Say whether the following is true or false and support your answer by a proof: The sum of any five consecutive integers is divisible by 5 (without remainder)?

1 Answer
Aug 25, 2017

See a solution process below:

Explanation:

The sum of any 5 consecutive integers is, in fact, evenly divisible by 5!

To show this let's call the first integer: n

Then, the next four integers will be:

n + 1, n + 2, n + 3 and n + 4

Adding these five integers together gives:

n + n + 1 + n + 2 + n + 3 + n + 4 =>

n + n + n + n + n + 1 + 2 + 3 + 4 =>

1n + 1n + 1n + 1n + 1n + 1 + 2 + 3 + 4 =>

(1 + 1 + 1 + 1 + 1)n + (1 + 2 + 3 + 4) =>

5n + 10 =>

5n + (5 xx 2) =>

5(n + 2)

If we divide this sum of any 5 consecutive integers by color(red)(5) we get:

(5(n + 2))/color(red)(5) =>

(color(red)(cancel(color(black)(5)))(n + 2))/cancel(color(red)(5)) =>

n + 2

Because n was originally defined as an integer n + 2 is also an integer.

Therefore, the sum of any five consecutive integers is evenly divisible by 5 and the result is an integer with no remainder.