Elementary POW, February 12-16, 1996
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****************************************************** Elementary Problem of the Week, February 12-16,1996 This week's problem was submitted by Mr. Grant Whitaker's class (Form 4 Set One), Underhill School, Maidstone, Kent, England. Here is a problem for you from my class: Find the dimensions of all rectangles that have area equal numerically to twice the perimeter, where both the length and the width are whole numbers. ***************************************************** This week's Bonus Puzzler was submitted by Sam Hahn, a student in Ms. McCarthy's 4th grade class at Center School, Stow, MA. Using the pogs shown below, can you make two even rows of four pogs each by moving only one pog? O O O O O O ******************************************************
******************************** Joyce Kilmer School, Mahwah, NJ Mrs. Bach's 5th grade math class Amy Killough (POW), Hara Rosen (POW), Kiren Khanduja (POW), Jason Kondo, Brett Freeman, and Dave Rounds (Bonus) Center School, Stow MA Mrs. Pensa's 3rd grade class Andrew Bawn (POW), Chris Wade (POW & Bonus), Kevin Maloney (Bonus), Kristin Squires (POW & Bonus), Ted Powers (POW) Center School, Stow MA Ms. McCarthy's 4th grade class Sam Hahn (POW & Bonus), Joey Morrissey (Bonus) Bagnall School, Groveland, MA Grade 4, Mrs. Sturtevant Rudianne Subatch and Suzanne Carroll (POW) Drexel Hill School of the Holy Child, Drexel Hill, PA Grade 6 - Mrs. Brennan Kim Fugok, Ryan Grace, Christine McGowan, and Amanda Tumminelli (POW & Bonus) Bagnall School, Groveland, MA Grade 4, Miss Seager Nick Stephenson IHM School, Philadelphia, PA Lisa Simmons, Grade 6 (POW) Munsey Park School Manhasset, NY Miss Duggan's fourth grade Dan, Matt, Ashley and Willy (POW) Georgetown Day School, Washington, D.C. Teacher: Paul Nass, Third Grade Aneil Baron Kerr Elementary, Pittsburgh, PA Fred Rimmel, Computer teacher - 4th grade Todd Feiler, Evan Mossman, Matt Spindler (POW) Leal School Urbana, IL Grade 4 Lucie, Stephen, Marie (Bonus)
***************************** Mrs.Bach's 5th Grade Math Class Joyce Kilmer School Mahwah, NJ Dear Ruth, This was really a hard problem. The first solutions were found by trial and error. None of the students were able to come up with a pattern to build on for further solutions, so they were stuck with trial and error. C. Bach Amy Killough Answer: 8 by 8 Area = 64 Perimeter = 32 32 + 32 = 64 I got 8 by 8 by first taking 2 by 2. I then kept adding a number above. For example next I would try 3 by 3, then 4 by 4, and so on. Sometimes between the numbers I would try a mix. Like, 3 by 4 so 4 by 5 and so on. Hara Rosen Answer: 20 by 5 Area =100 Perimeter = 50 50 + 50 = 100 I got 20 by 5 by first thinking 10 by 5. Then I noticed that I was10 off so I doubled 10 and got 20. I tried 20 by 5 and got my answer. Kiren Khanduja Answer: 12 x 6 Area = 72 Perimeter = 36 36 + 36= 72 I took Hara Rosen's answer and realized that both numbers were multiples of each other so I took 12 and 6 and got the answer above. Our names are Jason Kondo, Brett Freeman, and Dave Rounds. We're 5th graders in Mrs. Bach's math class in Joyce Kilmer School. To figure out the pog answer, we to take the bottom pog and put it on top of the upper left position. By doing that, we had two equal rows of 4 pogs each. Then we counted 2-1-1 down and 2-1-1 across. 000 2-000 0 0 0 0 0 0 ***************************** Andrew Bawn - Mrs. Pensa's 3rd grade class - Center School - Stow MA First I used tiles then I used my calcalater. I got 8X8 = 64. ****************************** Chris Wade - Mrs. Pensa's 3rd grade at Center School - Stow MA First I tried a 6 by 6 square and Mrs.Pensa said that 6 by 6 was too small so I tried 7by 7 and that was to small to. So I tried 8by8 and i found out that 8 by8 worked by using tiles. bonus puzzler: 1 2 3 4 5 6 Move # 3 to below number 6 and you will have two rows of four: 1, 4, 5, 6 and 4, 5, 6, 3! ****************************** Joey Morrissey Grade4 mrs. McCarthy Center School Bonus Puzzler I solved the problem by doing all the combinations possible and found that if I took the bottom pog and put it on top of the corner pog it worked. ****************************** Kevin Maloney - Mrs. Pensa's 3rd grade - Center School - Stow MA I tried all different ways of placing the pogs and finally came up with this plan: 1 2 3 4 5 6 Move # 3 to above # 1 so there are five pogs in the row. 3 1 2 4 5 6 You then have four in a row - 3, 1, 4, 5 and 6, 5, 4, 1 ****************************** Kristin Squires Mrs. Pensa's third grade.center school. Stow MA I solved the problem of the week by getting some graph paper and then i said if it was 1x1 there would be no inside so then I went on to 2x2 but that didn't work either so I kept on following this method until Igot to eight by eight and then I said that that would work so an 8x8 square is my answer. and my anser to #2 is that you would move #1 on top of number four. 4 5 6 3 2 1 ****************************** Sam Hahn - Miss McCarthy's 4th grade - Center School Stow MA How I solved the problem is I said that we could call width "w" and call length "l". First I tried to look at some of the combinations that might work but when that did'nt work. I then thought of a brut force way to do the problem I could start with