Question

Show that one of three consecutive odd integers is a multiple of 3

Answer 1

Since it is an odd number

So

1st number be 2n+1

Being consecutive odd

2nd number will be 2n+1+2=2n+3

And

3rd number 2n+5

2n is always an even number and condition will revolve around this only giving rise to 2 cases.

Case 1 : if 2n is divisible by 3

then 2n + 3 will always be divisible by 3.

Case 2 : if 2n is not divisible by 3

Then remainder is either 1 or 2

Case 2 (a). Remainder 1 implies 2n-1 is divisible by 3.

If multiple of 3 is added to 2n-1 that will also be divisible by 3

2n-1 +3 + 3

=2n-1 +2×3

=2n+5

So 2n+5 is divisible by 3

Case 2 (b) Remainder 2 implies 2n-2 is divisible by 3.

If multiple of 3 is added to 2n-2 that will also be divisible by 3

2n-2 +3

=2n-2 +1×3

=2n+1

So 2n+1 is divisible by 3

So from all the above case it os verified that out of 3 consecutive odd number one will always be divisible by 3.

Answer 2

Step-by-step explanation:

Let 3 consecutive positive integers be n, n + 1 and n + 2 .

Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2.

:

Therefore:

n = 3p or 3p+1 or 3p+2, where p is some integer

If n = 3p = 3(p) , then n is divisible by 3

If n = 3p + 1, then n + 2 = 3p +1 + 2 = 3 p + 3 = 3 ( p + 1 ) is divisible by 3

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3

Thus, we can state that one of the numbers among n, n+1 and n+2 is always divisible by 3

Hence it is solved.

THANKS

#BeBrainly.