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Prove the sum of three consecutive integers is a multiple of 3

Step by step, with the reason for every step · Direct proof · Discrete Math

Step by step

The question

Question: prove that the sum of three consecutive integers is a multiple of 3.

1 Write it with letters

Call the first number n, a whole number.

The next integers are 1 more each time: n + 1, n + 2.

the numbers = n, n + 1, n + 2

Careful!

An example shows one number.

The letters show every number at once.

Careful: one example is not a proof

2 Put them in

Add them up.

sum = n + (n + 1) + (n + 2)

3 Work it out

Open the brackets: n + n + 1 + n + 2.

= n + n + 1 + n + 2

Like terms together: 3n + 3.

= 3n + 3

4 Show the form

Every term has a 3 in it.

We take out 3: 3(n + 1).

= 3(n + 1)

n + 1 is a whole number, because whole numbers added and multiplied stay whole.

n + 1 is a whole number

3 times a whole number: so the sum is a multiple of 3.

So the sum is a multiple of 3. ∎

The square at the end means: the proof is done.

5 Check

An example: 3, 4, 5.

It gives 12.

3 + 4 + 5 = 12

12 is 3 times 4: a multiple of 3, as the proof says.

Right!

12 = 3 × 4

Let's look back.

First: write it with letters.

Then: put them in.

Then: work it out.

Then: show the form.

Last: check.

Answer
n + 1 is a whole number

Your turn

Try this one yourself, then check:

Show the answer
sum = 7(n + 3)

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