Question

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

explain this question briefly

Answer 1

Hey buddy here is ur answer ......


We applied Euclid Division algorithm on n and 3. 

a = bq +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 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.


hope answer will be helpful to u.....

plz mark it as brainalist one...

Answer 2

Step-by-step explanation:

Euclid's division Lemma any natural number can be written as: .

where r = 0, 1, 2,. and q is the quotient.

thus any number is in the form of 3q , 3q+1 or 3q+2.

case I: if n =3q

n = 3q = 3(q) is divisible by 3,

n + 2 = 3q + 2 is not divisible by 3.

n + 4 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.

case II: if n =3q + 1

n = 3q + 1 is not divisible by 3.

n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3.

n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 is not divisible by 3.

case III: if n = 3q + 2

n =3q + 2 is not divisible by 3.

n + 2 = 3q + 2 + 2 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.

n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) is divisible by 3.

thus one and only one out of n , n+2, n+4 is divisible by 3.

Hence, it is solved... mmm

THANKS