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Asked by | 11 Mar, 2009, 12:21: PM

We know that themultiples of 3 will differ by 3.

Also every multiple of 3 is either an odd number or an even number.

Suppose

out of the given numbers

n,n+2,n+4

If n is a multiple of 3, there's nothing more to prove.

So, let's assume that  n is not a multiple of 3.

Then n-3 is not a multiple of 3, so

either n-1, or n-2 must be a multiple of 3.

So next multiple of 3 would be either (n-1)+3= n+2

or

(n-2)+3 =n+1

So we see that if n is not a multiple of 3 then n+2 may be a multiple of 3 or n+1 may be a multiple of 3

Next,

if n+1 is a multiple of 3 , then (n+1)+3 must be a multiple of 3

i.e. n+4 must be a multiple of 3.

So we see that if n is not a multiple of 3 then either n+2 or n+4 is a multiple of 3.

Thus either of n, n+2 or n+4 is always a multiple of 3.

i.e. one out of n,n+2 or n+4 is divisible by 3.

Answered by | 11 Mar, 2009, 01:20: PM

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