Question

Show that 1 and only 1 out of n, n+1 and n+2 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

HEYA !!
HERE'S THE ANSWER ⏬

By using Euclid's Division Algorithm :

a = bq + r
For a = n and b = 3, we have,. { r= 0 , 1 , 2 }
n = 3q + r-------- (i)

Putting r = 0 in (i), we get 
n = 3q 
=>, n is divisible by 3. 

n + 2 = 3q + 2 
=> , n+2 is not divisible by 3

Now , n+ 4 = 3q + 4
again , n + 4 isn't divisible by 3

Putting r = 1 in (i), we get 
n = 3q + 1 
=>, n is not divisible by 3. 

n + 2 = 3q + 3 = 3(q + 1) 
=>, n + 2 is divisible by 3. 

n + 4 = 3q + 5
=> n+ 4 is not divisible by 3 .

Putting r = 2 in (i), we get 
n = 3q + 2 
=> n is not divisible by 3. 

n + 2 = 3q + 4 
=> n + 2 is not divisible by 3. 

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

Hence , it's true that only one out of n , n+2 , n+ 4 is divisible by 3 .

HOPE IT HELPS YOU ✌️!!