Question

Show that one of every 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.