?............answer plz .
Answers
Step-by-step explanation:
By Euclid's division lemma, we have
a = bq + r; 0 ≤ r ≤ b
Let a = n and b = 3 be any two positive integers.
Applying Euclid's division lemma, we have
n = 3q + r {0 ≤ r ≤ 3}
∴ n = 3q + 0, 3q + 1, 3q + 2
(i) When n = 3q:
n = 3q is divisible by 3.
Add '2' on both sides, we get
n + 2 = 3q + 2 --- Not divisible by 3 as it leaves remainder 2.
n + 4 = 3q + 4 ---- Not divisible by 3 as it leaves remainder 1.
Thus, n is divisible by 3.
(ii) When n = 3q + 1:
n = 3q + 1 ----- Not divisible by 3 as it leaves remainder 1.
n + 2 = (3q + 1) + 2
= 3q + 3
= 3(q + 1) ------ it is divisible by 3.
n + 4 = (3q + 1) + 4
= 3(q + 1) + 2 ------- Not divisible by 3 as it leaves remainder 2
Thus, (n + 2) is divisible by 3.
(iii) When n = 3q + 2:
n = 3q + 2 ------- Not divisible by 3 as it leaves remainder 2.
n + 2 = (3q + 2) + 2
= 3q + 4
= 3(q + 1) + 1 ----- Not divisible by 3 as it leaves remainder 1
n + 4 = (3q + 2) + 4
= (3q + 6)
= 3(q + 2) ------- divisible by 3.
Thus, n + 4 is divisible by 3.
∴ Therefore, for every value of r such that 0 ≤ r ≤ 3 only one out of n,n+2 and n + 4 is divisible by 3.
Hope it helps!