So, first of all, where can the powers of 11 be found in Pascal’s triangle? If you look at the first row of Pascal’s triangle, it’s 1.1. Let’s interpret this as 11. The second row is 1,2,1, which we’ll call 121, which is 11×11, or 11 squared. Going down to the third row, we get 1331, which is 11x11x11 or 11 cubed. And from the fourth row, we get 14641, which is 11x11x11x11 or 11^4. This information is summarized in the following diagram:

eleven
1 2 1
1 3 3 1
1 4 6 4 1

11 = 11^1
121 = 11^2
1331 = 11^3
14641 = 11^4

But what do we do from row 5 onwards? Row 5 is 1,5,10,10,5,1, but if you have a calculator, you can check that 11^5 is 161051, not 15101051. The pattern seems to stop working. However, we can actually apply it to rows 5 and up, since we can interpret 1,5,10,10,5,1 as 161051.

First, we need to understand why the pattern seems to have stopped working; then, we have the possibility to solve things. The reason is that in row 5, we suddenly have two-digit numbers (the 10’s). It’s easier if we think of the numbers in Pascal’s triangle fitting into spaces. In row 5, we’re squashing two digits into the same space.

To understand how to interpret 1,5,10,10,5,1, we need to think about exactly what we’ve been doing so far. When we looked at 1,2,1, for example, we put the first 1 in the hundreds column to mean 100, the two in the tens column to mean 20, and the last 1 in the ones column to mean 100. means 1. Now we can see that when you get a 10, for example in the hundreds column, this actually means 10x 100 = 1000. In other words, you just treat the ten as “0 leads 1” like when doing addition. This is shown for 1,5,10,10,5,1 below:

1 5 0 0 5 1
+.1 1 these 1 have been carried from 10
= 1 6 1 0 5 1

Surprisingly, therefore, we can quickly compute any power of 11 using Pascal’s triangle. This can occasionally help if you ever need to calculate a power of 11 quickly. However, the fun doesn’t end there: by modifying Pascal’s triangle, we can quickly calculate any number multiplied by a power of 11. For example, we could calculate 241 x 11^2. All we do is start with 2,4,1 as our first row. Since we are trying to multiply by 11^2, we have to calculate another 2 rows of Pascal’s triangle from this initial row. For this, we use the rules of adding the previous two terms as in Pascal’s triangle itself. This is shown below:

2,4,1
2,6,5,1
2,8,11,6,1

2 8 1 6 1
… 1
2 9 1 6 1

This is a great way to quickly calculate sums that involve multiplying by 11, so even if you’ve never been good at arithmetic, try it out on your friends or family and impress them with your lightning-fast calculations!

To show why this works, let’s take the number abcd (where a, b, c, and d are each a digit from 0 to 9) and multiply it by 11. We can break this multiplication into two bits, as in the diagram below. :

abcd x 11=abcd x 10 + abcd x 1

When multiplying a number by 10, you simply add a 0 to the end, so abcd x 10 is the same as abcd0. Now, we can add this to abcd x 1:

abc 0
+. to BCD

This gives an answer of a(+0) b+a c+b d+c 0+d. This may seem unwieldy, but wait a minute! It’s exactly the same as Pascal’s triangle sums! You can verify this using the following diagram.

…to BCD
(0+)a a+b b+c c+dd(+0)

=a(+0) b+a c+b d+c 0+d

A similar process can be applied for any number of digits. So, we can see why this clever little trick works, though this doesn’t make it any less spectacular and it’s definitely worth trying out with your friends.