prove that one of any 3 consecutive positive integers must be divisible by 3
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prove that one of any 3 consecutive positive integers must be divisible by 3
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Step-by-step explanation:
Let three consecutive positive integers be n, =n + 1 and n + 2
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 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
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
Answer:
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Let three consecutive positive integers be n, =n + 1 and n + 2
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 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
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.