Question

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



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Answer 1

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We know that any positive integer of the form 3q or, 3q+1 or 3q+2 for some integer q and one and only one of these possibilities can occur.

So, we have following cases:

Case-I: When n=3q

In this case, we have

n=3q, which is divisible by 3

Now, n=3q

n+2=3q+2

n+2 leaves remainder 2 when divided by 3

Again, n=3q

n+4=3q+4=3(q+1)+1

n+4 leaves remainder 1 when divided by 3

n+4 is not divisible by 3.

Thus, n is divisible by 3 but n+2 and n+4 are not divisible by 3.

Case-II: when n=3q+1

In this case, we have

n=3q+1,

n leaves remainder 1 when divided by 3.

n is divisible by 3

Now, n=3q+1

n+2=(3q+1)+2=3(q+1)

n+2 is divisible by 3.

Again, n=3q+1

n+4=3q+1+4=3q+5=3(q+1)+2

n+4 leaves remainder 2 when divided by 3

n+4 is not divisible by 3.

Thus, n+2 is divisible by 3 but n and n+4 are not divisible by 3.

Case-III: When n=3q+2

In this case, we have

n=3q+2

n leaves remainder 2 when divided by 3.

n is not divisible by 3.

Now, n=3q+2

n+2=3q+2+2=3(q+1)+1

n+2 leaves remainder 1 when divided by 3

n+2 is not divisible by 3.

Again, n=3q+2

n+4=3q+2+4=3(q+2)

n+4 is divisible by 3.

Hence, n+4 is divisible by 3 but n and n+2 are not divisible by 3.

Answer 2

Answer:

thus any number is in the form of 3q , 3q+1 or 3q+2. n is divisible by 3, ... n+4 = 3q+2+4 = 3q+6 = 3(q+2) is divisible by 3. thus one and only one out of n , n+2, n+4 is divisible by 3.31