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

by Euclid's division Lemma any natural number can be written as:   .

where r = 0, 1, 2,.........(a-1). and q is the quotient.

put a = 3: b = 3q+r and r = 0,1,2.

thus any number is in the form of 3q , 3q+1 or 3q+2.

case I: if n =3q

n is divisible by 3,

n+2 = 3q+2 is not divisible by 3.

n+4 = 3q+4 = 3(q+1)+1 is not divisible by 3.

case II: if n =3q+1

n = 3q+1 is not divisible by 3.

n+2 = 3q+1+2=3q+3 = 3(q+1) is divisible by 3.

n+4 = 3q+1+4 = 3q+5 = 3(q+1)+2 is not divisible by 3.

case III:  if n = 3q+2

n =3q+2 is not divisible by 3.

n+2 = 3q+2+2 =3q+4 = 3(q+1)+1 is not 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.

hope this helps you.

cheers!!

• 89

we know that any natural no. is in the form of 3q, 3q+1and 3q+2

if n=3q

=> n is divisible by 3 and the others are not.

similarly, if n=3q+1

only n+2 is divisible by 3 and the others are not

similarly, if n=3q+2

only n+4 is divisible by 3 and the others are not.

hence it is proved

QED

Hope this helps

So thumbs up pls :D

• 1

Let n=3k+a where a=0,1,2

now, if a=0

3k, 3k+2 or 3k+4

If a=1

3k+1, 3k+3 or 3k+5

If a=2

3k+2 , 3k+4 or 3k+6

Now, here number which are divisible by 3 are, 3k , 3k+3 or 3(k+1) and 3k+6 or 3(k+2)

Hence proved that in every situation only one number is divisible by 3.

Hope it helps

• 0
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