So far, I've shared three cool things about Pascal's triangle: what happens when you color odd triangles, patterns in the sum of rows, and how the sum of rows points to a mysterious mathematical fact. A thought struck me as I played with coloring Pascal's triangle. Below is a lovely pattern formed by coloring only the odd numbers.

Coloring the odd numbers means you are NOT coloring the even numbers. You are leaving alone numbers that can be divided by 2 without having a remainder. Or, numbers divisible by 2 are white while those with remainders after dividing by 2 get colored. What would happen if you tried this with numbers divisible by 3? Print a new triangle and test numbers to leave those divisible by 3 white and color those that are not divisible by 3.

sierpinski.pdf |

Usually, in the upper elementary grades, students are taught divisibility rules. Some call these tricks but, as I alluded to in my last post, I distrust mindless tricks that lack deep understanding. For now, I'll share the rule so that you can avoid all that division by three when you color the sheet. If the sum of the digits of a number is divisible by 3, then the number is also divisible by three. For example, 345 is divisible by 3. The sum of the digits is 3 + 4 + 5 = 12, and 12 is divisible by 3. So, 345 is also divisible by 3

Let's look at the largest number in Pascal's triangle: 24,310. The sum of the digits is 2 + 4 + 3 + 1 + 0 = 10, which isn't divisible by 3. Since 10 isn't divisible by 3, neither is 24,310. Since we are coloring numbers that aren't divisible by 3, we would color triangles with 24,310. The image below gives you an idea how to start this one in case you are confused.

Let's look at the largest number in Pascal's triangle: 24,310. The sum of the digits is 2 + 4 + 3 + 1 + 0 = 10, which isn't divisible by 3. Since 10 isn't divisible by 3, neither is 24,310. Since we are coloring numbers that aren't divisible by 3, we would color triangles with 24,310. The image below gives you an idea how to start this one in case you are confused.

If I were designing a math curriculum, I'd probably have kids practice divisibility rules by coloring Pascal's triangle to see what kind of Sierpinski triangles emerge. Not only is it self-checking, but students get to make lovely patterns!

Back to my pet peeves: sharing tricks with no reasoning to back them. Why is it that, when the sum of the digits of a number is divisible by 3, the number itself is also divisible by 3?

When I reason through these things, I start with the easiest scenario — in this case, numbers with only one digit: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. This is a bit obvious. The sum of the digits is the same thing as the number itself. For example, the sum of the digits of 6 is 6 because there is just one digit, 6. Both will be divisible by 3 or both will not divisible by 3 because the sum and the number are the same.

It gets a little trickier with two-digit numbers. It helps to use a real number first. Suppose we look at the number 52. The sum of the digits of 52 is 5 + 2, which is 7. The sum, 7, isn't divisible by 3 because 7 ÷ 3 = 2 with a remainder of 1. If we do long division of, 52 ÷ 3, we get 17 with a remainder of 1. So far, so good.

You could prove this to be true for every integer from 10 through 99. That would be so tedious, and it would take a long time unless you plugged it into a spreadsheet as I did below. Plus, it doesn't really help you see why nor does it help with really large numbers. It would literally take forever to check each and every number in existence. It only took me 15 minutes to craft this spreadsheet to prove this divisibility test works for two-digit numbers. Unfortunately, we must find a way to show the truth of the test for every number up through infinity. There is no spreadsheet that large!

Back to my pet peeves: sharing tricks with no reasoning to back them. Why is it that, when the sum of the digits of a number is divisible by 3, the number itself is also divisible by 3?

When I reason through these things, I start with the easiest scenario — in this case, numbers with only one digit: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. This is a bit obvious. The sum of the digits is the same thing as the number itself. For example, the sum of the digits of 6 is 6 because there is just one digit, 6. Both will be divisible by 3 or both will not divisible by 3 because the sum and the number are the same.

It gets a little trickier with two-digit numbers. It helps to use a real number first. Suppose we look at the number 52. The sum of the digits of 52 is 5 + 2, which is 7. The sum, 7, isn't divisible by 3 because 7 ÷ 3 = 2 with a remainder of 1. If we do long division of, 52 ÷ 3, we get 17 with a remainder of 1. So far, so good.

You could prove this to be true for every integer from 10 through 99. That would be so tedious, and it would take a long time unless you plugged it into a spreadsheet as I did below. Plus, it doesn't really help you see why nor does it help with really large numbers. It would literally take forever to check each and every number in existence. It only took me 15 minutes to craft this spreadsheet to prove this divisibility test works for two-digit numbers. Unfortunately, we must find a way to show the truth of the test for every number up through infinity. There is no spreadsheet that large!

How do you prove this true for two-digit numbers without checking each one?

It's quite slick.

The whole key to the proof is rearranging numbers. Think about our goal for a moment. We must show that, if the sum of the digits 7 + 8 is divisible by 3, then so is the number.

78 = 70 + 8

78 = 63 + (7 + 8)

Do you see it? By pulling out one of the sevens, you end up with the sum of the two digits 7 + 8.

78 = (9 x 7) + (7 + 8)

Why am I making it uglier? Just like any good story, it gets darkest just before the sun rises. We know that 9 x 7 is divisible by 3 because 9 is divisible by 3. So, if the sum of the digits, 7 + 8, is divisible by 3, then the number is divisible by 3.

Let's try this with another two-digit number.

56 = 50 + 6

56 = 45 + (5 + 6)

56 = (9 x 5) + (5 + 6)

Again, 9 x 5 is divisible by 3 because 3 is a factor of 9. However, the sum, 5 + 6, is not divisible by 3. Neither is the number.

Let's try it again to see if we can see the pattern.

84 = 80 + 4

84 = 72+ (8 + 4)

84 = (9 x 8) + (8 + 4)

As always, 9 x 8 is divisible by 3 as is the sum, 8 + 4. So, 84 is divisible by 3.

Can you apply this to other two-digit numbers? Can you put it into your own words? If so, you can write a proof for two-digit numbers. Mine is in the spoilers (or it will be soon).

Can you apply this idea to three-digit numbers? Can you put it into your own words? If so, you can write a proof for three-digit numbers.

It's quite slick.

The whole key to the proof is rearranging numbers. Think about our goal for a moment. We must show that, if the sum of the digits 7 + 8 is divisible by 3, then so is the number.

78 = 70 + 8

78 = 63 + (7 + 8)

Do you see it? By pulling out one of the sevens, you end up with the sum of the two digits 7 + 8.

78 = (9 x 7) + (7 + 8)

Why am I making it uglier? Just like any good story, it gets darkest just before the sun rises. We know that 9 x 7 is divisible by 3 because 9 is divisible by 3. So, if the sum of the digits, 7 + 8, is divisible by 3, then the number is divisible by 3.

Let's try this with another two-digit number.

56 = 50 + 6

56 = 45 + (5 + 6)

56 = (9 x 5) + (5 + 6)

Again, 9 x 5 is divisible by 3 because 3 is a factor of 9. However, the sum, 5 + 6, is not divisible by 3. Neither is the number.

Let's try it again to see if we can see the pattern.

84 = 80 + 4

84 = 72+ (8 + 4)

84 = (9 x 8) + (8 + 4)

As always, 9 x 8 is divisible by 3 as is the sum, 8 + 4. So, 84 is divisible by 3.

Can you apply this to other two-digit numbers? Can you put it into your own words? If so, you can write a proof for two-digit numbers. Mine is in the spoilers (or it will be soon).

Can you apply this idea to three-digit numbers? Can you put it into your own words? If so, you can write a proof for three-digit numbers.