Question

solution please for the question

Answer 1

Let 'n' be a positive integer which when divided by 3 gives quotient 'q' and remainder 'r'.
Then , using Euclid's division lemma,
n= 3q+r where r= 0,1 or2.
CASE-1
let r be 0.
Then n = 3q (which is divisible by 3)
let r =1
Then n= 3q +1 ( which is not divisible by 3)
let r = 2
Then n = 3q +2 ( which is not divisible by 3)

CASE 2 -
Let r = 0
Then n+1 = 3q +1 ( which is not divisible by 3)
let r= 1
Then n+1= 3q + 2 (which is not divisible by 3)
Let r=2
Then n+1= 3q + 3 ( which is divisible by 3)

CASE 3-
Let r =0
Then n+4 = 3q +4 (whuch is not divisible by 3)
Let r = 1
Then n+4= 3q + 5 ( which is not divisible by 3)
Let r = 2
Then n+4= 3q+6 (which is divisible by 3)

Hence, one and only one out of n , n+1 and n+ 4 is divisible by 3.

Hope it helps!