Solution:

let n be any positive integer and b=3
n =3q+r
where q is the quotient and r is the remainder
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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.

Introduction:
We are given three numbers: n, n^2, and n^4, where n is any positive integer. We need to show that only one out of these three numbers is divisible by 3.

Proof:
To prove that only one out of the three numbers is divisible by 3, we will consider two cases:

Case 1: n is divisible by 3.
If n is divisible by 3, it can be written as n = 3k, where k is an integer.

Subcase 1.1: n^2 is divisible by 3.
In this case, substituting n = 3k into n^2, we get:
n^2 = (3k)^2 = 9k^2

Since 9 is divisible by 3, n^2 is also divisible by 3.

Subcase 1.2: n^4 is divisible by 3.
In this case, substituting n = 3k into n^4, we get:
n^4 = (3k)^4 = 81k^4

Again, since 81 is divisible by 3, n^4 is divisible by 3.

Thus, if n is divisible by 3, both n^2 and n^4 are divisible by 3.

Case 2: n is not divisible by 3.
If n is not divisible by 3, it can be written as n = 3k + r, where k is an integer and r is the remainder when n is divided by 3. Since n is a positive integer, r can only be 1 or 2.

Subcase 2.1: n^2 is divisible by 3.
In this case, substituting n = 3k + r into n^2, we get:
n^2 = (3k + r)^2 = 9k^2 + 6kr + r^2

The first two terms, 9k^2 and 6kr, are divisible by 3 since they contain a factor of 3. Therefore, for n^2 to be divisible by 3, r^2 must also be divisible by 3.

Since r can only be 1 or 2, r^2 can only be 1 or 4, neither of which is divisible by 3. Hence, n^2 is not divisible by 3.

Subcase 2.2: n^4 is divisible by 3.
In this case, substituting n = 3k + r into n^4, we get:
n^4 = (3k + r)^4

Expanding the above expression using the binomial theorem, we get:
n^4 = 81k^4 + 108k^3r + 54k^2r^2 + 12kr^3 + r^4

The first term, 81k^4, is divisible by 3 since it contains a factor of 3. Similarly, the second and fourth terms, 108k^3r and 12kr^3, are also divisible by 3 as they contain a factor of 3. However, the third term, 54k^