Multiples of 9 and 3 with Cuisenaire rods
By: Alice Germain
Tags: Stem Education, Teaching Resources, Mathematics Education, Mathematics, STEM

You may have learnt at some point that multiples of 9 and 3 can be easily identified by checking whether the sum of their digits is itself a multiple of 9 and 3, respectively. But have you learnt why it is so? I, personally, didn’t. But it is a nice thing to look at, and it can be shown first with Cuisenaire rods before using algebra to generalise the observed pattern.

Let’s start by looking at the 5 first multiples of 9 with Cuisenaire rods in a 100-square, which is very useful when we want to focus on place value. We observe that adding 9 means increasing the number of tens by 1 and decreasing the number of units (or ones) by 1.

Multiples of 9

In the second row of 100-squares, the same numbers are shown, but this time thinking about these numbers as multiples of 9, that is, by placing a 9-rod wherever possible. We observe that for 18, we have a 9-rod and a 1-rod in the first column. We could do the same for the two full columns of 27, or, instead of two 1-rods, place a 2-rod. In that way, we see that the rod placed horizontally on top of the full columns is the number of tens. We see that for the whole number to be a multiple of 9, we need the number of tens (the horizontal rod) plus the number of units (the rod in the last column) to be a multiple of 9.

We can apply the same reasoning for the multiples of 3.

Consider the numbers 12, 24, 39, and 48 in a 100-square. First, we split them into tens and units as shown in the first row of 100-squares, and, in the second row, we place three 3-rods in each full column. Finally, in the third row, we show how the whole number is a multiple of 3 because the sum of the number of tens (given by the rod placed horizontally on top of the full tens columns) and the number of units is a multiple of 3.

Multiples of 3

What about multiples of 6?

The rule that the sum of the digits must be a multiple of 6 won’t work here as 10 cannot be split into a multiple of 6 + 1. However, as 6 is 2 × 3, a multiple of 6 is both a multiple of 3 and a multiple of 2. In other words, they are even multiples of 3 — so, the sum of their digits is a multiple of 3 and the last digit is 0, 2, 4, 6, or 8.

Linking with Algebra

These properties can be demonstrated with algebra, translating with equations the pattern we have observed here.

For the multiples of 9, we have used the fact that each ten is made of 9 + 1. Considering a two-digit number written ‘tu’ where t and u are digits (i.e., whole numbers between 0 and 9), we have ‘tu’=10t+u. We now replace 10 by 9 + 1, which gives ‘tu’=(9+1)t+u=9t+t+u. If t+u, the sum of the digits of the number ‘tu’, is a multiple of 9, then it can be written as t+u=9k, where k is a whole number. In that case, we can write

‘tu’=9t+9k=9(t+k),

which shows that ‘tu’ is a multiple of 9 since t+k , as the sum of two whole numbers, is a whole number itself.

For the multiples of 3, we have replaced each ten by three 3-rods plus 1 unit. So, we have ‘tu’=10t + u = (3 × 3+1)t + u = 3 × 3 t + t + u. If t+u, the sum of the digits of the number ‘tu’, is a multiple of 3, then it can be written as t+u=3k, where k is a whole number. In that case, we can write

‘tu’ = 3 × 3t + 3k = 3 (3t + k),

which shows that ‘tu’ is a multiple of 3 since 3t+k is a whole number.

We can generalise this to all whole numbers since any power of 10 (with a positive exponent) can be written as a multiple of 9 (and thus of 3) + 1:

100 = 9 × 11 + 1; 1000 = 9 × 111 + 1, etc.