Question

Show that one and only o e out of n,n+2,n+4 is divisible by 3,where n is any positive integer.

Answer 1

Sol : 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

Answer 2

Answer:

Step-by-step explanation:

Let X be any positive integer

and divisor (B)=3

then,by the Euclid division Lemma,

there exist two positive integer q and r such that,

x = 3q+r,r is greater than or equal to 0 and smaller than 3

where q is the questiont and r is the remainder

Then,possible remainders are= 0 or 1 or 2

Now when r= 0

x = 2q+0

x=2q ,2q=n

So,x=n

When r=1

x=2q+1

x=1(2q+1),n=2q+1

So,x=n

When r=2

x=2q+2,2q=n

So,x=n+2