From a set of 1 through 9 playing cards, I draw five cards and get cards showing 8, 4, 2, 7, and 5. I ask my 6th graders to make a 3-digit number and a 2-digit number that would yield the greatest product. I add,** “But do not complete the multiplication — meaning do not figure out the answer. I just want you to think about place value and multiplication.”**

I ask for volunteers who feel confident about their two numbers to share. This question brings out more than a few confident thinkers — each was so confident that he/she had the greatest product. (I’m noting here that I wasn’t entirely sure what what the largest product would be. After this lesson, I asked some math teachers this question, and I appreciate the three teachers who shared. None of them gave the correct answer.)

I say, “Well, this is quite lovely, but y’all can’t be right.” I ask everyone to look at the seven “confident” submissions and see if they could reason that one yields a greater product than another, then perhaps we might narrow this list down a bit.

Someone sees “easily” that #7 is greater than #6. The class agrees.

Someone says #7 is greater than #1 because of “doubling.” She says, “I know this from our math talk. Doubling and halving. Look at #1. If I take half of 875, I get about 430. If I double 42, I get 84. Both of these numbers [430 and 84] are smaller than what are in #7. So I’m confident #7 is greater than #1.”

Someone else says #5 is greater than #4 because of rounding, “Eight hundred something times 70 is greater than eight hundred something times 50. The effect of multiplying by 800 is much more.”

Someone says, “Number 2 is also greater than #1 because of place value. I mean the top numbers are almost the same, but #2 has twelve more groups of 872.”

But the only one that the class unanimously agrees on to eliminate is #6. Then I ask them to take 30 seconds to quietly examine the remaining six and put a star next to the one that they believe yield the greatest product. These are their votes.

I tell them that clearly this is a tough thing to think about because we’ve had a lot of discussion yet many possibilities still remain. And that’s okay — that’s why we’re doing this. We’ve been doing enough multiplication of 2-digit by 2-digit during math talks that it’s time we tackle something more challenging. So #3 gets the most votes.

I then punch the numbers into the calculator, and the kids are very excited to see what comes up after each time that I hit the ENTER key. Cheers and groans can be heard from around the room. Turns out #3 does has the greatest product (63,150) out of the ones shown.

**Ah, but then someone suggests 752 times 84**. **I punch it into the calculator and everyone gasps. It has a product of 63,168.**

Their little heads are exploding.

I give them a new set of five for homework: 2, 3, 5, 6, and 9. They are to go home and figure out the largest product from 3-digit by 2-digit multiplication. They come back with **652 times 93**.

The next day, we try another set: 3, 4, 5, 8, and 9. We get the greatest product by doing **853 times 94**. There is a lot — as much if not more than the day before — of sharing and arguing and reasoning about multiplication and place value.

Many of them see a pattern in the arrangement of the digits and are eager to share. They’ve agreed on this placement.

Then we talk about making sure we know we’ve looked at all the possible configurations. They agree that the greatest digit has to either be in the hundreds place of the 3-digit number or in the tens place of the 2-digit number. We try a simple set of numbers 1 through 5, and we agree that there are just 9 possible candidates that we need to test. The same placement holds.

Then we draw generic rectangles to remind us that we’ve just been looking for two dimensions that would give us the largest area.

I remember saying to the class, more than once, that this is tough to think about. To which Harley, sitting in the front row, says, “But it’s like we’re playing a game. It’s fun.” Oh, okay. :)

## 18 Comments

This is genius.

That they managed to formulate a theory is so powerful.

That it may not be the best, but was close, matches a lot of real world problem solutions.

Mathematically suboptimal is often the practically achievable goal, and the gain from going further is outweighed by the cost.

Fun problems and great discussion.

What do you make of the fact that many students weren’t convinced by their peers’ reasoning on which choices to eliminate? I’m really curious about how kids interpret evidence and respond to logical arguments. What they find compelling can be a fascinating window into what they do and don’t understand.

Separately, you didn’t write this, but I guess that you chose to show the area model (with that shading) to help them see how this is related to maximizing the product of two 2-digit numbers:

(a) it is pretty fast to see that the smallest of your 5 digits should go in the 1s place of a the 3-digit number

(b) use the other 4 digits to form the pair of 2-digit numbers that have the largest product A =(10*greatest + 4th greatest) and B= (10* 2nd greatest + 3rd greatest).

