Question

show that one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer.

Answer 1

On dividing n by 3,let q be the quotient and r be the remainder.
By euclid's division lemma,
n=3q+r where 0≤r<3
when,r=0
than n=3q+0=3q
here n is clearly divisible by 3
when,r=1
than n=3q+1
than n+2=3q+1+2=3q+3=3(q+1)
here n+2 is clearly divisible by 3
when,r=2
than n=3q+2
than n+4=3q+2+4=3q+6=3(q+2)
here n+4 is clearly divisible by 3
Therefore one and only one out of n,n+2,n+4 is divisible by 3,where n is any positive integer....
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Answer 2

If n is any positive integer, then it is either odd it even .

Case 1. When n is odd.

n = 2a + 1

For some whole number a.

The three numbers would be:

(2a + 1),(2a + 3), and (2a + 5).

When each of these is divided by 3 there can be three remainders:

1,2 and 0.

If 2a is a multiple of 3, then 3 divides n+ 4 but doesn't divides any of the rest of the numbers.

If 2a is not a multiple then either the remainder is 1 or 2.

If the remainder is 1, n is divisible by 3. The rest aren't.

If the remainder is 2 then n+4 is divisible but the rest aren't.

Case 2. When n is even.


n = 2y for some natural number y.

Then the numbers :

2y, 2y + 2 and 2y + 4.

If 2y is a multiple of 3 then n is divisible by 3 but the rest aren't.

If 2y is not a multiple of 3, then either the remainder is 1 or 2.

If the remainder is 1, then n+2 is divisible by three but the rest aren't.

If the remainder is 2 then n+4 is divisible by 3 but the rest aren't.