Add together all possible values for gcdn n3 as n varies ove

Add together all possible values for gcd(n, n+3) as n varies over the natural numbers.

Solution

The 4 numbers are n , n+1 , n+2 , n+3 they are of the form

4m ,4m+1 , 4m+2 , 4m+3

4m , 4m+2 have the gcd =2 , but 4m+1 , 4m+3 will not have 2 as its factor

hence all the 4 nos will not have any common divisor   

gcd ( n , n+ , n+2 , n+3) =1