prove that any Number Composed of three Consecutive Digits must be divisible by three

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To prove that any number composed of three consecutive digits must be divisible by three, we can use the following argument:

Let the number be represented by the digits a, b, and c, where a is the hundreds digit, b is the tens digit, and c is the ones digit. Then, we can express the number as 100a + 10b + c.

We can then factor out a 3 from this expression as follows:

100a + 10b + c = 99a + a + 9b + b + c
= 3(33a + 3b) + (a + 3b + c)

Since 33a + 3b is an integer and a + 3b + c is the sum of three consecutive digits, it must be an integer between 3 and 27 inclusive. Therefore, (a + 3b + c) is equal to one of the integers 3, 6, 9, 12, 15, 18, 21, 24, or 27.

Since (a + 3b + c) is divisible by 3, we can conclude that 100a + 10b + c is also divisible by 3. Therefore, any number composed of three consecutive digits must be divisible by three.