Question

Show that one and only one out of n ,n+1, n+2 is divisible by 3.

Answer 1

let n = 3q + r 
therefore by euclids division lemma, 0 =< r < 3
therefore r = 0 , 1 , or 2
therefore n = 3q or 3q + 1 or 3q + 2
case 1: ( n = 3q)
now n = 3q = 3 (q)
∴divisible by 3
n+ 1 = 3q + 1 = 3 (q) + 1
∴not divisible by 3
n+2 = 3q + 2 = 3q + 3 -1 = 3 (q+1) - 1
therefore not divisible by 3
∴here , only n is divisible by 3

Case2: (n = 3q + 1)
n = 3q + 1 = 3 (q) + 1
∴not divisible by 3
n+1 = 3q + 1 + 1 = 3q + 2 = 3q + 3 -1 = 3 (q + 1) - 1
∴not divisible by 3
n+3 = 3q + 1 + 2 = 3q + 3 = 3 (q + 1)
∴divisible by 3
∴here only n + 2 is divisible by 3

Case 3:(n = 3q + 2)
n = 3q + 2 = 3q + 3 -1 = 3 (q + 1) - 1
∴not divisible by 3
n+ 1 = 3q + 2 + 1 = 3q + 3 = 3 (q + 1)
∴DIVISIBLE by 3
n + 2 = 3q + 2 + 2 = 3q + 3 + 1 = 3 (q + 1 ) + 1
∴not divisible by 3
∴here only n+1 is divisible by 3

so only one out of them is divisible by 3
pls mark as brainliest

Answer 2

Answer:

All the best

Step-by-step explanation: