Question

?............answer plz .​

Answer 1

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!