Question

show that one and only one out of n,(n+1)and(n+3)is divisible by 3,where n is any positive integer​

Answer 1

Answer:

Step-by-step explanation:

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

Answer:

Step-by-step explanation:let a be any rational number ie,3q,3q+1 and 3q+2

Case 1=when n=3q

=>3(q)=3m,where m=q

Ie,divisible by 3

n+1=3q+1

=>3(q)+1=3m+1

Where m=q

ie,not divisible by 3

n+2=3q+2

=3(q)+2=3m+2

Where m=q

ie,not divisible by 3

Case 2= when n=3q+1

3(q)+1

Not divisible by 3

n+1= 3q+1+1= 3q+2

Not divisible by 3

3q+2+1=3q+3

Divisible by 3

Do the same for case 3,ie,n=3q+2

Hence,proved!