# Show that one and only one out of n,

Question:

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.

Solution:

Given numbers are n, (n+ 4), (n + 8), (n + 12) and (n + 16), where n is any positive integer.

Then, let n = 5q, 5g + 1, 5g + 2, 5g + 3, 5q + 4 for q∈ N                           [by Euclid’s algorithm]

Then, in each case if we put the different values of n in the given numbers. We definitely get one and only one of given numbers is divisible by 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, letq be the quotient and r be the remainder.

Thenn=5q + r, where 0

$\Rightarrow \quad n=5 q+r$, where $r=0,1,2,3,4$

$\Rightarrow \quad n=5 q$ or $5 q+1$ or $5 q+2$ or $5 q+3$ or $5 q+4$

Case I If $n=5 q$, then $n$ is only divisible by 5 .

Case II If $n=5 q+1$, then $n+4=5 q+1+4=5 q+5=5(q+1)$, which is only divisible by $5 .$

So, in this case, $(n+4)$ is divisible by 5 .

Case IV If $n=5 q+4$, then $n+16=5 q+4+16=5 q+20=5(q+4)$, which is divisible by 5 .

So, in this case, $(n+16)$ is only divisible by 5 .

Hence, 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.

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