Question

show that one and only one out of n, n+3, n+6, n+9 is divisible by 4

Answer 1

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

According to EUCLID's division Lemma,
we know, any number a is in the form of
bq + r where 0 ≤ r < b.
e.g a = bq + r ; where 0 ≤ r < b .

Let here n is divisible by 4 .
e.g n = 4q + 0 { we know, any number divisible by other numbers when remainder equals zero }

n = 4q => it is divisible by 4
n + 3 = 4q + 3 , it isn't divisible by 4
n + 6 = 4q + 4 +2 = 4(q +1) +2 , it isn't divisble by 4 .
n+9 = 4q + 9 = 4q + 8 +1 = 4(q +2) +1 , it isn't divisible by 4.

hence, if we let n is divisible by 4 then rest are not divisible by 4 . it means one and only one out of n , n +3 , n + 6, and n+9 is divisible by 4 .

hence proved//