There's a relatively quick way to multiple any number by 11. I had to multiply 11 today during work and that made me think of this old school trick I was taught. So I thought I would share it with all my fine friends here at RAFO.

So, basically, you can just ignore the 11 in this trick. Concentrate only on the other number you are multiplying. It's a pretty simple concept once you get the hang of it. Let's start with 11 X 12. As I said, ignore the 11. The answer always begins with the 1st digit of the number you are multiplying (1) and ends with the last digit of the number (2), with an exception I will get to in a moment.

So, 11 X 12 = 1_2 (begins with 1st digit, ends with last). To find the middle number, you simply add the two digits of the number (12, in this case) together (1+2). So the answer is 11 X 12 = 132 (begins with 1st digit, ends with last, add the 1&2 together).

11 X 14 = ?
1st digit is 1. The last is 4. 11 X 14 = 1_4. Add 1&4 for the middle number. 11 X 14 = 154.

Now, you might see the exception I mentioned earlier. What happens if the middle number is greater than 10?

11 X 29 = ?
1st digit is 2...or is it? Let's continue. Last digit is 9. 11 X 29 = 2_9. Add 2&9 together...wait, that equals 11! What do you do when you add a number that goes over 10? You carry over! Remember, the number always carries over to the left. So 2+9=11. Leave the 2nd digit of that answer in the middle of the answer (219) and carry the 1 over to the 2, which now makes it a 3. So the answer is 319.

Again, 11 X 29 = 2_9 (begins with 1st digit, ends with last). Add the 2&9 together. Since that equals 11, leave the 2nd digit (1) in the answer and carry the 1st digit (1) over to the 2. 2+1=3. 11 X 29 = 319!

Ok, that works with 2 digit numbers. But what about 3 digits? Same concept. Remember, everything moves to the left. Watch.

11 X 424 = ?
Begins with 1st digit (4) and ends with last (4). 11 X 424 = 4_ _4. Wait, why is there 2 missing slots? Everytime you add a digit, you increase the middle. Duh! Sheesh, moving on. Add the 2nd and last digit together first. 2+4=6. Leave the 6 in the last middle slot. Remember, we are always moving to the left. Now add the 1st and 2nd digit together. 4+2=6. Drop that into the 1st middle slot. Now you have your answer!

11 X 424 = 4664!

Ok, what happens again when the digits add to over 10?
11 X 268 = ?

Begins with the 1st digit (2) and ends with the last <8>. 11 X 268 = 2_ _8. Add the 2nd and last digit together. 6+8=14. Wait, it's over 10 again! Calm down, grab a paper bag and breathe slowly. Ready? Leave the 4 in the last middle slot. The 1 carries over to the 6, which now equals 7. Add the 1st and 2nd digits now. Remember, the 6 is now a 7. 2+7=9. Drop that in the 1st middle slot. There's your answer.

11 X 268 = 2948!

What if it was 11 X 368 instead of 268? That 6, which became a 7, would now be added to the 3, which would give you 10. So leave the 0 in the 1st middle slot, carry the 1 over to the 3, and now you answer became 4048. Get it?

In conclusion, if you wanted to keep adding digits, say 11 X 4325, the same concept applies. Increase the middle slots and add from right to left, carrying over the excess and leaving the last digit behind. What do you think of this trick? Easy to understand?