# Find the Four digit number

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Find the Four digit number
As an intro activity to Algebraic thinking, have the participants complete Handout 3. This will begin a discussion of language used in the classroom. Sample answer: 9837 Since the thousands digit is three times the tens digit, the tens digit can be 3, 2, or 1. Since the sum of the four digits is 27, and the greatest four-digit odd number with different digits is 9875 and its digits sum to 29, we will start with the tens digit as a 3 and the thousands digit as a 9. Since 9+3=12, then 12 of the allowed sum of 27 is used. Because the number is odd and the sum of the two digits remaining needs to be used in the ones place. The three digits used—9, 3, and 7–- sum to 19. Therefore, 8 is left for the hundreds digit, yielding All requirements of the problem have been met. On separate sticky notes write the digits 0 – 9. A special four-digit number has the following traits: All the digits are different The digit in the thousands place is 3 times the digit in the tens place. The number is odd. The sum of the digits is 27. What is the special four-digit number? NCTM (2006). Mathematics Teaching in the Middle School, Volume 11, number 6. Handout 3

Expressions & Equations
Let participants write answers to the statements one at a time. Discuss their responses. Click for the answer before going on to the next question. Discuss special situations about order of operation, etc. Would students agree that all of the balanced equations are correct? They often think that the ANSWER must follow the equal sign…such as they may not think = 12 – 2, because = 10. EXPRESSIONS for 10: – X ÷ 3 1 x ÷ ÷ 4 + 5 Write expressions representing 10. EQUATIONS equaling 10: 5 + 5= – 2= x 5= ÷ 3=10 1 x 6 + 4= ÷ 3= ÷ 4 + 5=10 Write equations equaling 10. Use the above to write balanced equations. BALANCED EQUATIONS: 5 + 5 = 12 – x 5 = 30 ÷ 3 1 x = ÷ ÷ = 5 + 5

Order of Operations and Balanced Equations
Use number and operation tiles to build “Balanced Equation Puzzles” Prior to class, you should print, laminate and cut the tiles provided. There should be 2 colors of tiles per each group of 2 or 4. Each color should include at least: 3 sets of numbers 0-9; 3 sets of operation symbols +, - X, and ÷; one set of ( , ) on separate cubes; and 4 equal signs. You should prepare enough for each person to have a set or at least a set for each pair of participants. Using a different color for each set is a good organizational method, because if one is on the floor, etc. you know where it goes. Follow the direction on the screen, additional directions are on the handout. What concepts do students gain from this? ---Correct use of operational symbols ---Balanced equations ---Order of operations 1 X 6 + 4 8 + 6 ÷ 3 = Separate the number tiles from the operation tiles. Draw 6 numbers and 2 operations. Put down an equation or draw until an equation is possible. SLE….NO , 2.7.3, 3.7.2, 2.8.1, 2.8.4, 3.8.2 6 5 Draw to replace the tiles played. If no tiles are played draw a tile from either pile and advance to the next person. 6 + 1 = 1 4 ÷ 2 2 3 8 = 8 - 7 1 6 + 2 x 3 = 6 ( 2 ) + 1 ÷ = 1 Handout 4

ALGEBRA…. Decisions, Decisions, Decisions
Ask participants to complete the problem on Handout 5. Discuss their strategies, the WHYS and HOWS of the outcome. Dear Bill, Today is my 55th birthday. I have decided to give away some of my money each year to my relatives. You may choose one of the following options: Option 1: \$100 dollars now, \$90 next year, then \$80 the year after, and so on. Option 2: \$10 dollars now, \$20 next year, then \$30 the year after, and so on. Option 3: \$1 now, \$2 next year, \$4 the year after and so on doubling each year. You will only receive money until I retire. Write to me soon and tell me how you want your money. With Love, Aunt Judy Handout 5

