2 = (A-B)(A+B) . Now we need to know the pair of factors of N which when multiplied gives N . Applying the symmetrical property of factors we need to look for factors upto sqtr(N).
Also both the factors must be either both even or both odd (as their sums must be even for integer values of A and B).
From the given options we get the pair of A+B and A-B
K(110) = (110, 1) , (55 , 2) ,( 5, 22), (11,10) .However for each pair the 2nd condition is not satisfied. So number of such pairs (A,B) = 0
Similarly for K(105) we get valid numbers (A,B) as (53,52),(19,16) and (13,8) = three
For K(216) we get four such pairs (satisfyingboth conditions)
K(384) we get 6 .
Hence the answer is K(384)
Let the two numbers be A and B with A > B
Now K(N) = A2 - B2 = (A-B)(A+B) . Now we need to know the pair of factors of N which when multiplied gives N . Applying the symmetrical property of factors we need to look for factors upto sqtr(N).
Also both the factors must be either both even or both odd (as their sums must be even for integer values of A and B).
From the given options we get the pair of A+B and A-B
K(110) = (110, 1) , (55 , 2) ,( 5, 22), (11,10) .However for each pair the 2nd condition is not satisfied. So number of such pairs (A,B) = 0
Similarly for K(105) we get valid numbers (A,B) as (53,52),(19,16) and (13,8) = three
For K(216) we get four such pairs (satisfyingboth conditions)
K(384) we get 6 .
Hence the answer is K(384)