Visualizing Divisibility Tests to Strengthen Numeracy

Visualizing Divisibility to Strengthen Numeracy

By Fred Harwood, SFU grad student, educational consultant, and (semi)-retired math educator

I am currently studying elementary and middle school numeracy at SFU. As a recently retired secondary math teacher, these studies give me new perspectives to look at how students develop their mathematical understanding. In secondary schools, we often lose sight of the underpinnings of numeracy and utilize patterns or rote memorization.

Marc Garneau provided me with a beautiful perspective that was very empowering. In a 2014 NCTM conference presentation he showed how the divisibility test for nine could be visualized. Picture the number 3213. For decades I taught the test of adding the digits to see if the total is divisible by 3 or 9 as this is what works. Here 3+2+1+3 = 9 so both 3 and 9 would divide into 3213. Marc said to look at the place values of 1000’s, 100’s, 10’s and 1’s.


You can think of 10x10x10 cubes, 10×10 squares, 10 strips and leftover 1’s. One less than any power of 10 is obviously divisible by 9.

Taking the one cube away from the 1000 cubes leaves 999 cubes. If students don’t see this is divisible by 9 then look at the fact there are 99 tens and 1 nine. Taking one from each of the 99 tens give us 99 nines and the 99 ones can be made into 11 sets of 9. This gives us 111 sets of 9.

If we take 1 piece from each massed set we would end up with groups that are all products of 9. In our number we would see three sets of 999, two sets of 99, one set of 9 and then all the leftover pieces loosely gathered that total 9 pieces. Every set of pieces is divisible by 9 so the whole number 3213 is also divisible by 9.

In the example 64 236, we could take 6 singles, one from each of the ten-thousands, a single from each of the 4 thousands, 2 singles from the hundreds, 3 singles from the tens and amass them with the 6 leftover pieces. There are now six sets of 9999, four sets of 999, two sets of 99, three sets of 9 and 6 + 4 + 2 + 3 + 6 = 21 left over pieces. All the complete sets are divisible by 9 and also 3 except for the 21 leftovers. Therefore 64 236 is not divisible by 9 but would be divisible by 3.

Now consider other divisibility tests. Powers of 10 are divisible by 2, 5 and 10. For example, 100 is 50 twos, 20 fives or 10 tens. Since every place value is based on a power of 10 then every place above the ones place is already divisible by 2, 5 and 10. Only the leftover 1’s place is left to check for divisibility. In 2 352, the two thousands, three hundreds and five tens are each divisible by 2, 5 and 10. So we check the two leftover 1’s. Only 2 goes into this so 5 and 10 won’t go into 2 352 but 2 will.

Which is numerically more powerful: Even numbers end in 0, 2, 4, 6, or 8 or seeing these numbers as one 2, two 2’s, three 2’s, four 2’s or five 2’s by rearranging little blocks?

Testing for 4 or 25 (or 100) we can have students memorize looking at the last two digits because 4 is 2 squared and 25 is 5 squared OR we can have them visualize that all powers of 10 greater than one are divisible by 4 and 25. For example, thousands and hundreds are both divisible by 4 and 25 so we only need to check the tens and ones to see if they too are divisible.

Eleven has many divisibility tests. Thinking about what numbers close to a power of 10 is divisible by 11, we see that 11 is 1 more than 10. 99 is one less than 100, 1 001 is one more than 1 000 and 9 999 is one less than 10 000 etc. Therefore the scenario alternates between having a place value needing one more for each holder or one less. Taking 12 837 as an example, we need to subtract 1 from the 10000, add 2 ones o the two thousands, subtract eight 1’s from the hundreds, add three 1’s to the tens to make each of these groups divisible by 11. If we were using flip chip colouring, we would have 1 + 8 + 7 whites and 2 + 3 reds making a total of 11 once the zero pairs are deducted. The alternating of digits positive and negative and summing to test for 11 is now obvious that it works. 872 146 822 would have 8 + 2 + 4 + 8 + 2 of one colour and 7 + 1 + 6 + 2 of the other. 24 – 16 = 8 so this big number could not be divisible by 11.

Having student teams develop their own tests for divisibility using their understanding of powers of 10 place values would be very powerful and involve much good mathematical reasoning together. Manipulatives could aid this reasoning and visualizing for numbers under 10 000. Given an existing divisibility test, trying to prove why they work instead of just demonstrating that they do would also be a valuable number theory activity for students from intermediate to post-secondary.


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