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

Explanation:

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

Case I : When n = 3q

In this case, we have,

n=3q, which is divisible by 3

n=3q

= adding 2 on both sides

n + 2 = 3q + 2

n + 2 leaves a remainder 2 when divided by 3

Therefore, n + 2 is not divisible by 3

n = 3q

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

n + 4 leaves a 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 a reaminder 1 when divided by 3

n is not divisible by 3

n = 3q + 1

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

n + 2 is divisible by 3

n = 3q + 1

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

n + 4 leaves a 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

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 divsible by 3

n = 3q + 2

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

n + 4 is divisible by 3

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

Answer 2

Answer:

let n be any positive integer and b=3

n =3q+r

where q is the quotient and r is the remainder

0_ <r<3

so the remainders may be 0,1 and 2

so n may be in the form of 3q, 3q=1,3q+2

$\underline{CASE-1}$

IF N=3q

n+4=3q+4

n+2=3q+2

here n is only divisible by 3

$\underline{CASE-2}$

if n = 3q+1

n+4=3q+5

n+2=3q=3

here only n+2 is divisible by 3

$\underline{CASE-3}$

IF n=3q+2

n+2=3q+4

n+4=3q+2+4

=3q+6

here only n+4 is divisible by 3

HENCE IT IS JUSTIFIED THAT ONE AND ONLY ONE AMONG n,n+2,n+4 IS DIVISIBLE BY 3 IN EACH CASE