Question

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

Answer 1

Answer:

Given numbers 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 є N

Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

Then n = 5q + r, where 0 ≤ r < 5.

⇒ 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.

So, in this case, n is only divisible by 5.

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

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

Case III If n = 5q + 2, then n + 8 = 5q + 2 + 8 = 5q + 10 = 5(q + 2) which is only divisible by 5.

So, in this case, (n + 8) is only divisible by 5.

Case IV If n = 5q + 3, then n + 12 = 5q + 3 + 12 = 5q + 15 = 5(q + 3) which is only divisible by 5

So, in this case, (n + 12) is only divisible by 5.

Case V If n = 5q + 4, then n + 16 = 5q + 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 of n, n + 4, n + 8 n + 8, n + 12 and n + 16 is divisible by 5 where any positive integer is an integer.

Answer 2

see the difference in each term it is 4 not 5 so one can easily conclude

n=1,

1,5,9,13,17,21

so only 5 is divisible

pls mark it brainlist thanks