Question

Show that exactly one of the numbers n, n+2,n+4 os divisble by 3​

Answer 1

Answer:

Step-by-step explanation:

Since n is a positive integer taking b =3

We can write n = 3q + r , where q is some integer

n = 3q , n = 3q + 1 , n = 3q + 2

Case 1 = When n = 3q, n + 2 = 3q + 2 and n + 4 = 3q + 4 clearly only 3q is divisible by 3

Case 2 = When n = 3q + 1, n + 2 = 3q + 3 and n + 4 = 3q + 5 . Here also only n + 2 = 3q + 3 = 3(q + 1) is divisible by 3. Other two namely n and n + 4 are not divisible by 3.

Case 3 = When n = 3q + 2, n + 2 = 3q + 4 and n + 4 = 3q + 6 and in this case, only n + 4 = 3(q + 2) is divisible by 3

Hence only one out of n, n + 2 and n + 4 is divisible by 3 for any positive integer n .

Answer 2

Step-by-step explanation:

let n= 1

so the numbers are 1,2 and 3 and 3 is divisible by 3

let n= 2

so the no. are 2,4 and 6 and 6 is divisible by 3

let n= 3

so the no. are 3,5 and 7 and 3 is divisible by 3

let n= 4

so the no. are 4,6 and 8 and 6 is divisible by 3

so it proves that exactly one of the numbers n,n+2,n+4 is divisible by 3

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