Decisions, Decisions, Decisions
Dear Bill, Today is my 55th birthday. I have decided to give away some of my money each year to my relatives. You may choose one of the following options: Option 1: \$100 dollars now, \$90 next year, then \$80 the year after, and so on. Option 2: \$10 dollars now, \$20 next year, then \$30 the year after, and so on. Option 3: \$1 now, \$2 next year, \$4 the year after and so on doubling each year. You will only receive money until I retire. Write to me soon and tell me how you want your money. With Love, Aunt Judy 1. Bill starts to figure out how much he will receive from the three options If Judy retires at age 59 which option do you think Bill should choose? Justify your answer. 2. Bill thinks that his Aunt Judy is likely to wait until she is 65 years old to retire. What is his best choice option for her retirement at age 65? Justify your answer 3. Since Bill is not sure what age Aunt Judy will retire, how do you think Bill should reply to Aunt Judy’s letter? Explain clearly which option you think Bill should choose. Show your reasoning. Name: _________________________________

BULGING BACKPACKS Some questions to ponder…

Calculator Instructions for TI 84
To clear data in all lists: Press 2nd MEM. Choose ClrAllLists. Press Enter. Press Enter. You will see DONE Choose 7 – Reset Choose 2 – Default Choose 2 - Reset. Enter your data in lists: Press STAT. Enter the estimates in L1. Press ENTER after each number. Enter actual weights in L2. Press ENTER after each number.

To create a scatter plot:
Press 2nd Y= (STAT PLOT). Press ENTER for Plot 1. Press ENTER for On. Arrow down to Type: Press ENTER for Scatter Plot. Arrow down to Xlist: Press 2nd STAT and choose L1. Arrow down to Ylist: Press 2nd STAT and choose L2. Arrow down to Mark: Press ENTER for the box. Press WINDOW. Set the appropriate window for this problem. Press TRACE and use the right and left arrow keys.

To create two box-and-whisker plots:
Press 2nd Y= (STAT PLOT). Press ENTER for Plot 1. Press ENTER for On. Arrow down to Type: Press ENTER on the box and whisker icon. Arrow down to Xlist: Press 2nd STAT and choose L1. Press ENTER. Set Freq to 1. Press WINDOW. Set the appropriate window for this problem. Press TRACE and use the right and left arrow keys to see the values for each quartile. Repeat steps 1-7 choose Plot 2 and L2 .

Some questions to ponder…

BULGING BACKPACKS Overview: Students will estimate and find the actual weight for each of their backpacks. The class constructs a scatter plot to display some of the data; then each student creates a personal scatter plot. This lesson gives students experience plotting points on a Cartesian plane and interpreting those points in terms of a real-world situation. The function y=x is used to describe the meaning for points above and below this line. With numerous similar experiences, students develop an intuitive understanding of function and correlation. Prerequisite Concepts and Vocabulary: Finding explicit and recursive rules for patterns Graphing points Scaling axes for specific data Understanding the vocabulary function, scatter plot, coordinate, variable, and axis Time: This lesson takes approximately 90 minutes. Materials: Bathroom scales Student backpacks Large sheet of graph paper for class scatter plot 1 sheet of graph paper per student Graphing Calculator (optional)

BULGING BACKPACKS Lesson Plan: Discuss plan with the class Each student makes a table and records his or her own data (Table included). Students make estimates for only three backpacks, then find their actual weight, to help students get some general idea of how much backpacks weigh. Students use the bathroom scale to weigh each of the backpacks to the nearest half-pound. For the class data have each student record their estimate and actual weight on a table on the board. Label the x-axis actual weight and the y-axis estimates.(This allows us to talk about the points above the line y=x being overestimated and the points below being underestimated as well as anyone on the line as being perfect) Next assign the individual scatter plots for their estimates of all backpacks. The next three can be done as a class on the class scatter plot or assigned for each individual to write about. Draw the function y=x+2 and discuss what that tells us. Draw the function y=x-3 and discuss what that tells us. Draw the function y= 7 and discuss what that tells us. Extension Using the graphing calculator, create a box and whisker plot for estimates in Plot 1 and another box and whisker plot in Plot 2. Display both plots in the same window. Discuss the differences in the range of the estimates and the actual weights. Check for outliers and compare the statistics for estimates and actual weights using the trace feature of the calculator. Adapted from: Lawrence, Ann, and Charlie Hennessy. Lessons for Algebraic Thinking, Grades 6-8. Math Solutions Publications, 2002.