one and go up to twenty and if there were no answers I would go to two and start all over agan. When I found out how long it was going to take me, I asked my dad if we could make a program that would check all the combinations from one to one hundred and this is what we worked out together. // Sam's and his Dad's program to solve the area and perimeter problem #include#include #include void main( void ) { // let w represent the width // let l represent the length int w, l; // lets pick values for w starting at 1 to 100 // and for each value of w we will pick a value for // l from 1 to 100 for (w=1;w<=100;w++) for (l=1;l<=100;l++) // using these values for width and length compute the area // compute the perimeter times 2 and if the results are equal // we have an answer so print it out if ( (w*l) == (4 * (w+l)) ) printf("length %d width %d area=%d perimeter=%d\n",l,w,l*w,2*(w+l)); } // the answers are: length 20 width 5 area=100 perimeter=50 length 12 width 6 area=72 perimeter=36 length 8 width 8 area=64 perimeter=32 Bonus Puzzler was submitted by Sam! ****************************** Ted Powers - Mrs. Pensa's 3rd grade class - Center School - Stow MA I tried different numbers for the length and the width. I tried 1s, 2s, 3s, 4s, 5s, 6s, 7s , etc as a rectangle and couldn't figure it out. Then I tried using the numbers as a square and when I got to 8 x 8 it worked because 8 + 8 + 8 + 8 = 32 and 8 x 8 = 64 and 64 is twice as large as 32. I also tried 9 x 9 and 7 x 7 but they didn't work. ****************************** Rudianne Subatch and Suzanne Carroll Grade 4, Mrs. Sturtevant Bagnall School, Groveland, MA We got 2 answers. One rectangle with a length of 8 and a width of 8 with a area and twice the perimeter equaling 64. Also a rectangle with a length of 0 and a width of 0 with an area and twice the perimeter equaling 0. We solved this by drawing a picture on graph paper. We tried different numbers and counted the perimeter of each and to blocks inside to find the area. ***************************** Kim Fugok, Ryan Grace, Christine McGowan, and Amanda Tumminelli Grade 6 - Mrs. Brennan Drexel Hill School of the Holy Child, Drexel Hill, PA We were only able to find one solution to the problem. If the length is 8 and width is 8 then the area is 64 sq units and the perimeter is 32 units. We used a guess and check method to find our answer. Bonus Solution: 1 5 6 2 3 4 We numbered each pog. We used several sets of dice to represent the pogs. After moving the pogs around we noticed that at least one pog would be in both rows. So we started moving the pogs to different positions. We finally found our solution by moving pog number 4 and placing it on top of pog number 1. By doing this you have two rows with 4 pogs. Row one has pog numbers 4, 1, 2, and 3 and row two has pog numbers 4, 1, 5, and 6. *********************** Nick Stephenson Grade 4, Miss Seager Bagnall School, Groveland, MA One square, 8x8 has an area and twice the perimeter equaling 64 and one rectangle, 12 x 6 has anarea and twice the perimeter equaling 72. That's all I found. First I used guess and check. Then I learned if the perimeter and the area are the same, just double the length and the width and the new area is equal to twice the new perimeter. Meghan L. Grade 4, Miss Seager Bagnall School, Groveland, MA Puzzler: O O O Call them 1 2 3 O O 4 5 O 6 Move number 5 on top of number 1. Then both rows will have 4. ************************* DEAR MRS.CARVER, HERE IS MY ANSWER TO THE PROBLEM OF THE WEEK, ________20__________ I I I I 5 I___________________ I 20+20+5+5= 50 perimeter 20 *5= 100 area 100-50=50 Lisa Simmons IHM School, Gr 6 Philadelphia, PA ********************** Hi! this Is Dan, Matt, Ashley and Willy,we are in Miss Duggan's fourth grade class in Munsey Park. We are answering your problem. At first it was very hard to read it because it had so many hard words, like numerancy, but we finally understood it. First we started working with things that are not rectangles, but then we read the problem again and found out that it had to be a rectangle. Sometimes we thought we had it but it wasn't a rectangle. The answer is: an 8x8 square. Your math solvers, Dan,Matt,Ashley and Willy ************************ Let a,b be the sides of the rectangles. ab = 4 (a+b) which is 1/b=1/4 - 1/a But a and b must be whole numbers. Therefore a = 5,6,8,12,20, and in this order b = 20,12,8,6,5 Aneil Baron Georgetown Day School, Washington, D.C. Third Grade Teacher: Paul Nass *********************** Todd Feiler, Evan Mossman, Matt Spindler 4th Fred Rimmel, Computer teacher Kerr Elementary Pittsburgh, PA We used guess, test and revise. We found only two answers: an 8 x 8 square and a 6 x 12 rectangle. Fred Rimmel ********************** X-Sender: Leal-Students@lynx.ed.uiuc.edu (Unverified) Date: Thu, 22 Feb 1996 09:58:57 -0600 To: ruth@mathforum.org From: Leal-Students@lynx.ed.uiuc.edu (Lucie) Subject: Elementary POW Dear Ms. Carver, Our names are Lucie, Stephen, and Marie. We are in 4th gr. at Leal School in Urbana, IL. We have found the awnser to the Elementary POW of January 26. It is: You move the last pog in the vertical row and you put it on the pog that connects both rows, so it makes each row have four pogs. Sincerely, Lucie, Stephen and Marie *********************
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