Question

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

Answer 1

Answer:

how answer of 4 HCF divisibility by 408

Step-by-step explanation:

748, 408, 512

Answer 2

SOLUTION:-

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

CASE 1: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

=n+2 is not divisible 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 2: When n=3q+1: in this case,we have

n=3q+1

=n leaves remainder 1 when divided by 3

=n is not 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 3: 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

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

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