Show that one and only one out of `n, n + 4, n + 8, n + 12 and n + 16` is divisible by 5, where n is any positive integer.

Explanation

Given number are n,(n+4), (n+8) , (n+12) and (n+16), where n is any positive integer. Then let n=5q, 5q+1, 5q+2,5q+3,5q+4 for `q in N` [By Euclid's algorithm] Then, in each case if we put the different values of n is the given number. We definetely get one and only of given number is divisible y 5. Hence, one and only one out of n, n+4, n+8, n+12 and n+16 is divisible by 5.
Alternate Method
On dividing on n by 5, let q be the quotient and r be the remainder.
Then, n=5q+r, where `0 le r lt 5`
`Rightarrow n=5q+r, where r=0,1,2,3,4
`Rightarrow n=5q or 5q+1 or 5q+2 or 5q+3 or 5q+4`
CASE I If n=5q, then n is only divisible by 5.
CASE II If n=5q+1, then n+4 n+4=5q+1+4=5q+5=5(q+1), which is only divisibly by 5.
So, in this case, (n+4) is divisible by 5.
CASE III If n=5q+3, then n+2=5q+3+12=5q+15=5(q+3), which is divisible by 5.
So, in this case (n+12) is only divisible by 5.
CASE IV If n=5q+4, then n+16=5q+4+16=5q+20=5(q+4), which is divisibly by 5.
So, in this case, (n+16) is only divisible by 5.
Hence one and only one of n, n+4, n+8, n+12 and n+16 is divisible by 5, where n is any positive integer.