# A challenging arithmetic / number theory problem

While visiting MIT on Thursday I stumbled on this really interesting problem:

Show that any positive integer $n$ has a (positive) multiple which has only the digits 1 and 0 when represented in base 10.

This is – obviously – not a problem intended for kids, but I thought that it would be fun to talk through it with them since the ideas in the solution are actually accessible to kids.

Here’s the introduction to the problem and a few initial thoughts from my kids:

After their initial thoughts we moved on to talking about the solution. The first step involves looking at numbers that only have 1’s. This step gives us a nice opportunity to talk about remainders.

Also, finding a way to describe the patter in the remainders is a nice challenge for the kids.

Now that we’ve understood a little bit about the remainders, we move to the key idea in the solution – as we look through the numbers 1, 11, 111, and etc, we’ll eventually find two numbers that have the same remainder when we divide by $n$.

This is a challenging idea for kids to understand, but we make some nice progress towards understanding that idea. It is a surprise application of the Pigeonhole Principle.

Now for the last step in the solution – once we find a two numbers in the list 1, 11, 111, 1111, and etc with the same remainder when we divide by $n$, how do we find a multiple of $n$ with only 1’s and 0’s?

Finally, we wrap up by exploring a few other similar ideas – for example, do all positive integers $n$ have a multiple consisting of only 3’s and 0’s? How about one consisting of only 2’s and 3’s?

So, a really neat (though advanced) math idea that kids can understand. Working through this problem is a nice way to talk about remainders and also about place value. I don’t go through problems like this one very often, but with the right problem it is neat to be able to talk about some advanced ideas with kids!

# Building place value and number sense with letter problems

I saw this wonderful problem from the 2009 AMC 8 last night:

Problem #24 from the 2009 AMC 8

Here’s the problem:

The letters A, B, C, and D represent digits. If AB + CA = DA and AB – CA = A, what is the value of D?

Many questions similar to this one appear on math contests for younger kids. At first these problems didn’t really stand out from all of the other problems, but lately I’ve come to see them as a neat way to get kids thinking about place value and number sense in general.

I also find the difference in approach between my older son (6th grade) and younger son (4th grade) to be fascinating.

Here are their approaches from last night.

My older son’s initial approach is to try to calculate the value of the individual digits:

After he finds the value of the various digits, I asked him to find a different approach that didn’t require so much calculation. Knowing the answer already helped, of course, but moving away from calculation also allowed him to see the place value ideas more clearly:

Now for my younger son – his initial approach involves much less calculation. He doesn’t have quite as much mathematical sophistication as his older brother (since he’s 2+ years younger) and he struggles a little to communicate the ideas that he’s seeing:

At the end of the last video he arrived at the idea to try out a few numbers. Once he starts down that path the place value ideas sort of emerge from the shadows and he finds his way to the end of the problem relatively quickly:

So, hopefully a nice example of how kids approach this type of arithmetic problem. Hopefully the example also shows how this type of problem can help kids think about place value and build number sense.

# What a kid learning math can look like – place value edition

My older son left some of the work from his math club on the kitchen table and one of the problems looked like it would be a nice challenge for my younger son.

What I like listening to my son here is the struggle he has searching for the words to describe what he’s seeing with the numbers. Finding the right way to describe what you are seeing / doing is an important piece of learning math.

The problem challenges you to fill in the digits of a number puzzle. Here’s his initial reaction to the problem until we were interrupted by a phone call!):

So, there was actually no one there when I answered, so we moved on! I like the way he winds his way to the solution. The one little trick in the problem – that the letter O didn’t have to be a unique number – I left for the middle school kids. There was enough here to keep track of for a 4th grader.

# A decimal misconception

Last week I started the chapter on fractions in decimals in our Introduction to Number Theory book with my younger son. Yesterday we were talking about repeating decimals like 1/3 and found that he had a little misconception about decimals:

He thought that since 1/3 = 0.3333…., that 10/3 would be 0.3030303030….. because when you multiply each of the 3’s in 1/3 by 10 you’d get a 30. Interesting idea, and it took me a little bit to see what he was doing. We spent a little time in yesterday’s video talking about ways to see that 10/3 was not equal to 0.30303030…., but I wanted to dig in a little deeper today. In particular, I wanted to show him where his thinking was right.

We started by reviewing the decimal expansion for 1/3 and what it means to write “1/3 = 0.33333…..”

