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## Pascal’s Triangle and Powers of 11

So, first of all, where do you find the powers of 11 in Pascal’s triangle? If we look at the first row of Pascal’s triangle, it is 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. Moving to the third row, we get 1331, which is 11x11x11 or 11 cubes. And from the fourth row, we get 14641, which is 11x11x11x11 or 11^4. This information is summarized in the following diagram:

1 1

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? 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 apply this to rows 5 and beyond, 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 got two-digit numbers (the 10s). It’s easier if we think of the numbers in Pascal’s triangle fitting into the spaces. In row 5, we are crushing two digits in the same space.

to understand how to interpret 1,5,10,10,5,1, you have to think about exactly what we’ve been doing so far. When we saw 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 1. Now we can see that when we get a 10 in, say the hundreds column, this actually means 10x 100 = 1000. In other words, just treat the ten as “0 takes 1” like when you’re doing addition. This is shown for 1,5,10,10,5,1 below:

1 5 0 0 5 1

+.1 1 these 1 were taken from the 10th

= 1 6 1 0 5 1

Amazingly, therefore, we can quickly calculate 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 the 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. To do this, we use the rules of adding the two previous terms just like 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 calculate sums that involve multiplying by 11 quickly, so even if you’ve never been good at arithmetic, try this on your friends or family and impress them with your lightning speed calculations.

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

abcd x 11 = abcd x 10 + abcd x 1

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

abcd 0

+. abcd

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

… a… b… c… d

(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, although that doesn’t make it any less spectacular and still worth trying out with your friends.

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