Question

Prove that own and only one out of any three consecutive positive integers is divisible by 3.

Answer 1

Hint: To prove the given question we need to know the definition of consecutive numbers or integers and ordered triplet, also Euclid’s division lemma which is given by \[a = bq + r\] where \[0 \leqslant r \leqslant b\] .

Complete step-by-step answer:
Let any three consecutive positive integers is of the form \[n,n + 1,n + 2\] where n is any natural number that is 1, 2, 3 and so on
We know that from division lemma
[ division lemma states that if two positive integers \['a'and'b'\], then there exists a unique integer \['q'and'r'\] such that which satisfies the condition \[a = bq + r\] where \[0 \leqslant r \leqslant b\]].
\[n = 3q + r\], \[r = 0,1,2\]
Let us consider three cases that is
Case (1) \[r = 0\]
\[ \Rightarrow \]\[n = 3q\]
\[ \Rightarrow \]\[n + 1 = 3q + 1\]
\[ \Rightarrow \]\[n + 2 = 3q + 2\]
Here \[n = 3q\] , n is divisible by 3 but it is not divisible by \[n + 1\] and \[n + 2\]
Case (2) \[r = 1\]
\[ \Rightarrow \]\[n = 3q + 1\]
\[ \Rightarrow \]\[n + 1 = 3q + 2\]
\[ \Rightarrow \]\[n + 2 = 3q + 3\]
Here \[n + 2 = 3q + 3\]is divisible by 3 but it is not divisible by\[n\]and \[n + 1\]
Case (3) \[r = 2\]
\[ \Rightarrow \]\[n = 3q + 2\]
\[ \Rightarrow \]\[n + 1 = 3q + 3\]
\[ \Rightarrow \]\[n + 2 = 3q + 4\]
Here \[n + 1 = 3q + 3\]
\[ \Rightarrow \]\[3(q + 1)\]
\[\therefore \]\[n + 1\] is divisible by 3 but it is not divisible by \[n\]and \[n + 2\]
Hence one among the \[n,n + 1,n + 2\]is divisible by 3
For example,
Consider the ordered triplet \[(a,b,c)\]=\[(n,n + 1,n + 2)\]
At n=1, \[(a,b,c) = (1,2,3)\]
At n=2 \[(a,b,c) = (2,3,4)\]
At n=3 \[(a,b,c) = (3,4,5)\]
This process continues. As we observe that each triplet consists of a number that is divisible by 3.
Hence any one of the consecutive positive integers is divisible by 3.

Note: Consecutive integers are those numbers that follow each other. They follow in a sequence or in order.
Consecutive means unbroken sequence
In consecutive integers follow a sequence such a way that each subsequent number is one more than the previous number.