Question

show that of the number n, n+2 ,n+4 only one of them is divisible by 3.​

Answer 1

Step-by-step explanation:

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

CASE-1

IF N=3q

n+4=3q+4

n+2=3q+2

here n is only divisible by 3

CASE 2

if n = 3q+1

n+4=3q+5

n+2=3q=3

here only n+2 is divisible by 3

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

Plz mark as brainliest

Answer 2

:

Mark Me As Brainliest And Thank Me If The Answer Is Useful.

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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.