Show that exactly one of the numbers n, n+2 or n+4 is divisible by 3
Answers
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.
Let n = 3k, 3k + 1 or 3k + 2.
(i) When n = 3k :
n is divisible by 3.
n + 2 = 3k + 2
⇒ n + 2 is not divisible by 3.
n + 4 = 3k + 4
= 3 (k + 1) + 1