Next we checked to see if we could use our knowledge of decimal expansions to understand what the decimal expansion of 10/3 would be. Right away we saw a bunch of 30’s – that opened the door to explaining why my younger son’s initial thoughts about 10/3 were nearly right. For it to be right, you need a digit that represents the number 30. That number, which I write in the video, isn’t the same as .30303030…., though.

Next we looked more carefully at the series 10/3 = 30/10 + 30/100 + 30/1,000 + 30/10,000 . . . . We saw that the right hand side could be simplified to become: 3 + 3/10 + 3/100 + 3/1000 . . ., which is exactly 3 + 1/3.

Finally, a little bit of an unexpected ending to the project. After I turned off the camera after the first movie my older son asked what the number .3030303030… was actually equal to. Good question. We set out to answer that question here. The approach I took was similar to the “standard” proof that 0.99999… = 1, though the boys took the proof in a pretty clever direction:

So, a fun little morning project. My younger son’s thoughts about 10/3 were really interesting to me. Hopefully this bit of extra work today helped him get a better understanding of place value and arithmetic.

# A cool coincidence with the 2014 Putnam and an old blog post

A few weeks ago I wrote up a fun exercise that my younger son and I had worked through:

That exercise explored a fairly straightforward situation: what two digit numbers are equal to 4 times the sum of their digits? There are a few: 12, 24, 36, and 48. My son saw the pattern that the first digit had to be twice the second digit, so I asked him: what about the number Fifty Ten?

My question is sort of silly, I know, but I wanted to get him thinking about place value. Fifty Ten would be the same as 60, but the sum of the digits of these two “numbers” is different.

This morning I got quite a surprise reading problem B1 from the 2014 Putnam Exam:

The 2014 Putnam exam at the Art of Problem Solving site

Essentially the question asks about this situation: if you allow 10 to be a digit in base 10, some positive integers will have a unique representation in the new number system and some won’t. For example, the number ten could be written in the usual way as 10, or in a new way as the single “digit” (10). The number 19, however, can only be written one way in this new number system. What are the positive integers will have a unique representation in this new number system?

What a cool coincidence – a question where fifty ten is actually a number!

How fun to have an old project with my younger son sort of overlap with a Putnam question 🙂

Updated to include talking through the problem with my son:

# What learning math sometimes looks like part 3: Multiplying in binary

We’ve spent the last week or so in our number theory book talking about arithmetic in bases other than 10.  One of my favorite activities to help kids learn about other bases is using Duplo blocks to model arithmetic in binary.  This activity has seemed to help both of the boys get a little bit better understanding of place value.

The section we were covering in the book today was multiplication in other bases.  Unfortunately what I thought was going to be one last short example in subtraction took more time than I expected.  That problem made our discussion of multiplication shorter than it needed to be.    When I got home tonight I thought it would be good to do revisit multiplication so we took out the Duplo set to work through examples of multiplication in binary.

We got off to an interesting start when my son choose the example $101_2$ times $100_2$.  He recognized that multiplication by $100_2$ just added zeros to the original number.  This was an interesting observation since we’d not talked about that specific idea this morning.  I wanted to try out another example to see what would happen when this trick wasn’t there to help.  Turned out that the trick was getting in the way a little:

I think my son was quite surprised to see that his method at the end of the last movie didn’t work.  One of the things about multiplying in other bases is that you lose your number sense a little bit and it isn’t easy to see when you’ve arrived at the wrong answer.  That’s at least part of what makes these exercises in other bases such a nice way to build up the ideas of place value – that’s really the only thing you can focus on in these problems!

We looked at the problem again and tried to figure out where things had gone wrong the last time.  Going through it a bit more slowly helped see that several numbers last time were accidentally combined into one.   Having found our way through this problem, I gave him one last problem to work through.  He seemed to have a little better sense of the multiplication process by the end of the exercise:

This was a really interesting process to watch.  A little trick that he learned was limiting his ability to understand how to multiply.  It was hard for him to see that this trick wasn’t helping, though, since the wrong answers aren’t so easy to see when you are working in different bases.   When we walked through the problem the second time it was a little easier for him to see what went wrong since he knew that *something* had gone wrong.  It was nice to see him work through the last problem completely on his own after all of this work.

Finally, just for completeness, here are two videos where we do addition and subtraction in binary with duplo blocks:

# Adding in binary with Duplo blocks

Several years ago I was talking about adding in binary with my older son.  On a whim we started using Duplo blocks to see how a “binary adding machine” would work.  It was a really fun exercise and I returned to it today with my younger son when we started the section in our book about adding in other bases.

