solution:-Let a and b are any two positive integer
then a=n and b=3
now, according to Euclid division algorithm
a=bq+r ,where 0≤r<b
n=3q+r ,where 0≤r<3 ...........(A)
value of r=0,1,2 (because given range is called value r is equal to 0 , 1 and but not equal to 3 )
so we can take in three case
for case-1 take r=0
(A)⇒ n=3q (n is divisible by 3)
added both side 2 in LHS and RHS
n+2=3q+2 (n+2 is not divisible by 3)
again,added both side 4 in LHS and RHS
n+4=3q+4 (n+4 is not divisible by 3)
we can see that one and only one out of n,n+2 and n+4 is divisible by 3
for case-2 take r=1
(A)⇒ n=3q+1 (n is not divisible by 3)
added both side 2 in LHS and RHS
n+2=3q+1+2
n+2=3q+3
n+2=3(q+1) (n+2 is divisible by 3)
again,added both side 4 in LHS and RHS
n+4=3q+1+4
n+4=3q+5 (n+4 is not divisible by 3)
we can see that one and only one out of n,n+2 and n+4 is divisible by 3
for case-3 take r=2
(A)⇒ n=3q+2 (n is not divisible by 3)
added both side 2 in LHS and RHS
n+2=3q+2+2
n+2=3q+4 (n+2 is not divisible by 3)
again,added both side 4 in LHS and RHS
n+4=3q+2+4
n+4=3q+6
n+4=3(q+2) (n+4 is divisible by 3)
we can see that one and only one out of n,n+2 and n+4 is divisible by 3
hence proved in three case and we see each case one and only one out of n n+2 n+4 is divisible by 3 . ans
