Welcome to Everyday Math! Riverside Elementary Math Night February 10, 2009.
Everyday Math and Algorithms
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Everyday Math Everyday Math and and
AlgorithmsAlgorithms
A Look at the Steps in A Look at the Steps in Completing the Focus Completing the Focus
Algorithms Algorithms
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Partial SumsPartial SumsAn Addition An Addition AlgorithmAlgorithm
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268+ 483
600Add the hundreds (200 + 400)
Add the tens (60 +80) 140Add the ones (8 + 3)
Add the partial sums(600 + 140 + 11)
+ 11751
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785+ 6411300Add the hundreds (700 + 600)
Add the tens (80 +40) 120Add the ones (5 + 1)
Add the partial sums(1300 + 120 + 6)
+ 6
1426
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329+ 9891200 100
+ 18
1318
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The partial sums algorithm for The partial sums algorithm for addition is particularly useful for addition is particularly useful for adding multi-digit numbers. The adding multi-digit numbers. The partial sums are easier numbers to partial sums are easier numbers to work with, and students feel work with, and students feel empowered when they discover that, empowered when they discover that, with practice, they can use this with practice, they can use this algorithm to add number mentally.algorithm to add number mentally.
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An alternative subtraction An alternative subtraction algorithm algorithm
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When subtracting using this algorithm, start by going from left to right. 9 3 2
- 3 5 6 Ask yourself, “Do I have enough to subtract the bottom number from the top in the hundreds column?” In this problem, 9 - 3 does not require regrouping.
12
13
Move to the tens column. I cannot subtract 5 from 3, so I need to regroup.
12 8
Now subtract column by column in any order
5 6 7
Move to the ones column. I cannot subtract 6 from 2, so I need to regroup.
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Let’s try another one together
7 2 5
- 4 9 8
15
1211 6
Now subtract column by column in any order
2 7 2
Start by going left to right. Ask yourself, “Do I have enough to take away the bottom number?” In the hundreds column, 7- 4 does not need regrouping.
Move to the tens column. I cannot subtract 9 from 2, so I need to regroup.
Move to the ones column. I cannot subtract 8 from 5, so I need to trade.
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Now, do this one on your own.
9 4 2
- 2 8 7
12
313 8
6 5 5
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Last one! This one is tricky! 7 0 3
- 4 6 9
13
9 6
2 4 3
10
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Partial Products Algorithm Partial Products Algorithm for Multiplicationfor Multiplication
Focus Algorithm
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Calculate 50 X 60
67X 53
Calculate 50 X 7
3,000 350 180 21
Calculate 3 X 60
Calculate 3 X 7 +Add the results 3,551
To find 67 x 53, think of 67 as 60 + 7 and 53 as 50 + 3. Then multiply each part of one sum by each part of the other, and add the results
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Calculate 10 X 20
14X 23
Calculate 20 X 4
200 80 30 12
Calculate 3 X 10
Calculate 3 X 4 +Add the results 322
Let’s try another one.
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Calculate 30 X 70
38X 79
Calculate 70 X 8
2, 100 560 270 72
Calculate 9 X 30
Calculate 9 X 8 +Add the results
Do this one on your own.
3002
Let’s see if you’re right.
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1. Create a grid. Write one factor along the top, one digit per cell.
2. Draw diagonals across the cells.
3.Multiply each digit in the top factor by each digit in the side factor. Record each answer in its own cell, placing the tens digit in the upper half of the cell and the ones digit in the bottom half of the cell.
4. Add along each diagonal and record any regroupings in the next diagonal
0
62
4
1
8
08
32
2
4
Write the other factor along the outer right side, one digit per cell.
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0
62
4
1
8
08
32
2
4
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35
15
10
4
92
1
1
4
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The lattice algorithm for multiplication has been traced to India, where it was in use before A.D.1100.
Many Everyday Mathematics students find this particular multiplication algorithm to be one of their favorites. It helps them keep track of all the partial products without having to write extra zeros – and it helps them practice their multiplication facts
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Partial QuotientsPartial QuotientsA Division AlgorithmA Division Algorithm
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The Partial Quotients Algorithm uses a series of “at least, but less than” estimates of how many b’s in a. You might begin with multiples of 10 – they’re easiest.
12 158There are at least ten 12’s in 158 (10 x 12=120), but fewer than twenty. (20 x 12 = 240)
10 – 1st guess
- 12038
Subtract
There are more than three (3 x 12 = 36), but fewer than four (4 x 12 = 48). Record 3 as the next guess
3 – 2nd guess- 36
2 13 Sum of guesses
Subtract
Since 2 is less than 12, you can stop estimating. The final result is the sum of the guesses (10 + 3 = 13) plus what is left over (remainder of 2 )
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Let’s try another one
36 7,891100- 3,600
4,291Subtract
100- 3,600
7 219 R7
Sum of guesses
Subtract
69110- 360
3319 - 324
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Now do this one on your own.
43 8,572100- 4,300
4272Subtract
90-3870
15199 R 15
Sum of guesses
Subtract
4027- 301
1012 - 86