Question

Show that one and only one out of n, n+3, n+6 and n+9 is divisible by 4 for some positive integer n.​

Answer 1

Step-by-step explanation:

Apply Euclid Division algorithm on n and 4.

a=bq+r on putting a=n and b=4

n=4q+r , 0<r<4

i.e n=4q or

or n=4q+1

or n=4q+2

or n=4q+3

Positive remainders are 0,1,2, and 3 because r varies from 0 to 3

n=4q is divisible by 4

or n+3=4q+1+3=4q+4 also divisible by 4

or n+6=4q+2+6=4q+8 is also divisible by 4

or n+9=4q+2+9=4q+12 is also divisible by 4

Hence n,n+3,n+6,n+9 are divisible by 4.

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