Question

show that of the numbers n,n+2,&n+4,only one of them is divisible by 3

Answer 1

We applied Euclid Division algorithm on n and 3.
a = bq +r  on putting a = n and b = 3
n = 3q +r  , 0<r<3
i.e n = 3q   -------- (1),n = 3q +1 --------- (2), n = 3q +2  -----------(3)
n = 3q is divisible by 3
or n +2  = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.

Answer 2

HEY FRIEND

HERE IS YOUR ANSWER

===============================

LET n=3k,3k+1 and 3k+2

WHEN n=3k
__________

n=3k
n is divisible by 3

n+2=3k+2
n+2 is not divisible by 3

n+4=3k+4
n+4 is also not divisible by 3

WHEN n=3k+1
___________

n=3k+1
n is not divisible by 3

n+2=3k+1+2
n+2=3k+3
n+2=3(k+1)
n+2 is divisible by 3

n+4=3k+1+4
n+4=3k+5
n+4 is not divisible by 3

WHEN n=3k+2
___________

n=3k+2
n is not divisible by 3

n+2=3k+2+2
n+2=3k+4
n+2 is not divisible by 3

n+4=3k+2+4
n+4=3k+6
n+4=3(k+2)
n+4 is divisible by 3


SO ONLY OF THE NUMBERS AMONG n, n+2 and n+3 is divisible by 3.

==============================

HOPE THIS HELPS YOU