Question

prove that one of every three cosecutive integer is diviseble by 3

Answer 1

Assume the three consecutive integers as :-
                
      → n , n + 1 and n + 2 respectively 

⇒ If suppose any number is divided by 3 , then it's remainder will be either 0 ,1 or 2 .

⇒ According to Euclid's algorithm-
     
            $a= bq+r$ { where ''a'' is n , ''b''  is 3 and ''r'' will be less than 3

Hence , for the three consecutive integers we get as follows - 

⇒$n=3p+0 = 3p$  

⇒$n=3p+1$  

⇒$n=3p+2$

→ If suppose n = 3 p - Then
 
                ''n'' will be divisible by 3  
    
→ If suppose n = 3 p + 1 - Then ,
 
               $n+2= 3p +1 +2$
    
                           =  $3p + 3$
          
                           = $3 ( p + 3)$ → Which is divisible by 3

→If suppose n = 3 p + 2 , Then ,
 
             $n+1= 3p+2+1$
 
                         = $3p+3$
 
                         = $3(p+1)$ → Which is divisible by 3

⇒ Hence it is proved that one of every three consecutive integer ( i.e n , n + 1 ,     n + 2 ) is divisible by 3.

Answer 2

Step-by-step explanation:

Let 3 consecutive positive integers be n, n + 1 and n + 2 .

Whenever a number is divided by 3, the remainder we get is either 0, or 1, or 2.

:

Therefore:

n = 3p or 3p+1 or 3p+2, where p is some integer

If n = 3p = 3(p) , then n is divisible by 3

If n = 3p + 1, then n + 2 = 3p +1 + 2 = 3 p + 3 = 3 ( p + 1 ) is divisible by 3

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3

Thus, we can state that one of the numbers among n, n+1 and n+2 is always divisible by 3

Hence it is solved.

THANKS

#BeBrainly.