Question

7. show that one and only
one out of
m, hall, (n+2) is divisible by 3.
where s' is
itive
1
any posi
integer​

Answer 1

Step-by-step explanation:

Then, n = 3q +r , where<br><br>m<br> Case I If n = 3q then n is clearly divisible by 3. <br> Case II If n= ( 3q +1) then ( n+2) = (3q +3) = 3(q +1) , which is clearly divisible by 3. <br> In this case, ( n+2) is divisble 3. <br> Case III If n = ( 3n +2) then (n+1) = ( 3q +3) = 3(q +1) , whihc is clearly divisible by 3. <br> In this case, (n+2) is divsible by3 . <br> In this case , (n +1) is divisible by3. <br> Hence, one and only one out of n,( n +1) and (n+2) is divisible by 3.

Answer 2

ANSWER

Since n, n+1, n+2 are three consecutive integers then there must be one number divisible by 3 at least.

If the remainder at dividing n by 3 is 1, then n+2 must be divisible by 3 and if the remainder at dividing n by 3 is 2, then n+1 must be divisible by 3. Similarly for n+1 and n+2.

Let n be divisible by 3.

3

n+1

=

3

n

+

3

1

Now, n is divisible by 3 but 1 is not. So we get n+1 not divisible by 3. Similarly,n+2 will not be divisible by 3 as well if n is divisible by 3.

3

n+2

=

3

n

+

3

2

In the same way, if n+1 is divisible by 3 then n and n+2 can't be divisible by 3. If n+2 is divisible by 3 then n and n+1 cannot be divisible by 3.