Prove that one of any three consecutive integers is divisible by 3
Answers
Answer: just start from 3. Add three to it. We get another number that is divisible by 3 (that is, 6). Start from 4 and and three. We get 7, and 6 is included in the intervening numbers. Take 5 and aad 3. We get 8. We see that 6 is present in these intervening numbers too. The next number is 6 itself. This way, your given condition is proved.
Step-by-step explanation:
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.