Name: _____________________________
BULGING BACKPACKS…STUDENT WORKSHEET Name Actual Weight Estimate

How much does each shape weigh? Explain.
Do handout 9. Have cylinders, spheres, and cubes available to use as manipulatives. If you do not have these you can use 3 different colored counters or linking cubes to represent each shape. Teachers need to be encouraged to use the manipulatives, this way they will encourage the students to use them. After they finish take time to discuss each problem. Click and……. Example 1: Most weight is the cylinder; least weight is the cube 2 spheres = 1 cylinder….and….3 cubes = 1 sphere….therefore….6 cubes = 1 cylinder Example 2: 1 cube = 2 cylinders …so… 1 cube + 1 cylinder = 3 cylinders 3 cylinders = 6 spheres…therefore….1 cylinder = 2 spheres 1. Most weight is the cylinder; least weight is the cube 2 spheres = 1 cylinder….and 3 cubes = 1 sphere….therefore 6 cubes = 1 cylinder 2. What will balance two spheres? Explain. 1 cube = 2 cylinders …so… 1 cube + 1 cylinder = 3 cylinders 3 cylinders = 6 spheres…therefore 1 cylinder = 2 spheres From Teaching Student Centered Mathematics P. 280 Handout 9

From Teaching Student Centered Mathematics P. 280
3. Manipulatives: Wooden cubes, spheres, cylinders (or three other types of items to represent these) Example 3: 2 cubes + 1 sphere = 8 and 3 spheres = 12 therefore 1 sphere = 4 and 2 cubes + 4 = 8, 1 cube =2 Example 4: 1 cylinder + 1 cube = 13 and 2 spheres + (1 cylinder + 1 cube) = 21 therefore substitution 2 sphere + 13 = 21, 2 spheres=8, 1 sphere = cube + 2 spheres = 14, 1 cube + 8 = 14, 1 cube = cylinder + 6 = 13, 1 cylinder = 7 Example 5: Equation 1: 1 sphere + 1 cylinder = Equation 2: 1 sphere + 1 cube = 6 Since 7 is one more than 6 and the spheres are the same, then 1 cylinder = 1 cube + 1 Equation 3: 1 cylinder + 1 cube = Substitute: (1 cube + 1) + 1 cube = 6, 2 cubes + 1 = 9, 2 cubes =8, Therefore 1 cube = 4 Equation 2 Substitute: 1 sphere + 4 = 6, 1 sphere =2 Equation 1 Substitute: cylinder = 7, 1 cylinder = 5 2 cubes + 1 sphere = 8 and spheres = 12 therefore 1 sphere = cubes + 4 = 8, cubes =4, cube = 2 4. 1 cylinder + 1 cube = spheres + (1 cylinder + 1 cube) = 21 Use substitution: 2 spheres + 13 = 21, spheres = 8, spheres = 4. 1 cube + 2 spheres = 14, cube + 8 = 14, cube = 6. 1 cylinder + 6 = 13, cylinder = 7. 5. Equation 1: 1 sphere + 1 cylinder = Equation 2: 1 sphere + 1 cube = 6 Since 7 is one more than 6 and the spheres are the same, then 1 cylinder = 1 cube + 1 Equation 3: 1 cylinder + 1 cube = Substitute: (1 cube + 1) + 1 cube = 6, cubes + 1 = 9, 2 cubes =8, Therefore 1 cube = 4 Equation 2 Substitute: 1 sphere + 4 = 6, sphere = 2 Equation 1 Substitute: cylinder = 7, cylinder = 5 From Teaching Student Centered Mathematics P. 280

