## Multiples of 3 and Consecutive Numbers

Numbers that are multiples of 3 are 3, 6, 9, 12, ….

###### Problem

Prove that numbers that are multiples of 3 can be expressed as sum of three consecutive numbers.

###### Solution (2 out of 5 marks)

3 = 0 + 1 + 2

6 = 1 + 2 +3

9 = 2 + 3 + 4

12 = 3 + 4 + 5

15 = 4 + 5 + 6

Comment: Generating examples means you understood the statement that multiples of 3 can always be expressed as sum of consecutive numbers. That’s why you get 2 marks. However, even if you listed hundreds of multiples of 3  and expressed each as sum of consecutive numbers you will still get 2 out of 5 marks. Showing that you understood the statement is not the same as proving it. Same with generating cases unless you can list all of them. I might consider giving an additional half-mark if you can list a million of them:-).

###### Solution (full mark)

Let n, n + 1, n + 2 be the three consecutive numbers, n is an integer

Sum is n + n + 1 + n + 2 = 3n + 3 = 3 (n + 1)

Since 3 is a factor in 3(n + 1), 3(n + 1) it is a multiple of 3.

###### Something to think about:

Can we always express a number as sum of at least two consecutive numbers?

No.

###### Proof?

The number 4.

Comment: One counterexample is enough to disprove a statement.

## 4 thoughts on “Multiples of 3 and Consecutive Numbers”

1. The one who can lists ten million of consecutive examples, showing thus he/she understands the statement, is then granted one additional half a mark… If I understand the statement 😉

f(w) = 2 + (1/2) w(v), with ‘v’ being a non discrete variable, and w(v) = [1, 2, 3] when v supegal [ (10)^6; (10)^7; (10)^8 ].

Right, Mistress?

2. Upper-written formula does NOT help at knowing whether a given integer is or isn’t a muliple of 3.

The work up here states that the sum of three consecutive integers is a multiple of the integer 3. Good job! But [I think] it does not help backwards – when the sum is the given integer to analyse – at surveying @ its divisibility (or not) by 3.

3. aisha says:

Sum is n + n + 1 + n + 3 = 3n + 3 = 3 (n + 1)

—–
surely the answer to the first part of this sum should be 3n + 4, as you have 3 n’s and 1+3 = 4 ?

correct me and explain to me if i am wrong. it may just be a typo on your part. Please explain.

4. It’s a typo. It should be n + n + 1 + n + 2 = 3n + 3 = 3 (n + 1). Thanks. I will correct it.