Question

Show that one and only one out of n, n+2,n+4 is divisible by 3 , where n is any positive integer

Answer 1

heya!!!!☺️

⏬⏬ANSWER⏬⏬

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We apply Euclid Division algorithm on n and 3.

a = be + r on putting a = n and b = 3

n = 3q + r 0 < r < 3

I.e., n = 3q ------ (1),

n = 3q + 1 --------(2),

n = 3q + 2 -------(3)

n = 3q is divisible by 3

or n + 2 = 3q + 1 + 2 = 3q + 3 is also divisible by 3

or n + 4 = 3q + 2 + 4 = 3q + 6 is also divisible by 3

Hence, n, n + 2, n + 4 are divisible by 3.

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hope it helps ^,^

Answer 2

$\huge\mathfrak{\underline{\underline{Solution:-}}}$

Let n be an arbitrary number.

Thus, on dividing n by 3, we get some quotient (let it be q) and remainder (let it be r)

•°• By Euclid's Division Lemma, we have

n = 3q + r, where r is greater than or equal to 0 and less than 3 i.e., 0 ≤ r < 3

•°• The possible remainders are 0, 1, 2.

=> n = 3q;

=> n = 3q + 1;

=> n = 3q + 2

Case 1 :

If n = 3q, then n is divisible by 3

Case 2 :

When n = 3q + 1, then (n + 2) = 3q + 3 = 3(q + 1)

So, here, (n + 2) is divisible by 3

Case 3 :

When n = 3q + 2, then (n + 4) = 3q + 6

= 3(q + 2), i.e., it is divisible by 3

Thus, (n + 4) is divisible by 3

Hence, one and only one among n, n + 2, n + 4 is divisible by 3.