Is There a Better Algorithm for Longhand Multiplication?

[Embison]

How come on step 3 it says to carry a 2nd 4? and What does it mean when it says that 40 is greater than 9? Where did the number 9 come from? What is it referring to?


http://www.webmath.com/cgi-bin/k8ipcalc.cgi?op=mult&n1=808&n2=745&back=k8ipmult.html

Thank you.

 

[Doc Al]

How come on step 3 it says to carry a 2nd 4?

Because 5 x 8 gives 40, a two-digit answer.

and What does it mean when it says that 40 is greater than 9? Where did the number 9 come from? What is it referring to?

You only need to carry if you end up with a two-digit number. Any number greater than 9 is two-digit.

 

[Embison]

But when you end up with 40 it's in the last sequence of multiplication for the first number 5. So you just write 40 underneath. And it even says there are no more digits to carry over..but then it says to add the number 4 anyways. What purpose does adding a 4 next to the number 8 serve?

Thank you.

 

[Doc Al]

It serves no real purpose since you know you have no more digits to muliply. Putting the 4 next to the 8 doesn't take much extra time--perhaps it's good to get into the habit of doing that, this way you'll never miss a carry. Don't over analyze it!

 

[Embison]

Don't you think that would be a bad habit to carry the last number like that..Because then you have the potential to add that number the way you're supposed to when you carry a number?

 

[Doc Al]

And if you did add it, what would you get? I don't see a problem.

 

[Embison]

44 instead of 40

 

[Doc Al]

44 instead of 40

How did you get that? You multiply 5 X 8 and get 40. It's two digits, so you don't think, you always carry. You write down the 0(first digit), carry the 4. You go to the next step: 5 X ?... hey, you're done, no more digits--treat it as 5 X 0 = 0. Put down the 0, add the 4 that you carried: 0 + 4 = 4(next digit). You get 40.

Of course, after some experience you will learn to peek ahead. If you are at the end of the line of digits, no point in carrying! But if you did--no problem, you still get the right answer.

 

[symbolipoint]

I barely looked at the responses; but just realize that each digit in the factors is counting by place value. If you could transcribe the factors into expanded form, you will see clearly what numbers in each factor are being multiplied by which numbers of the other factor. You could literally draw a picture to illustrate this, and I HAVE done this in order to help students. This can be an excellent way to illustrate the meaning of the Distributive Property of multiplication over addition.

 

[Embison]

How did you get that?

I didn't. I know how to do long handed multiplication. Who doesn't. And the answers are there anyways so obviously it's 40. I was saying...Don't you think it's a bad habit to carry the last 4 and put it beside the 8 because when you carry a number you're supposed to add it after multiplying the two numbers together...Which would be 5 X 8 + 4 = 44. I've never heard of carrying any numbers when you're done multiplying like it shows in this example. And it's completely pointless and would probably be confusing for a kid who's just learning. There is the potential for adding the last carried number to the last two numbers multipled. And you would come up with the wrong answer. I wasn't ever taught once to carry the last remainder so when I saw it in this example it looks like a mistake more than anything. Why would you carry the 4 when you just write 40 at the bottom anyways? It doesn't makse sense to go and add a 4 up there at all. How does that help?

 

[Doc Al]

Don't you think it's a bad habit to carry the last 4 and put it beside the 8 because when you carry a number you're supposed to add it after multiplying the two numbers together...Which would be 5 X 8 + 4 = 44.

You are doing the carry operation incorrectly. When you multiply a digit you add whatever you carried from the previous multiplication. So when you multiply 5 X 8 = 40 you add the previous carry, if any. In this case, you haven't carried anything, so it's just 5 X 8 + 0 = 40.

There's nothing wrong with automatically writing down the 0 and carrying the 4. Of course, since there are no more digits to multiply*, you just get 5 X 0 = 0. Now add the 4 that you are carrying and you get 5 X 0 + 4 = 4. And you're done, with the correct answer.

*(Note that 808 can be written as 0808.)

 

[ice109]

does anyone know a better algorithm?