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!

Join the John Ehret Patriots in New Orleans, Lousiana, USA, in Noticing and Wondering about this Problem of the Week!

The Hopi Indians invented Totolospi, a game of chance. The game is played with three cane dice, a counting board, and a counter for each player. Each cane die can land round side up (r) or flat side up (f).

The moves of the game are determined by tossing the three cane dice with these rules:

Toss	Move
three round sides up (rrr)	player advances 2 lines
three flat sides up (fff)	player advances 1 line
any other toss of the three cane dice	player doesn't advance

This week’s Pre-Algebra Problem of the Week, “Happy New Year Wish,” was so much fun! It’s based on the true wondering that our colleague Suzanne had when she wished her son, Specialist Lee Alejandre, “Happy New Year!” while he was stationed in Seoul, South Korea. She wondered why the time was 14 hours earlier in Seoul, and what it had to do with the longitudes of Philadelphia and Seoul.

I had so much fun reading students’ work as they connected their understanding of math to their thoughts about time zones, globes, and longitudes. There were many different solution methods, and students talked a lot about how they got started on the problem… whether it was talking to friends, getting out a globe, drawing a picture, Googling, or thinking of a simpler related problem. Here are some of my favorite quotes! I wonder if this problem was extra “juicy” because it was about a real puzzle…

Student eleven from Caughlin Ranch ES brought up an important point: what happens if we thought about going east from Seoul to Philadelphia, or west from Philadelphia to Seoul? Why can’t we base our calculations on the 157º difference. Here’s what Student eleven had to say, “i also remembered a time when a book was answering a question involving the international dateline. this told me that the same thing would work going west but you’d go +1day. so goin east would be less complicated (i assumed).”

Julia R. from Birch Wathen Lenox School helped me think about how to decide which way to round in this problem when she said, “Although the numbers werent exact, 210 degrees meant that the time was passing through the 14th hour of longitude, and aproaching the 15th. It was closer to the 14th hour though.” Did you think about why, if 203º of difference is 13.53333… hours apart, we round to 14 hours instead of 13? Did looking at a globe help you decide? Read More→

I was going to write this blog post about moving from guess and check to an algebraic solution method, and why you might want to use algebra.

Then I was reading through some of the Guess and Check submissions we got, and thinking, “wow, that looked pretty effortless.” One student, Jack M. from Rosemont School of the Holy Child, pointed out that the solution has to be a multiple of 3 between 20 and 40. If you start your guess and check with some of that reasoning, you know you won’t have to do very many guesses. And if you’re good at adjusting a guess up or down based on your results, you can get the answer pretty quickly. Jack only needed two guesses: he started with the first multiple of 3 above 20, which is 21, and got to 24 on his second guess!

Rashmi R. from West Woods Upper Elementary School got the answer in 5 systematic guesses (he didn’t focus on the multiple of 3 idea, but was still very efficient). He guessed 10 CDs were sold, then 20, then 30, then 25, then 24. Each time he used the data from the previous guess to think about if he needed to increase his guess or decrease to something between the previous two guesses.

Adam S. from Highlands Elementary School may have had a lucky first guess or he may have thought hard about a reasonable starting number… he doesn’t say. Either way, he started with 25 as first guess, realized it was too high, adjusted to 23 for his second guess, realized it was too low, and got to 24 on his third guess. Take a look at his work, below:

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In this week’s gift-themed Pre-Algebra Problem of the Week, students had to figure out an original cost given some percentages of the original cost. Specifically, they knew that five children contributed to pay for one big gift for their mother. Two of the (relatively broke) children paid $75 each. Of the remaining three, one paid 20% of the cost, one paid 25% of the cost, and the last paid 1/3 of the cost. Then the question was, how much did the whole thing cost? (A more realistic question might have been, “if the whole thing cost $692.31, did the last person to pay really contribute 1/3 of the cost like she planned to?” but it’s more algebraic reasoning this way).

