| Math Mojo - Making Math Meaningful |
This was the question:
I was told that multiplication can be performed from left-to-right. Does anyone know about this?
Professor Homunculus' answer:
Sure. Eventually, there will be a lot about this up on this site. In the meantime, here is the easy way (without doing it in your head).
Let's say you have a three-digit number times a three-digit number, say 458 x 392.
The answer is going to have as many digits as the multiplication of the hundreds (3 x 4 = 12, which is 2 digits) plus as many digits as are in the rest of both numbers. There are 4 more digits (the 5,9,8, and 2). 2 + 4 = 6, so there are going to be a total of 6 digits in the answer. That tells you that the first digit of your answer is going to go in the hundred-thousands column
| hundred thousands |
ten
thousands |
thousands |
hundreds |
tens |
Ones |
|
| 4 |
5 |
8 |
||||
| x |
3 |
9 |
2 |
|||
| 1 |
2 |
|
|
|
|
Multiply the hundreds by the hundreds and write the answer in so that the last digit of the answer will end up in the ten-thousands column.
|
| |
3 |
6 |
|
|
|
Multiply the hundreds of the multiplicand (the 4) by the tens of the multiplier (the 9), getting 36. Write this number so its final digit ends up in the thousands column.
|
| |
1 |
5 |
|
|
|
Multiply the tens of the multiplicand (the 5) by the hundreds of the multiplier (the 3), getting 15. Also write this number so its final digit ends up in the thousands column.
|
|
|
|
|
8 |
|
|
Multiply the hundreds of the multiplicand (the 4) by the ones of the multiplier (the 2), getting 8. Write this number so its final digit ends up in the hundreds column.
|
| |
|
4 |
5 |
|
|
Multiply the tens of the multiplicand (5) by the ones of the multiplier (the 9), getting 45. Also write this number so its final digit ends up in the hundreds column.
|
| |
|
2 |
4 |
|
|
Multiply the ones of the multiplicand (8) by the hundreds of the multiplier (3), getting 24. Also write this number so its final digit ends up in the hundreds column.
|
| |
|
|
1 |
0 |
|
Multiply the tens of the multiplicand (the 5) by the ones of the multiplier (the 2), getting 10.) Write this number so its final digit ends up in the thousands column.
|
| |
|
|
7 |
2 |
|
Multiply the ones of the multiplicand (the 8) by the tens of the multiplier (the 8), getting 72). Also write this number so its final digit ends up in the thousands column.
|
| 1 |
6 |
Multiply the ones in the multiplicand (8) by the ones in the multiplier (2) and get 16. Write this number so the final digit ends up in the ones column.
|
||||
| 1 |
7 |
9, |
5 |
3 |
6 |
Add the above numbers. You should also
do this left-to-right. If you can't add left-to-right, try back here in
awhile. There should be a lesson on it soon. If you need to know about
it sooner, e-mail me and request
a lesson on it. |
Go over what you have read above. It is not easy to "get" the first time you read it. It will make a lot of sense if you press on, though.
Want to know why it works? Check out the lesson on expanded notation.
|