How much does each shape weigh? Explain.
Name: _______________________________________ Student Worksheet How much does each shape weigh? Explain. Do handout 9. Have cylinders, spheres, and cubes available to use as manipulatives. If you do not have these you can use 3 different colored counters or linking cubes to represent each shape. Teachers need to be encouraged to use the manipulatives, this way they will encourage the students to use them. After they finish take time to discuss each problem. Click and……. Example 1: Most weight is the cylinder; least weight is the cube 2 spheres = 1 cylinder….and….3 cubes = 1 sphere….therefore….6 cubes = 1 cylinder Example 2: 1 cube = 2 cylinders …so… 1 cube + 1 cylinder = 3 cylinders 3 cylinders = 6 spheres…therefore….1 cylinder = 2 spheres 1. 2. What will balance two spheres? Explain. From Teaching Student Centered Mathematics P. 280 Handout 9

From Teaching Student Centered Mathematics P. 280
Manipulatives: Wooden cubes, spheres, cylinders (or three other types of items to represent these) Example 3: 2 cubes + 1 sphere = 8 and 3 spheres = 12 therefore 1 sphere = 4 and 2 cubes + 4 = 8, 1 cube =2 Example 4: 1 cylinder + 1 cube = 13 and 2 spheres + (1 cylinder + 1 cube) = 21 therefore substitution 2 sphere + 13 = 21, 2 spheres=8, 1 sphere = cube + 2 spheres = 14, 1 cube + 8 = 14, 1 cube = cylinder + 6 = 13, 1 cylinder = 7 Example 5: Equation 1: 1 sphere + 1 cylinder = Equation 2: 1 sphere + 1 cube = 6 Since 7 is one more than 6 and the spheres are the same, then 1 cylinder = 1 cube + 1 Equation 3: 1 cylinder + 1 cube = Substitute: (1 cube + 1) + 1 cube = 6, 2 cubes + 1 = 9, 2 cubes =8, Therefore 1 cube = 4 Equation 2 Substitute: 1 sphere + 4 = 6, 1 sphere =2 Equation 1 Substitute: cylinder = 7, 1 cylinder = 5 3. 4. 5. From Teaching Student Centered Mathematics P. 280

Arkansas Benchmark Questions
For the four Benchmark questions, find them in the FACILITATORS’ FOLDER and print them four to a page. Pass them out to the participants, making sure that at least two people have the same question. Everyone can have more than one question, but there probably won’t be time for everyone to do all the problems. Give them time to work their problem(s), individually. Let the two that have the same problem explain how they worked it. (Often they will have different strategies and this is important to see). 7th grade: answer is D. 8th grade: answer is C. 7th grade 8th grade Handout 10

7th grade Open Response 2007 1. 3x + 2x + x=180 3x + 2x + x=180
Let participants that worked this problem explain their strategies. Allow further discussion. Point out that it is good to divide the work area into 3 sections. Steps will appear as you click. See Handout 11 for more explanations. 1. 3x + 2x + x=180 3x + 2x + x=180 6x = 180 x = 30 Angle x = 300 Angle 2x = 2(30) = 600 Angle 3x = 3(30) = 900

8th grade Open Response Pam and Greg are each building a pyramid of tiles. The number of tiles needed is represented by the rule , where n is the number of levels in the pyramid. The pattern for the pyramid is shown below. Let participants that worked this problem explain their strategies. Allow further discussion. After discussions click and steps will appear. See Handout 11 for more explanations. n(n+1) 2 Draw the next pattern in the sequence. How manytiles would be in a 10-level pyramid? Show your work. 3. Greg has 4-inchtiles, and Pam has 2-inchtiles. They are each going to build a 24-inch tall pyramid. Greg predicts he will need half as many tiles as Pam since his tiles are twice as large. Compare the pyramids to see why Greg is incorrect. 55 See handout 10 See handout 10

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