By Sunday, October 17th, please respond to the post with:
Your name and the name of your group members.
What operation that you all analyzed.
Your notes on what each student did to answer the question.
An explanation of why the student’s process works and how it relates to the standard algorithm.
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Group 2:
https://docs.google.com/document/d/18IDJUCv1Y8FJNp1632UJB6fzo3G5D-eLb46ZReedhfY/edit
Group 3
https://docs.google.com/document/d/1n8Dsebc_fShBKImUikHx08v2Byu2a3pt_HG6xhORi2U/edit?usp=sharing
Group 1 Link
https://docs.google.com/document/d/1z8kjie1bsgJg1MFDuup4QBW5DXygZK9ArPfkL8m5qh0/edit?usp=sharing
https://docs.google.com/document/d/1z8kjie1bsgJg1MFDuup4QBW5DXygZK9ArPfkL8m5qh0/edit?usp=sharing
https://docs.google.com/document/d/1_VzAzpK86jf_LJ09SU_ucjmLQxD9l47UtkJwaeiIRvs/edit?usp=sharing
Dawn Kebrich, Danny Zhang, Maria Turcios
We analyzed Multiplication.
Dawn: Emily broke it down using the distributive property which is extremely helpful and works because if the problem is too difficult or confusing you can simplify the problem as much as you need to. It relates to standard algorithm because distributive property is like problem-solving which is just like the algorithm theory. Emily broke it down using distributive property because visually it is better to understand. First she took 12 and separated it by 10 and 2 which then she did 10 times 13 which is 130 and 2 times 13 which is 26 which equals 156. 43 she broken down to 20 and 20 and 3 then times everything by 62. Which gave her 1240, 1240 and 186 which gave her total of 2666. Growing up I always used the distributive property like Emily because it was easier for me to solve equations and problems.
Maria: Tabitha used Lattice to obtain her answer. She set the first problem on the top of the box and then the second set of numbers on the right side placed vertically. Each digit of the number placed near a specific box either it on the very top or on the right side of the box, demonstrating a use of regrouping. Tabitha Multiplied the 2×1= 2, and 2×3=6. The 2, having to be multiplied with each digit on the right side. When reaching to number that a single one’s digit, she places a 0 on top of that first individual box. She then moved on to the left column of the box 1 and multiplied with the numbers on the right side. She continued to carry the rule, as she placed the 0 on top on the individual box. After multiplying each set the number pertains in, she then added all the numbers going diagonally.
Danny: Sasha used the partial product algorithm. For the first problem, she split 13 into 10 and 3 and 12 into 10 and 2. She then multiplies 3 with 10 and 2 and then multiplies the 10 from 13 with the 10 and 2 from 12. Finally, she adds all her products up to come to the answer of 156. She essentially did the same thing for the second question, where she splits up 43 into 40 and 3 and 62 into 60 and 2. She then multiplies 2 with 3 and 40 and then multiplies 60 with 40 and 3. Lastly, she adds all her products up to come to the answer of 2666. Compared to the standard algorithm, the partial product algorithm is very similar. Both algorithms split the multiplicand up into separate place values, but only the partial product algorithm splits the multiplier in addition.
Group 4: Rebecca Dhanpaul, Alison Huang and Jasmine Santiago
Operation: Division
https://docs.google.com/document/d/1iBWRwbR7GUsz3pLgDARt-Il1qUuV7QynXVM5037fGS8/edit?usp=sharing