I like talking about adding in binary with Duplo blocks for a couple of reasons.  First, it helps reinforce the idea of place value.  Second, it shows that you can add numbers in bases other than 10 without first converting them back to base 10.  Finally, both of these nice features happen in a setting that is pretty fun and surprising for the kids.

Our project from this morning went pretty well:

So well, in fact, that my son asked if we could do more tonight, so we did this second project.  This time we added 4 binary numbers, or was it 100 numbers for you binary fans 🙂

We’ll cover subtraction the same way, too, which is even more fun – you just need a new color to represent -1.  Can’t wait for that!

# A neat coincidence with place value problems

Over the last few days I’ve been working through some pretty difficult place value problems with my younger son.  These problems come from the “Algebra with Integers” section of Art of Problem Solving’s Introduction to Number Theory book.  As I wrote yesterday part of the difficulty is that we haven’t really covered any algebra yet.

Despite the lack of familiarity with algebra, my son seems to really like these problems.  He is absolutely fascinated when we arrive at the answers and is amazed that some of the problems have more than one answer.  The problems are all like the one in the video below:

Tonight I tried to throw in a little curve ball – how many two digit integers are 11 times the sum of their digits?  I wasn’t sure if he would see that this situation was impossible right away.  He didn’t, which was lucky because I really wanted to see how he would react to the algebra telling him there were no solutions.  Initially that math gave him a little trouble, but I loved seeing him realize that something was wrong without being quite sure what it was:

Definitely a fun couple of days.  I like these problems both for the easy introduction to algebra and for the place value review.   They also serve as a surprise example of a “low entry / high ceiling” problem.   I hadn’t really thought about this feature of these problems until I saw this challenge problem involving digits and decimal representation from Christopher Long on Twitter yesterday:

as well as a few fun follow ups like this one (and check Chris’s twitter feed for plenty more):

Pretty sure that the first time I saw a problem of this type was in the book Lure of the Integers by Joe Roberts which shows this interesting result:

The problems on Twitter are obviously a little harder than the ones I’m covering with my son, but I think older kids would probably like them a lot.   Even just showing some of the solutions might catch a kid’s interest since It is pretty surprising to see how large the solutions are.   I love simple to state problems that turn out to be a little more difficult than you might initially think!

Always happy to see the math I’m covering with the boys overlap in some small way with the math people are sharing on Twitter!

# A neat place value exercise: what about fifty ten?

My younger son and I started a new chapter in our Introduction to Number Theory book today – “Algebra with Integers.”  It is big step up in difficulty from the prior chapters, and even more of a step up because I have really not covered any algebra with him at all.  Even though the discussions are a little longer now, many of the problems and examples remain accessible to him.  Sadly, that’s not going to remain true for much longer, so our little tour through introductory number theory will come to an end soon.

One of the first problems that we looked at today focused on place value:

Find a two digit integer that is equal to three times the sum of its digits.

Even getting going with this problem is a challenge if you’ve not had algebra, but a few concrete examples like 34 and 87 helped get the ball rolling.  Eventually he was able to write down an equation that would help solve the problem.  If the 10’s digit of the number is A and the 1’s digit of the number is B, we know that:

(1) 10*A + B = 3*(A + B),

and so 7A must be equal to 2B.  It was really interesting for me to see the leap from “the number is AB” to “the number is equal to 10A + B” happen right in front of me.

Solving this equation is no small task  and really is only possible after you recognize that A and B have to be single digit positive integers.  Once you do recognize that important piece, though, it is not super hard to see that the solution to the original problem is A = 2 and B = 7.  Thus, the number we are looking for is 27.

With that problem as a warm up we were ready for another challenge: find all two digit positive integers that are equal to four times the sum of their digits.

If we proceed as above we find that the equation we need for this problem is:

(2) 10*A + B = 4*(A + B),

which simplifies (a little more than the first one) to 2A = B.  Instead of just one solution here, there are several:  12, 24, 36, and 48 come to mind quickly once you know that B = 2A.

At this point my son told me that he thought it was interesting that these numbers were all multiples of 12 but then dropped that thought to tell me that he thought that there were no more solutions.

“Why not?”

“Because the next one would start with 5, but double 5 is 10 and 5(10) isn’t a two digit number.”

“What would it be if it was a number?”

“510”

“But the 5 is the tens digit, not the hundreds digit.”

” . . . . fifty ten . . . ?”

“Interesting.  What do we usually call that number?”

“60.”

“What is the sum of the digits of fifty ten?”

“15”

“What is 4 times 15?”

“60 . . . hey, it works!”