The Take-Away Game
- Write 23 X’s on a piece of paper.
- On your turn you can erase or take away 1, 2, or 3 of the X’s.
- Turns alternate. You cannot skip your turn.
- The person who erases or takes away the last X wins.
I’ve always felt a connection to Punxsutawney Phil because my birthday is on February 2nd, the same day folks watch to see if Punxsutawney Phil sees his shadow! One year, I noticed:
We talk a lot about the problem-solving process here at the Math Forum and try to develop resources that will help teachers help their students get better at problem solving. We discuss how to encourage students to share their thinking (such as through Noticing and Wondering) and how to cultivate classrooms that value those thoughts and ideas as much as answers. But if we take a look at our own “problem solving” product, the Problems of the Week, we have to acknowledge that there isn’t so much support for process, starting with the “Compose Answer” button that appears at the bottom of each problem. Oops!
We have considered a number of possibilities, including an option (chosen by the teacher) to show just the scenario for a problem and then have fields in which students can submit their Noticings and Wonderings. That sort of thing would require some significant programming time, so while we are working on putting it in place (I’ll blog about it more before we get too far), we are first going to support the PoW process through some wording changes in the submission process. We’ve come up with some possibilities and wonder if anyone has alternative ideas.
On a problem page, it says, “Compose Answer”, which of course implies you have “an answer”. We’re thinking of changing that to “Submit Ideas”, which seems a bit more welcoming to submissions that might not actually contain an answer yet (or ever).
Once you get to the “submission” page, there are four spots we’re suggesting alternative wording:
What do you think? Would these sorts of changes convey “process” to your students? Do you have any other suggestions?
During the February 26th MoMath Masters Tournament, @MoMath1 tweeted, “No googling – how many sides on an enneagon?” We thought, “Hey! We know enneagons!” If you don’t, maybe this scenario from a problem we first used in 1998 will give you some hints (as well as some ideas for something you could do with one).
Extend the sides AB and ED of the regular enneagon ABCDEFGHI until they intersect.
In the Math Fundamentals problem Frog Farming, the goal is to make at least four different rectangular pens, each of which uses 36 meters of fence. Many students thought of this the same way I did, which was to consider half the necessary perimeter as the sum of two adjacent sides.
Rachel B, Seven Bridges Middle School
I know that the problem was finding perimeter. I know first you divide 36 by 2 and get 18. Then you find addends of 18 and they are the length and width. I added 12 plus 6 which equals 18 and 12 times 2 plus 6 times 2 equals 36.
Sarah G, Laurel School
First I decided come up with a length and width for a rectangle that would equal 18 because 18 is half of 36 and you have to multiply that number by two to get the perimeter. I decided on 2 and 16. I checked it by doing 16+16+2+2= 36. One could by length=16, width=2.
Rachel and Sarah and I were thinking about perimeter, in the context of this problem, like this:
Another way I thought of this was as 2(L + W). Hmm…..
Then I was mentoring a few students in this problem and noticed that they were thinking about the problem differently.
Ethan Z, Lorne Park Public School
I thought of a rectange which has 4 sides and 2 sides are equal and the other 2 sides are equal because Farmer Mead wants a rectangular pen that uses 36m of fencing. First I got the answer by thinking of 2 equal numbers that add up to less than 36. Then, the last 2 equal numbers are the difference of 36 to the first 2 equal numbers. That’s how I got all the numbers of the first question.
Emily G, Laurel School
I used 2 numbers, and doubled 1 number by two (ex. 6×2=12). 36-12 is 24. 24 is an even number that can be split into 2, which is 12 (ex. 24÷2=12.) 24+12 is 36!
Maybe because I had “seen” the problem differently, it took me a few minutes to figure out what these other kids were doing. Then I realized they were “seeing” the problem like this:
This seems more like 2L + 2W! These students tended to use more of a Guess and Check strategy to find solutions, whereas kids who used the first method were a little more systematic from the start. But it was fun to me to see these two different methods to what is a pretty simple idea. I like when simple things are done different ways.
I wonder how you “saw” the problem when I first described it. And how did your kids tend to see it?
