This week, we asked Pre-Algebra students to solve a problem that turned out to be really tough. Students were asked to think about scoring in a game used to practice place value skills in the context of scientific notation. We used some examples to illustrate how the scoring worked, and many students used the examples to start teaching themselves place value in scientific notation.

It was pretty awesome to have a window into how students interpreted the examples we gave. Some students were able to make sense of what was happening. Other students got correct answers, but some of their words made me wonder if they were getting the right answer without really understanding what was going on. And some students saw things in the examples that I never would have thought of… that might lead to problems down the road! The whole experience could be a microcosm of what students see when teachers work out examples on the board.

Here are the examples we gave:

If a card contains the number 3.38 x 10^14 and Mrs. Nasamy picks an 8, the card is worth 8 x 10^12 points. If she picks a 3, it’s worth 3.3 x 10^14 points since 3 appears in two places, you must add both values:

(3 x 10^14) + (3 x 10^13) = (3.3 x 10^14) points

We include this example so you can think about how the game is played when the digit pulled appears in more than one place; note that the process would be the same if the digit appeared on two different cards.

And here are some ways students interpreted those examples:

Student 1 thinks about how the exponent and place value of the target digit are manipulated together:

When the 3.38×10^14 card was there the next time they showed it had gone from 3.38×10^14 to 8×10^12. It was like this because you cross out everything but the number pulled. If the number pulled is the first in back of the decimal the exponent needs to take 1 away. If it is the second back you do the same thing but take two away from the exponent.

Student 2 has an AHA! moment relating to the distributive property:

first off, i had to do a lot of time figuring out what the question is. it took me a while, but an AHA! was when i realized that these cards were like three #’s in sci. notation mashed together for example, (1.92×10^9) is (1×10^9)+(9×10^8)+(2×10^7). and if a 2 is pulled, your points are (2×10^7).

Student 3 provides a third example to check his/her understanding:

my interpretation of the problem is this. The teacher pick a card between 1 and 9. Whatever number she draws, you find the number in your scientific notation. for example, if she drew a 7, and one of your notations was 7.37 X 10^³, it would turn into 7.07 X 10^³. then you would solve the exponent, which is 10^³, and that is 1,000. Now your problem is simply 7.07 X 1,000. When you solve that simple multiplication problem, you get 7,070. That is your answer for that card.

Student 4 applies the examples to the main problem, finding the point score if Ms. Nasamy pulls an 9 and you have cards with 1.92 x 10^9 and 6.59 x 10^8. I am not sure if Student 4 has a misconception or not, because they do eventually come to the correct answer, however, I suspect something’s wrong after very careful reading:

I…started with 1.92*10^9. I then need to cross out the place values that aren’t 9, like the example showed us. Now the problem is 0.9*10^10. The exponent changed because I took the 0.02 place value away. Now, since there are 10 zeros instead of 9, the exponent changes to 10. To solve the problem, I first calculated 10^10, which is 10,000,000,000. Then I did 10,000,000,000*9, which is 90,000,000,000. Then I did the last one which is 6.59*10^8. Next I used the same process for the last card and changed the problem to 0.09*10^8. The exponent didn’t change because the 0.09 is the last place value there. After that, I did the same process as the first card and and solved the problem. I ended up with 900,000,000. Then I added the points together and got 90,900,000,000. I then converted that number into a decimal. I did that by counting the number of place value zeros and the turning that number into the exponent next to ten. I got 9.09*10^8.

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A middle-school math teacher, Ms. Alcala, shared the activity, “My Favorite No” on the Teacher Channel. I highly recommend watching the video. The idea is to collect work from students and then share, anonymously, one incorrect solution that you really like. You might like it because it’s a different approach or because of how much was correct about it or because it’s a common mistake. The idea is to share some work that’s not perfect in a very positive, non-judgmental way, and engage students in thoughtfully critiquing the work. Students get to work on the mathematical practice, “Construct Viable Arguments and Critique the Reasoning of Others” while also sharing and strengthening their content knowledge.

This week, reading student submissions to “Teeter Trio” I was struck by this awesome, but ultimately incorrect, submission:

Seesaw Balance POTW
A seesaw can balance with more than two people on it. The product of each person’s weight and distance from the fulcrum contributes to the balancing. If the sum of those products on one side equals the sum of the products on the other side, balance is achieved.
-As far as I know balance can be achieved when both sides are equal when the products are added in other words, wd=wd, when w is weight and d is distance from the fulcrum.
Shareef and his two little sisters, Marshay and Janeka, are playing on a seesaw. Shareef weighs 30 pounds more than Marshay and 35 pounds more than Janeka, so Shareef sits on one side to balance the two girls on the other.
-If Shareef is trying to balance the other two girls on the other side than we need to know everyone’s weights to fill in the equation of wd=wd.
Shareef=s
Marshay=m
Janeka=j
-Since Shareef is 30 pounds more than Marshay and 35 pounds more than Janeka than we have two different equations for her.
s=30+m
s=35+j
-Since we know this information we can find Marshay’s weight by putting these two equations together like so.
35+j=30+m     Subtract 30 on both sides
5+j=m
-For Marshay’s weight we get m=5+j and Janeka’s weight can’t be found so it will just be represented by the letter j.
Shareef is sitting 6 feet from the fulcrum and Janeka is sitting 4 feet from the fulcrum. If the seesaw is balanced, find a function that expresses Marshay’s distance from the fulcrum in terms of her weight.
-Now that we know Janeka’s distance and Shareef’s distance from the fulcrum but we need to know Marshay’s distance from the fulcrum. Her distance will be represented by the letter d. now that we have all the pieces of information than we can fill in the equation of wd=wd. (The right side will be Shareef and the left will be Marshay and Janeka)
wd=wd
6(35+j)=d(5+j)+4j       subtract 4j on both sides
6(35+j)-4j=d(5+j)        divide (5+j) on both sides
6(35+j)-4j/5+j=d         distribute the 6 and combine like terms
210+2j/5+j=d              Simplify
2(105+j)/j+5=d
Marshay’s distance from the fulcrum is 2(105+j)/j+5=d.

To help you think about what might be incorrect, it’s important to know that in the follow-up you will be given Marshay’s weight and asked to help calculate how far from the fulcrum she should sit.

My questions to you are:

1) What do you find awesome about this submission?

2) How would you fix it without redoing the problem? Is there a way to tweak this answer to make it better match what was asked?

Some Teeter Trio links in case you are interested:

• The problem [requires a Math Forum PoW Membership].
• Information about accessing Teeter Trio (and a selection of all our PoWs) for 21 days with a free Math Forum trial account.
• Information about becoming a Math Forum Problems of the Week Member. Consider starting with a $25 membership, which gives you access to all of this year’s Current PoWs — and now you can create 36 student logins as well!

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:

• The problem [requires a Math Forum PoW Membership].
• Information about accessing Frog Farming (and a selection of all our PoWs) for 21 days with a free Math Forum trial account.
• Information about becoming a Math Forum Problems of the Week Member. Consider starting with a $25 membership, which gives you access to all of this year’s Current PoWs — and now you can create 36 student logins as well!

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:

Renuka D

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 problem [requires a Math Forum PoW Membership].
• Information about accessing Voulez Vous de Voussoirs? (and a selection of all our PoWs) for 21 days with a free Math Forum trial account.
• Information about becoming a Math Forum Problems of the Week Member. Consider starting with a $25 membership, which gives you access to all of this year’s Current PoWs — and now you can create 36 student logins as well!