The last post about **multiplying by 11** only works for two-digit numbers. There’s also a simple trick for multiplying by 11 for larger numbers, and it’s very easy! Let’s have an example of 236 × 11.

First, we write the problem like this:

0236 × 11

We write a zero in front of the number we are multiplying. You’ll see why in a moment. To find the product when multiplied by 11, we simply add each digit to the digit on its right, starting with the units digit.

The units digit is 6, but there is no digit to the right of 6, so we add 0:

6 + 0 = 6

So the last digit of our answer is 6. Let’s write that down:

0236 × 11

0006

Next is the tens digit, the digit 3. We add it to the digit on its right, which is 6:

3 + 6 = 9

So the tens digit of our answer is 9.

0236 × 11

0096

Continue in the same way:

2 + 3 = 5

0 + 2 = 2

So here’s the final answer:

0236 × 11

2596

If we hadn’t written the 0 in front of the number, we might have forgotten the 0 + 2.

Let’s try another problem. This time we will have to carry digits. For example, let’s try 471 × 11. We setup the problem as before:

0471 × 11

Then we start with the units digit and the tens digit:

1 + 0 = 1

7 + 1 = 8

So far, our calculation looks like this:

0471 × 11

0081

Now when we add the hundreds digit:

4 + 7 = 11

So we write down the last digit and carry. Here, we write down 1 under the 4 and carry 1. So far it looks like this:

0471 × 11

0181

Then we do the final step:

0 + 4 + 1 (from the carry) = 5

So our final calculation looks like this:

0471 × 11

5181

**Why this trick works**

The secret behind the trick is the alignment when multiplying vertically. Notice that when we multiply by 11, we are actually multiply by 1 and by 10 by distributive property. That is, 236(11) = 236(10 + 1). Observe how this works in the vertical multiplication below.

236

×11

236

2360

2596

The steps come from the fact that when you add down vertically, each digit in the product is the sum of the digit and the digit to its right.