Some Frog Farming links in case you are interested:
Our last Geometry Problem of the Week, Voulez Vous des Voussoirs?, deals with building arches using trapezoidal blocks. Maybe you’ve had a chance to build such an arch at a museum. I’ve done it most recently with my sister’s family at the Minnesota Children’s Museum. In our problem, if you’re told the angle measure of the obtuse angles of the “trapezoid”, how can you figure out how many identical blocks are needed to build a semi-circular arch?
What is most interesting to me about this problem is that different people seem to “see” it in different ways. That is, they get one particular model in their head of the situation. When I first solved it, I “saw” it like this:
A few students saw it that way, though it wasn’t the most common method we saw. Here’s an excerpt from one solution:
I knew if the obtuse angles were 96 degrees, the supplement would be 84. Because of symmetry, the two sides of the voussoir must meet at the center of the semi-circle. In this way a triangle is formed, so therefore the angle at the center of the circle must be 180 – (84+84).
Here’s the second idea I “saw” when I solved this:
This was the basis of the most common method that students used this week, and is explained in excerpts from two solutions:
Gavin T, Highlands Elementary School
I knew that if all the angles were 90 degree angles, the arch wouldn’t be an arch, it would go straight up. 96 degrees meant that each angle angled the arch 6 more degrees.
Jed M, Waterford Elementary School
If there is a 6° increase on each angle. (I know that because there 96° angles, and if it was a 90° angle it would go straight up.) And there’s two angles on each voussoir. So theres a total of a 12° increase. I’ll use math sentences to get a total of 180 and what ever number times twelve equals 180 is my answer.
Then there’s the image conjured up by a group of students from Conners Emerson School in Maine. They made it a problem about angles of regular polygons. That had never occurred to me, so it was exciting to get their submission. Here’s my version of what they “saw”:
Here are some excerpts from their solution, including a hint and their picture:
Xingyao C, Tarzan M, and Tom G, Conners Emerson School
We drew a diagram of two of the voussoirs adjacent to each other. If the voussoirs make it all the way around, they would form a regular polygon.
The two adjacent acute angles of the trapezoids make an interior angle of the regular poylgon.
Hint: Try not to look at the voussoirs as solitary pieces, but as part of a convex polygon. You are not trying to find the information of each individual voussoir, but the information of the arch made by many.
(I can’t help but point out that they named their picture “French Words in Math”, which I think is awesome!
I wonder how you think your students “saw” this situation. Was one solution method more common, or was there a variety of models used in your classroom? How did you see the situation? Please let us know!
Some Voulez Vous de Voussoirs? links in case you are interested:
The last Math Fundamentals Problem of the Week was a puzzle of sorts. You’re told there are three chips with a number on each side. The six numbers are consecutive one digit positive numbers. One way to throw the chips gives you 6, 7, and 8, for a sum of 21. You’re also given four other sums between 16 and 23 that are possible. Aside from adding three numbers together, there isn’t any other math concept that’s central to this problem. It isn’t about “fractions” or “proportional reasoning” or “measurement”. It’s about understanding what’s going on, and coming up with a method to figure out what numbers are on the other sides of the chips.
Since solving a problem like this often lends itself to running around in circles for a bit before you hit upon the answer, the resulting submissions are usually of two different types. The first is the solution that states the answer, and then confirms that the answer is correct. Here’s an example:
On the back of the 6 there is a 5, on the back of the 8 there is a 4, and on the back of the 7 there is a 9
for 16 i added: 7 plus 5 plus 4.
for 17 i added: 6 plus 7 plus 4.
for 20 i added: 6 plus 5 plus 9
for 23 i added: 9 plus 6 plus 8
That example doesn’t give me any idea how the student approached the problem, what sorts of things they tried or did, whether they made any mistakes and found them, or anything that seems systematic and efficient. We could do some work, using the sums that they provided, to figure out the correct combinations. But the student doesn’t provide any insight into the process they used to find the answer.
It’s not uncommon to see a lot of submissions that confirm results. Sometimes it’s hard to keep track of the different paths you traversed on your way to the answer. Good recordkeeping is a must. I have to say, however, that reading the solutions for this problem, I was excited to see so many students telling the “story” of how they solved the problem. Most used some form of the Guess and Check strategy. Some used Logical Reasoning. Here are a couple of examples:
Told 6+7+8=21. Number are consecutive, and to add to 23, one number has to be 9, for 6+9+8. 6 and 8 stay- 7 and 9 are on the same chip. 4 and 5 are the other numbers. To add to 16 you have 4+5+7. 17 needs 4+6+7. 5 and 6 have to be on the same chip. 8 and 4 are the remaining numbers and are on the same chip.
