Question

Show that one and only one out of n, n + 3, n + 6 or n + 9 is divisible by 4.​

Answer 1

Answers

Let n be any positive integer and b=4.then by euclid's division lemma,x=4q+r.

positive remainders are 0,1,2and 3,since 0<=r<4.then x can be 4q/4q+1/4q+2/4q+3.

1. n=4q,it is divisible by 4

2. n=4q+1, n+3=4q+1+3=4q+4

=4(q+1),it is divisible by 4

3. n=4q+2, n+6=4q+2+6

=4q+8

=4(q+2),it is divisible by 4

4. n=4q+3, n+9=4q+3+9

=4q+12

=4(q+3) ,it is divisible by 4

therefore one and only one out of n,n+1,n+2,n+3 is divisible by 4

i hope it will help you

Answer 2

$\bf\large\underline\green{Answer:-}$

Let n be any positive integer and b=4.then by euclid's division lemma,x=4q+r.

positive remainders are 0,1,2and 3,since 0<=r<4.then x can be 4q/4q+1/4q+2/4q+3.

1. n=4q,it is divisible by 4

2. n=4q+1, n+3=4q+1+3=4q+4

=4(q+1),it is divisible by 4

3. n=4q+2, n+6=4q+2+6

=4q+8

=4(q+2),it is divisible by 4

4. n=4q+3, n+9=4q+3+9

=4q+12

=4(q+3) ,it is divisible by 4

therefore one and only one out of n,n+1,n+2,n+3 is divisible by 4