Question

Q. Show that exactly one of the number n, n+2 or n+4 is divisible b.y 3

Answer 1

Using Euclid's division lemma, we know that every integer can be represented as n = 3q + r where r = 0,1,2.

Case 1 : When n= 3q
 n= 3q        ...............(divisible by 3)
 n+1= 3q+1    ................(not divisible)
 n+2= 3q +2 .......................(not divisible)

Case 2: When n=3q +1
n=3q+1...................(not divisible)
n+1= 3q +2 ...............(not divisible)
n+2 = 3q+3...................(divisible)

Case 3:
n= 3q+2...................(not divisible)
n+1=3q+3...............(divisible)
n+2= 3q+4...............(not divisible)

Thus, we observe that one out of all the possible values of n are 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.