Prove that one of every three consecutive integers is divisible by 3.
Answers
Dear Student, Let the first positive integer be x. So, the next two numbers are x+1 and x+2. Using Euclid's Division lemma, x when divided by 3 can leave either 0, 1 or 2 are remainder. Let r be the remainder left by x when divided by 3. (r = 0 or 1 or 2) => r+1 is the remainder left by x+1 when divided by 3. and r+2 is the remainder left by x+2 when divided by 3. Now, if x is divisible by 3 i.e. r = 0, we are done with the proof that one of the three consecutive positive integers is divisible by 3. However, if r = 1, then x+2 leaves 3 as remainder when divided by 3. But as 3 is itself divisible by 3 => x+2 is divisible by 3. similarly, if r = 2, then x+1 leaves 3 as remainder when divided by 3. But as 3 is itself divisible by 3 => x+1 is divisible by 3. Therefore, one of the three consecutive positive integers is divisible by 3.
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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.