First I thought, since the toss of 23 is larger than the 6, 7, 8 toss, it must contain a number greater than 8. Therefore 9 is needed. The toss must be 9, 8, 6 because no other combination gives 23. So the 7 and 9 are on the same chip. This also means the other two numbers must be 4 and 5. Next I thought the toss of 16 must be made of 4, 5, 7. Then, the toss of 17 couldn’t be 4, 5, 8 because it is not possible since the 7-9 chip is missing.
So 4 must be on the 8 chip to have all 3 chips in the toss. This gives the 3 chips: 5/6, 4/8, 7/9.
Then I concluded that the 20 roll is possible with 5, 7, 8 so it all checked out.
However they did it, we really love reading stories about students’ problem solving adventures. How do you support your students in telling this story? Is it something that gets reported orally in the classroom a lot (which is great practice for eventually writing it down)? Do they read aloud to others? Do you just keep prodding them to say more about their process? We’d love to hear from you, since all the kids submitting to this problem didn’t do a good job by luck or accident!
Some Let the Chips Fall… links in case you are interested:
In the last Geometry Problem of the Week, Lotsa Popcorn!, we presented the World’s Largest Popcorn Ball (as of 2006), and explained that it weighed 3,415 pounds and was 8 feet in diameter. It was claimed to be “almost 50,000 times larger than the normal popcorn balls distributed for retail consumption.”
We went on to ask students to find the diameter of a popcorn ball whose diameter is 50,000 times smaller than this, as well as the diameter of a ball whose volume is 50,000 smaller. We also asked about the weight. We received a number of solid answers to this, including these two:
For #s 1 and 3, I simply took the original value (big ball) and divided by 50,000. For #2, I first found the volume of the big popcorn ball, then divided by 50,000 and used the formula for the volume of a sphere in reverse to find the radius. Then I multiplied that value by 2 to get the diameter.
1. 8/0,000= 0.00016 ft, which did not seem reasonable
2. Volume of original (big ball):
V=267.946667 ft3 (roughly)
volume of small:
V= 267.946667/50,000= 0.00535893 ft3
d=0.217154 ft or 2.6″ which did seem reasonable
3. 3,415/50,000= 0.0683 lbs, which, unless my perception is off, seems reasonable enough.
1. 0.00192 inches. No, this would not be a realistic size for a normal popcorn ball. 2. 2.605848 inches. This is a realistic size for a normal popcorn ball. 3. 1.0928 ounces. This is not a realistic weight for a normal popcorn ball.
1. Multiply eight feet by twelve inches to find how many inches are in eight feet. Then, divide by 50,000.
The radius is half the diameter, which would make it 4 feet.
V=267.94667 cubic feet
Then divide by 50,000 for 0.00535893 cubic feet.
By using this new volume in the above equation, the radius can be solved for.
Since the diameter equals 2r, it is 0.217154 feet. Multiply this by 12 to get 2.605848 inches.
3. 3415 multiplied by 16 equals the number of ounces in 3415 pounds:
This divided by 50,000 equals the answer, 1.0928 ounces.
Obviously we don’t need to give that much direction to students. We just wanted to make sure they would do something reasonable, and we figured we might need to support them in that, especially if they were working on their own without the benefit of a class or partner conversation. What we really did was make all the really interesting math decisions for the students and make it easier for us to read the solutions. We should have just presented them with the facts of the giant popcorn ball, and asked them if the 50,000 times bigger thing was reasonable. What would they compare? I’m thinking back to the So Many Salmon problem we used in Math Fundamentals earlier this year. We presented students with some facts and then asked a question, but let them determine how they would answer that question.
It makes me wonder how many people used the Scenario Only, which did only present the facts, and how many used the whole problem with the three specific questions. What would you have done? What do you think your students, whatever age they are, would have done with just the facts and the question of whether or not the 50,000 times bigger statement was reasonable?
