So, firstly, where can the powers of 11 be found in Pascal’s triangle? If we look at the first row of Pascal’s triangle, it is 1,1. We are going to interpret this as 11. The second row is 1,2,1, which we will call 121, which is 11×11, or 11 squared. Moving 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 summarised in the diagram below:

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 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 appears to stop working. However, we can in fact apply it to rows 5 and onwards, as we can interpret 1,5,10,10,5,1 as 161051.

Firstly, we need to understand why the pattern appears to have stopped working – then we stand a chance of sorting things out. The reason is that in row 5, we have suddenly got two digits numbers (the 10s). It is easier if we think of the numbers from pascal’s triangle fitting into spaces. In row 5, we are squishing 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 have 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 units column to mean 1. Now we can see that when get a 10 in, for example the hundreds column, this actually means 10x 100 = 1000. In other words, you just treat the ten as “0 carry 1” like when you are doing an addition sum. This is shown for 1,5,10,10,5,1 below:

1 5 0 0 5 1

+.1 1 these 1s have been carried from the 10

= 1 6 1 0 5 1

Amazingly, therefore, we can quickly calculate any power of 11 using Pascal’s triangle. This can help occasionally if ever you have to calculate a power of 11 quickly. However, the fun doesn’t stop here: 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. As we are trying to multiply by 11^2, we have to calculate a further 2 rows of Pascal’s triangle from this initial row. For this, we use the rules of adding the two terms above 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 involving multiplying by 11 quickly, so even if you never been good at arithmetic try this out 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, c and d are each a digit 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 multiplying a number by 10, you just add a 0 onto the end of it, so abcd x 10 is the same as abcd0. Now, we can add this to abcd x 1:

a b c d 0

+. a b c d

This gives an answer of a(+0) b+a c+b d+c 0+d. This may look unwieldy, but hang on a minute! It is the exactly the same as the sums from Pascal’s triangle! You can check this using the next diagram.

… a… b… c… d

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

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

A similar process can be applied for any number of digits. Therefore, we can see why this clever little trick works, although this makes it no less spectacular and is still definitely worth trying out on your friends!