Question

show that exactly one of the no. n,n+2,n+4 is divisible by 3

Answer 1

let b = 3 

and n is any positive integer  
acc. to euclid lemma
0 < r<b
0<r<3
r can be 0, 1 and 2
 in 1st case of n

when n = 3q
n = 3q ( which is divided by 3)
when n = 3q +1 ( it is not divisible by 3)
when n = 3q + 2 it is also not divisible by 3

in 2nd case of n +2
when n = 3q
n+2 = 3q +2 it is not divisible by 3
when n = 3q +1 
n+2 = 3q +1 +2 
3q +3 it is divisible by 3
when n = 3q+2
n +2 = 3q +2 +2 
3q + 4 it is not divisible by 3

in 3rd case we have n +4 
when n  = 3q 
n +4 = 3q +4 it is not divisible by 3
when n = 3q +1 
n+4 = 3q +1 +4
3q +5 it is not divisible by 3 
when n = 3q +2 
n +4 = 3q +2 +4
3q +6 which is divisible by 3

so it proves that in each case we found only 1 number to divide them 

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

THANKS

#BeBrainly.