Question
By direct proof, how can you prove that “The sum of any three consecutive even integers is always a multiple of 6”.
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Answer to a math question By direct proof, how can you prove that “The sum of any three consecutive even integers is always a multiple of 6”.
Esmeralda
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Let’s denote the three consecutive even integers as n, n+2, and n+4.
The sum of these three integers is n + (n+2) + (n+4) = 3n + 6.
We can factor out a 3 from this expression to get 3(n + 2).
Since n is an even integer, n + 2 is also an even integer. Therefore, the expression 3(n + 2) is a multiple of 6 because it’s the product of 3 and an even integer.
Hence, we have proved that the sum of any three consecutive even integers is always a multiple of 6.
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