Question

Show that only one of the numbers n, (n+2) and (n+4) is divisible by 3 where n is any positive integer.

Answer 1

Let me be any positive integer divided by 3 quotient q and remainder r
n=3q+r
r=0,1,2
Case -1 (when r=0)
1. n=3q+0
"n=3q"
2. n+2=3q+2
3. n+4 = 3q +4
( n is divisible by 3)
Case-2(where r=1)
1. n=3q+1
2. n+2=3q+3
3(q+1)
3. n+4=3q+1+4
= 3(q+1)2
( n+2 is divisible by 3)
Case-3(where r =2)
1.n=3q+2
2.n+2 =3q+3+1
3(q+1)+1
3. n+4=3q+2+4
3q +6
3(q+2)
(n+4 is divisible by 3)

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.