Anyway, for me, the one reading the student solutions to the problem, the interesting challenge was to figure out what went wrong when students got answers other than $692.31. Were there conceptual struggles? Problems thinking up methods to get to the answer? Or problems executing the chosen paths? I always look for common wrong answers, because those usually show me conceptual struggles. This week, the common wrong answer was $681.82 (or sometimes just $680, or $681). I thought, “huh, I wonder why students are getting that answer…” Why do you think they might be? Read More→

I had such fun with this the table gallery we made with the Falling Leaves PoW, that I thought I’d do another one. This time, rather than share my thoughts, I’m curious to hear what you notice and wonder about each table. Students worked to solve the logic problem “Happy Thanksgiving” (to view the problem, you can sign up for a free Math Forum trial account, or if you’re already a member, just log in).

Enjoy!

From Jakub Z. at IS 141Q The Steinway [his classmates Ricardo M. and Ahbab A. also submitted great versions of similar tables]

Grandchildren	Who they can work with	Who they want to work with
Travis	Michael,Vinolia,Nicholas,Jen,Clark	anybody
Michael	everybody except for Stephen	will work with anybody if Vinolia helps too
Vinolia	Jen,Nicholas,Vinolia,Michael	everybody except for Jen
Nicholas	everybody	wont work without Funmi
Jen	everybody except for Funmi,Stephen,and Vinolia	anybody
Clark	everybody	Stephen and Nicholas only if stephen works
Stephen	everybody	everybody except for Travis,Michael,Vinolia and Jen
Funmi	everybody	Travis,Jen and Vinolia unless Stephen works too.

From Shayden S. and Avery B. at W. P. Sandin

	T	M	V	N	J	C	S	F
T	A	A	A	A	A	A	A	A
M	A	A	O	A	A	A	A	A
V	A	A	A	A	X	A	A	A
N	A	A	A	A	A	A	A	O
J	A	A	A	A	A	A	A	A
C	A	A	XUN	XUS	A	A	A	A
S	X	X	X	A	X	A	A	A
F	X	OOC	XUS	A	X	OOM	A	A

Legend:

A- yes

X- no

O- needs person to work

XU(initial)- no unless (initial helps)

OO(initial)- needs person or someone else

From Gaetano I. at IS 141Q The Steinway

funmi	clark	nicholas	stephan
micheal + clark	nicholas +stephan	funmi	micheal
just clark	just stephan	travis	vinolia
vinolia	vinolia	jen	funmi
stephan	stephan	clark	clark
nicholas	funmi	vinolia	nicholas
vinolia	clark	micheal
	travis
	jen	stephan

From Jacob D. at Patriot Elementary

Washing Dishes	Michael	Vinolia	Jen	Nicholas	Funmi	Clark	Stephen	Travis
Travis>	+	+	+	+	+	+	+	+
Michael		+
Vinolia			-
Jen	+	+	+	+	+	+	+	+
Nicholas					x
Funmi	x					x
Clark		@		/@			/
Stephen	-	-	-					-

From Jogdand A. at Seminary Hill School

The Algebra Problem of the Week that just finished up was about that classic playground game of trying to balance two different-sized people on a teeter totter. How many of you had an “aha!” moment on the playground one day when you realized that by sitting farther away from the fulcrum (the point where the seesaw balances on the base) means you can balance with a kid a lot heavier than you? A grown-up even? Or that you could leave your equal-sized friends dangling off the ground by sitting as far away from the fulcrum as possible?

In this PoW we asked students to work with the mathematized version of the situation. Instead of borrowing some small children and heading out to the playground to find a balancing point, we gave submitters a relationship between weights and distances from the fulcrum:

(weight 1)*(distance 1) = (weight 2)*(distance 2)

Most submitters chose to work with the equation we gave them in some fashion. Henry S. from Stony Point Elementary used guess and check to find the missing distances:

M	32*36 = 1,152	32*30 = 960	32*20 = 640	32*100 = 3,200	32*80 = 2,560	32*90 = 2,880
L	40*18 =  720	40*12 = 480	40*2 = 80	40*82 = 3,280	40*62 = 2,480	40*72 = 2,880

Other students, like Natalia N. from Sekolah Ciputra, thought of the problem as having two variables:

x=3/2+y
balance:
32x=40y
32(3/2+y)=40y
48+32y=40y
48=8y
y=6+t
x=7.5+t

I don’t know what the +t represents in Natalia’s solution. Perhaps it has to do with the fact that we don’t know how long the teeter-totter is. What do you think she might be using the t for?

