Question

Prove show that one and only one out of n n + 2 or n + 4 is divisible by 3 where n is any positive integer

Answer 1

Hi Mate !!


Let a be any positive integer Which when divided by 3 gives q as quotient and r as remainder .

By Euclid's division lemma

a = bq + r

0 ≤ r < b


n = 3q + r

So , r = 0 , 1 , 2

n = 3q

n = 3q + 1

n = 3q + 2


Now ,

☢ Case :- 1 ( a )

n = 3q

It is divisible by 3 leaving remainder 0.

☣ Case :- 1 ( b )

n = 3q

n + 2 = 3q + 2

It is not divisible by 3 because it leave 2 as remainder.

☣ Case :- 1 ( c )

n = 3q

n + 4 = 3q + 4

It is not divisible by 3 because it leaves 4 as remainder.

☢ Case :- 2 ( a )

n = 3q + 1

It is not divisible by 3 because it leaves 1 as remainder.

☣ Case :- 2 ( b )

n = 3q + 1

n + 2 = 3q + 1 + 3

n + 2 = 3q + 3

It is divisible by 3 because it leaves remainder 3 which can further divide to give 0.

☣ Case :- 2 ( c )

n = 3q + 1

n + 4 = 3q + 1 + 4

n + 4 = 3q + 5

It is not divisible by 3 because it leaves 5 as remainder.


☢ Case :- 3 ( a )

n = 3q + 2

It is not divisible by 3 because it leaves 2 as remainder.

☣ Case :- 3 ( b )

n = 3q + 2

n + 2 = 3q + 2 + 2

n + 2 = 3q + 4

It is not divisible by 3 because it leaves 4 as remainder.

☣ Case :- 3 ( c )

n = 3q + 2

n = 3q + 2 + 4

n = 3q + 6

It is divisible by 3 because it leaves 6 as remainder which further can divide by 3 to get 0 as remainder.

Hence proved only one of them are divisible by 3 .

Answer 2

Hey !!

Here's your answer.. ⬇⬇

By Euclid's Algorithm..

a = bq + r
a = n
b = 3

n = 3q + r ---- ( whereas 0 < r < 3 )

taking r = 0, 1, and 2

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

n = 3q

n + 2 = 3q + 1 + 2
= 3q + 3

n + 4 = 3q + 2 + 4
= 3q + 6

3q, 3q + 3, and 3q + 6 all are divisible by 3.

hence, n, n + 2 , and n + 4 all are also divisible by 3.


HOPE IT HELPS..

THANKS ^-^