Question

prove that one of every three consecutive positive integers is divisible by 3​

Answer 1

Answer:

Let three consecutive positive integers be n, n + 1 and n + 2. ... 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.

Step-by-step explanation:

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Answer 2

Answer:

LET'S CHECK:

$\frac{1 + 2 + 3}{3} = \frac{6}{3} = 2 \\ \\ \\ \frac{4 + 5 + 6}{3} = \frac{15}{3} = 3 \\ \\ \\ \frac{7 + 8 + 9}{3} = \frac{24}{3} = 8$

YES SUM EVERY POSITIVE CONSECUTIVE INTEGER IS DIVISIBLE BY 3.

OR

LIKE 1,2,3 3 IS DIVISIBLE BY 3

4,5,6 6 IS DIVISIBLE BY 3.