(c) now, you have to decide whether to form 10*A + least or 10*B + least as the 3-digit number. The area diagram helps you see that this comes down to a choice of whether you get least * A or least * B in your final product. Since A > B, you get your final answer.

It only takes one student to not be convinced of a classmate’s reasoning for the choice to stay on table. I want to do this to honor a kid who’s being honest that he/she doesn’t quite understand the reasoning behind an argument — which may very well be completely logical to you and me and to all those around her. Some of my kids have trouble multiplying a number by 2 or dividing a number by 2 mentally. Some still struggle with place value. Thanks, Joshua.

This is such great stuff! Stop making me miss 6th grade. :(

This is a terrific problem to play around with. And such great arguments from the students!

For some reason, I found myself approaching this by trying to

improvea given arrangement by swapping two numbers. For example, you can increase the product of 845 * 72 by swapping the 4 and the 5. And some interesting things happened!That might be a fun wrinkle to explore next time. Which swap will increase the product the most? Under what circumstances would a swap not change the product? Is there a swapping algorithm that will always produce the maximum product?

Thanks for sharing the fun!

Here is the solution in ruby:

puts [8,4,2,7,5].

permutation.

map{|xs| [xs.join(‘, ‘), xs[0,3].join.to_i * xs[3,2].join.to_i]}.

sort_by{|e| -e[1]}[0,5]. # 5 first biggest elements

map{|e| “#{e[0]} => #{e[1]}”}

Thanks, pinouchon, but do you have the solution in sapphire? :p

I do wish I could do what you do there. Looks fun!

What a great activity. Love the focus on what multiplication is rather than how to do it (over and over again) with the standard algorithm.

This activity also proves that it is more about the question and how it inspires curiosity rather than how “real” or “relevant” the task is.

Thanks for sharing.

Problems like this are the glory of arithmetic. Who knew that multiplication could be so involved? I especially like that you discussed maximizing the area. These students will have a better understanding of optimization problems.

Also, who is Student 4 on mathtalks? He/she has excellent intuition; is it Gauss?

Hey Shawn. I have a few “Student 4” on mathtalks.org. Which number math talk in particular?

My colleague told me that a student whom I had last year said during her class’s math talk share this year, “I Gaussed it.” :)

Great post. FYI, I verified by computer that the digit ordering you suggest always gives the maximal product, even if the digits aren’t all distinct.

Actually, I think I have a counterexample to that…

Say the digits are 8, 8, 6, 5, and 2. Then we have 862*85=73270, which is not as large as 852*86=73272.

Fawn,

Like others, I was also wondering if your students’ generalization would work in every situation collection of 5 non-repeating digits. My intuition said no, but Joseph’s counterexample included a repeated digit while your game seems to imply that is not allowed.

I ran a Python program over every permutation of 5 digits and your students’ solution is correct exactly 75% of the time. You can see the results . I called the digits of the 3-digit factor ABC and the digits of the 2-digit factor DE:

ABC

x DE

I am curious if anyone can tease out an explanation for how to predict whether it is better to have the largest digit in place A or in place D.

Hi Tyler. Right, we are not allowed repeated digits. Wish I knew Python, and I love it so much that you and others are checking out this problem. Brian Hayes wrote a wonderful post about this post and he used Python also: http://bit-player.org/2015/mrs-nguyens-prestidigitation. Thanks for dropping in, Tyler.

First, thought: did any of the kids use the aooriach of ‘lets solve a simpler problem and see how that helps- e.g. a two digit x one digit number? If not then perhaps after kids have wrestled the problem to the ground they can try that and discover the power of simplifying the question.

Second, comment: I’ve learned a lot from watching your work and posts- but mostly how important it is to focus on kids learning rather than teachers teaching- and how hard it can be to unlearn ‘teaching’, especially when some parents etc. demand their kids are taught to pass exams by recognising problems and applying methods almost by rote.

Hi David. Sorry I didn’t catch your question until now. If anyone tried a simpler problem of 2-digit by 1-digit, then I wasn’t aware of it. But as I mentioned in post, we’d been doing a lot of 2-digit by 2-digit multiplication during math talks, and most of them remember to find two numbers that have the least difference — this resulted from our exploration of quadrilaterals with a fixed perimeter. Thank you, David.

Marvelous idea!!! I have to try it on my 4-grades course in China :) It should be fun!

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