Prove that one of every three consecutive integers is divisible by 3.

Hint: Consecutive integers means integers one after another. Integers here can be either positive or negative. Here we will consider this consecutive integers we will check their divisibility by 3 and then we will prove our statement

Complete step-by-step answer:
Suppose we have three consecutive integers as \[a,{\text{ }}a + 1,{\text{ }}a + 2\]
We know by theorem that integers are of the form \[3q,{\text{ }}3q + 1{\text{ }},{\text{ }}3q + 2\]
Now, we have the following cases
Case I –
Suppose n= 3q
Here n is divisible by 3 but other terms \[n + {\text{ }}1\] and \[n + {\text{ }}2\] are not divisible by 3
Case II –
Suppose n = 3q+1
In this case \[n + 2\] will become \[3q + 1 + 2{\text{ }} = {\text{ }}3q + 3\] is divisible by 3
It is observed that the other two terms $n$ and \[n + 1\] are not divisible by 3.
Case III –
Suppose n = 3q+2
Now replacing $n$ by \[n = {\text{ }}3q + {\text{ }}2\]
We get n+1 = 3q+ 2+1 = 3q+3 which is divisible by 3
Another term \[n + 2\] and $n$ are not divisible by 3
Therefore after considering the above three cases we can conclude that out of every three consecutive integer numbers, one of them is exactly divisible by 3 and the rest of two are not.

Note: Above question can be solved using directly any three consecutive integers like 6, 7, 8.
For checking divisibility of any number by 3 simply add all digits and see if the sum of digits is completely divisible by 3 or not .
Example – 81 sum of digits will be 8+1=9 and 9 is completely divisible by 3.
Here also you can notice 6 is divisible by 3 whereas 7 and 8 are not.
If there were three consecutive odd integers it would be n,n+2, n+4 but if there it is asked about consecutive even integers then they will be also n,n+2, n+4 .