Most submitters immediately thought of writing the two distances as x and x + 1.5, which impressed me because I don’t always think to ask, “how else could I write this quantity? What relationships could I use to write it in terms of another quantity?”

All of the submissions above mathematized the situation in the way we had suggested. But those weren’t the only kinds of responses. Some submitters did a lot of hard thinking about the physics involved, without getting into the mathematization, like Aspen K. from East Middle School:

They can both be balanced on it even though they don’t weigh the same. They can do this by the girl that is heavier sit on more of the end of the seesaw. So then the girl on the other end who must be lighter will sit closer to the middle. By doing that they should both be balanced out.

If I were Aspen’s teacher, I don’t think I would start my conversation with the mathematizing. Aspen is wrestling with the physics involved, and so I would head out to the playground or get out some see-saw models and start trying to balance lighter and heavier things. Once we had established the indirect relationship between weights and distances, then it would be time to start asking, “what are the quantities involved? How do they seem to be related?”

Finally, there were the students who chose to mathematize the problem in unexpected ways. These led to the solutions that were the most interesting and challenging for me, forcing me to look at the problem in totally new ways!

For example, Jeremy Z. and Steven M. from Cold Springs Elementary School used ratios to think about the problem, and never mentioned the equation with weights and distances. They may have used it to help them think of why if the weights are in a 4:5 ratio, the distances must be in a 5:4 ratio, but they didn’t explain that step of their reasoning so I’m not sure…

the ratio of the weights of them are Marnie:Lex=4:5. So the distance must be 5:4. since Marnie is sitting 1 1/2 feet further away from the middle of the seesaw then Lex. if 1 1/2 feet is 1 in difference in the ratio, then it is 90:72(inches).4*90=5*72. They both equal 360.

There were also some submissions that mathematized with some inaccuracies, or different theories about how the problem was working. Taylor H. from East Middle School focused on where to position students on a board of a given length so that one is 1.5 feet closer to the center. That was some good thinking! Several students focused on the length and how to arrange the students so they are 1.5 feet from the middle, ignoring the weight information. As Taylor worked the weight information in, I got a little confused. I think it’s neat that Taylor was trying to account for all of the quantities in the problem, and I would love to hear more from Taylor about if the weights were especially hard to think about and what they noticed about the weight information in the story:

Lets say the seesaw was 15 ft. long i subtracted 1 1/2 from that and got 13.5…. Then i had to divide that by two to get 6.75… It says that the girls weights are 32, and 40… I had to divide those by 6.75…… for Marnie it was 270.00 in. ……….. and finally for Lex it was 216.00 in. ………

Last but certainly not least, is this submission from Karrie P., also at East Middle School.

How i figured out this problem is I knew that Marine is 32 pounds and Lex is 40 pounds….. wich is a 8 pound difference. So, If Marine is 1 1/2 away from the fulcrum, than if you subtract 1/8, because there is a 8 pound differnce, from 1 1/2 it would be 1/2 away from the fulcrum for Lex. So…… over all, Marine is 1 1/2 away from the fulcrum and Lex is 1/2 away from the fulcrum.

I think it’s really neat that Karrie has the idea of reciprocal fractions (an 8 pound difference should result in a 1/8 difference in the distance) and I wonder where she got that idea. Many students wonder about the different ways to compare two quantities (should I subtract or divide? Why?) and I think that would be a neat conversation to have with Karrie (“how can you compare 32 pounds and 40 pounds? What’s the difference?”) Clearing up the fact that Marnie is 1 1/2 feet further away from the fulcrum seems like the last step to me, although it is still related to the whole idea of comparing numbers and the language we use with different types of comparisons.