Some Lotsa Popcorn! links in case you are interested:
Ratios is a concept that doesn’t always go smoothly in the classroom, and I know many teachers who grumble about the fact that their students don’t understand it better. I’d suggest that, as often is the case, this is at least partly attributable to the fact that we focus on procedures before the students have an intuitive idea of what ratios really represent.
I was reminded of this when I was reading the submissions to recent Math Fundamentals Problem of the Week, Anthony’s Famous Butter Rolls. Students are asked to figure out how many rolls Anthony has to make based on the instructions his mother gave him. One instruction is, “His mom wants him to make enough so that there will be 3 rolls for every 2 adults and 4 rolls for every 3 children.”
A ratio, you say! We can use proportions to do something! Yup, we could. But the problem doesn’t mention ratios, and only one kid even used the word “ratio” in their submission. Nobody used “proportion”. Instead, they solved the problem from a completely conceptual perspective, using four different methods. Below are excerpts from nine student solution.
By far the most common method was what these three kids did:
Mitchell Z, Highlands Elementary School
Since there will be 12 adults and each 2 adults gets 3 rolls, you first divide 12 by 2 because the answer is the number of the groups of adults. The answer to that is 6. Then you multiply 3 rolls by 6 groups of 2 adults. That would be 18 so now you know that Anthony needs to make 18 rolls for the adults.
Elisabeth M, Fred W. Miller School
[T]o find out how many rolls the adults will need, 12 adults in all,3 rolls for every 2 adults, I divided 12 by 2 and got 6 groups of adults. Then, I multiplied 6 by 3 and got a total of 18 rolls for adults.
Evan S, Fred W. Miller School
I figured out how many groups of 2 adults there were in 12 people.
12 divided by 2 = 6. Now that I know there are 6 groups of 2 and each group gets 3 rolls, there must be 18 rolls for the adults. (6×3 = 18)
Slightly less popular was the idea of making a table:
Molly M, Highlands Elementary School
In the problem, it said that for every two adults, Anthony should make three rolls. I wrote down numbers to twelve counting by two’s to start my graph, because there is twelve adults and for every two of them, Anthony needed to make three rolls. Here is my adult chart with my work.
Adult Rolls Adults Rolls 2 3 4 6 6 9 8 12 10 15 12 18
I knew now that the adults needed 18 rolls.
Andrei C, Lorne Park Public School
Since there were 3 rolls for every 2 adults and there were 12 adults:
3 rolls= 2 adults
3 rolls= 2 adults
3 rolls= 2 adults
3 rolls= 2 adults
3 rolls= 2 adults
3 rolls= 2 adults
18 rolls 12 adults
Another method that is really close to making a table is to draw a picture. We didn’t get any actual pictures from submitters (other than some yummy-looking pictures of rolls!), but some students did a nice job of explaining their picture. Here are two examples:
Adabelle W, Steele Elementary
I made a chart with 27 circles. There is 27 because 12 adults and 15 children = 27 people all together. I put an “A” in 12 circles for Adult and an “C” in the other 15 for children. Then I put a big circle around 2 of the adults and wrote a 3 with them for how many rolls they needed. I did that until there were no more adults left to group. Then I did the same thing with the children except instead of 3 rolls for every 2 adults I grouped 4 rolls for every 3 children. Anthony needed to make 18 rolls for the adults. 3*6=18. He needed to make 20 for the children.
Annabel L, Steele Elementary
I first of all, drew a picture of 12 adults and 15 children. I then split the adults into groups of 2 and put a 3 beside them because they’re 12 adults and they’re 3 rolls for every 2 adults. I then did the same thing to the children, only I split them into groups of 3 and put a 4 beside them because they’re 15 children and they’re 4 rolls for every 3 children. I then, added all of the rolls up and got 38.
The final method we saw wasn’t used by very many submitters, but it works just fine! These two students figured out how many rolls needed to be made for each individual adult (and child).
Nick L, Highlands Elementary School
I knew that for every 2 adults there should be 3 rolls so I divided 3 by 2 to get 1.5 rolls per adult. Next, I knew that there was 12 adults so I multiplied 12 by 1.5 to get 18 rolls for all the adults.
Elise B, Elgin Street School
3 rolls every 2 adults. Each adult will get 1 1/2 rolls
One more roll for each person: [that's part of the directions]
An adult will get 2 1/2
A child will get 2 1/3
For 12 adults:
24+6=30 rolls for all adults
These students aren’t solving this problem using a procedure that someone has taught them. They’re solving it by doing something that makes sense to them. They understand the concept of what it means for there to be three rolls for every two adults, and they’ve come up with methods that will find the correct answer every time. Two of these methods (the second and third) aren’t something you would want to use with a big number of people, because it would take a while, but the students who used those methods are ready for a conversation about what it might look like if you had 120 people. The other two methods (the first and the fourth) could be used for any number of people. All four methods were used in the context of the problem.
What sorts of methods would your students use to solve a problem like this? How would you move students from this sort of conceptual understanding to developing the efficient procedures that they’ll eventually be expected to execute? We’d like to know!
Some Anthony’s Famous Butter Rolls links in case you are interested:
In the recent Math Fundamentals problem Count the Discount, students calculated the price of a jacket after it had been on the sale rack for five days. Each day on the sale rack, the price was discounted an additional 20%. Submitters used four different methods to figure out the new price each day.
The most common method by far was to multiply the current price by 0.2 to find the amount of the discount, and then subtract that from the current price to get the new price. (Many students also knew that 20% of 100 is 20, without doing any arithmetic.)
Charlotte C, Steele Elementary School
On the first day the jacket is $100.
Percent means out of a hundred, since it was 20% of 100, It was obviously $20. To make sure I multiplied 100 by 0.20 to figure out the percent. On the first day $20 was taken off of the jacket. 100-20=80
On the second day the jacket costs $80. Since The jacket isn’t $100 anymore My first method is harder so I will stick to the second method. 80 multiplied by 0.20=16. Now the jacket costs $64 because the discount was $16. 80-16=64
Breanna C, East Middle School
If the coat is originally $100, and you are taking 20% off the item, then you will use multiplication. To make the equation, you will end up having to convert the percent into a decimal. The equation will be $100 times .20 which equals $20.00. Then you will subtract that amount from the total.
Some students calculated the 20% by first finding 10% and then doubling the result. I wonder if the students who used this method didn’t have a calculator and thought this would be a little easier to do in their heads.
Natalie V, Seminary Hill School
- I found 10% of $100.00 ($10.00), and multiplied it by 2.($10.00 x 2=$20.00)
- I subtracted the first day’s discount ($20.00) from $100.00, and the answer was $80.00.( $100.00 – $20.00= $80.00.)
- I found 10% of $80.00 ($8.00) and multiplied it by 2. ($8.00 x 2 =$16.00) That is the second day’s discount.
- I subtracted $16.00 from $80.00 and the answer was $64.00. ($80.00 – $16.00 = $64.00)
Ria D., Laurel School
The steps i used to find the percents were
- first i divided the numbers by 10
- next i multiplied that # by 2 to get 20%
-then i subtracted the 20% from the original #
Another method to find 20% is to divide by 5, which this next student did because he wasn’t sure how else to figure it out. (Aside: What experiences do we need to provide students so that they develop the understanding that multiplying by 0.2 and dividing by 5 are the same thing? I don’t mean teaching procedures for converting from percents to decimals to fractions, but really developing true understanding.)
Praeek D, Highlands Elementary School
I subtracted $20 from $100, because the text said the price would go down 20% every day, and 20% of $100 is $20. I got $80, so I put it down under price, and beside 1st day, because it was the price on the first sale day.
Next I divided $80 by 5, because I didn’t know how to get 20% of $80, but I knew 100% dived by 5= 20 %. I got $16.00, so I subtracted $16.00 from $80, because the price had to go down 20%, and 20% of $80 was $16.
The last method we saw is the one that I would use, since it involves one step instead of two to find the new price each day.
Zoe B, Birch Wathen Lenox School
To solve the problem, I multiplied the total by .8 each ‘day.’ I did this because after the 20% discount, the item would be 80% of the price.
Developing facility with percents gets a lot of emphasis in elementary school, and it would be super if kids could use all these methods equally well and decide which one they like best, or which is best suited for a specific situation. If your students tended to use one of these methods, what would they think of the others? Did you have any conversations with them about different methods and how they’re related? What method is most popular in your classroom?
Some Count the Discount links in case you are interested: