1. Multi-digit Numerical Long Division 1 © 2013 Meredith S. Moody

2.  Divide numbers with 2 or more digits using a variety of methods for long division  Divide numbers with 2 or more digits using the standard algorithm for long division 2 © 2013 Meredith S. Moody

3.  Division is determining how many groups of one number can be made out of another number  For example, I have the number 15 and I want to make 3 groups; how many will be in each group? The answer would be 5  That is the same as dividing 15 by 3 3 © 2013 Meredith S. Moody

4.  What if there are not a whole number of groups?  Let’s say I have 15 cookies and I want to make 4 bags (equal groups) of cookies.  If I divide 15 into 4 equal groups, I would have 3 cookies in each bag, but I would have 3 cookies left over.  3 cookies would ‘remain’  In other words, 3 is my remainder if I want to divide 15 by 4 4 © 2013 Meredith S. Moody

5.  Mathematical operations come in pairs  Which operations do you think are pairs?  Addition and subtraction are a pair  Multiplication and division are a pair  In order to divide, you have to understand multiplication 5 © 2013 Meredith S. Moody

6.  Multiplication is repeated addition ◦ 3 x 5 = 15 ◦ 3 + 3 + 3 + 3 + 3 = 15  Division is repeated subtraction ◦ 15 ÷ 5 = 3 (3 groups of 5, none left over) ◦ 15 – 5 – 5 – 5 = 0 6 © 2013 Meredith S. Moody

7.  If I have 15 cookies and want to make 5 equal bags of cookies, there must be 3 cookies in each bag  I can make 5 bags of 3 cookies.  5 x 3 = 15  15 ÷ 3 = 5  15 ÷ 5 = 3  Division and multiplication are inverse operations 7 © 2013 Meredith S. Moody

8.  What happens if the numbers are too large to divide mentally?  What if I want to divide 487 by 32?  How could I do that?  I could use a calculator, yes, but what if I don’t have one?  Let’s look at three different methods of dividing by hand ◦ Repeated subtraction ◦ Standard algorithm ◦ Scaffold division 8 © 2013 Meredith S. Moody

9.  Division is actually repeated subtraction  How many times can I subtract 32 from 487?  487–32=455–32=423–32=391–32=359 359-32=327-32=295-32=263-32=321 321-32=199-32=167-32=135-32=103 103-32=71-32=39-32=7  How many times did we subtract 32? 15  How many is left over? 7  Wow! That took a long time. Is there another way? 9 © 2013 Meredith S. Moody

10.  An “algorithm” is a step-by-step procedure for calculations  We can use a division algorithm for multi- digit division  In this method, there are specific parts with universal names  Knowing these names are important so everyone can discuss division without becoming confused 10 © 2013 Meredith S. Moody

11.  The division bracket is the “box” into which we put the dividend © 2013 Meredith S. Moody 11

12.  487 is the dividend, it goes in the “box”  32 is the divisor, it goes outside the “box”  The answer is called the “quotient”  The left over amount is called the “remainder” 12 © 2013 Meredith S. Moody

13.  The most efficient way to divide multi-digit numbers by hand is called ‘long division’  How many groups of 32 are in the number 4? 0. 32x0=0. subtract 4-0=4. ‘Bring down’ the next digit (8)  How many groups of 32 are in the number 48? 1. 32x1=32. subtract 48-32=16. ‘Bring down’ the next digit (7)  How many groups of 32 are in the number 167? 32x5=160. subtract 167-160=7 13 © 2013 Meredith S. Moody

14.  Wow, that standard algorithm doesn’t make sense to me  Is there another way?  Yes  Instead of trying to divide 487 by 32, we can break up our steps into smaller chunks  This is called scaffold division 14 © 2013 Meredith S. Moody

15.  We can break up large numbers using the place value system  487 becomes 400+80+7  How many groups of 32 can I make out of 400? Well, I know 3x4=12; I should be able to make about 12 groups of 32 out of 400  Well, if I make 12 groups of 32, how much of the 400 have I ‘used’? 12x32=384  How much of the 400 do I still have to ‘use’? 400-384=16; I have 16 ‘left over’ 15 © 2013 Meredith S. Moody

16.  Now I work with the number 80  How many groups of 32 can I make out of the number 80?  I know 3x3=9, but 90 is too much; I should be able to make 2 groups of 32 out of 80  If I make 2 groups of 32, how much of the 80 have I ‘used’? 32x2=64  How much do I have left to ‘use’? 80-64=16; I have 16 ‘left over’ 16 © 2013 Meredith S. Moody

17.  Now I have to look at the number 7  How many groups of 32 can I make out of 7?  None  Let’s use our ‘leftovers’  I had 16 left over from the 400, 16 left over from the 80, and 7 left over from my original work  16 + 16 + 7 = 39  How many groups of 32 can I make out of 39?  I can make 1 group of 32 out of 39, with 7 left over 17 © 2013 Meredith S. Moody

18.  Now I just add my groups together:  I had 12 groups in the 400  I had 2 groups in the 80  I had 1 group in the ‘leftovers’  12+2+1=15  I have 7 ‘left over’ now, so the answer to my problem: what is 487÷32, is 15 remainder 7  That was a little hard to follow; is there an easier way to write this?  Yes 18 © 2013 Meredith S. Moody

19.  Let’s put our scaffold method into an easy- to-read structure:  487 = 400 + 80 + 7  400÷32 = 12 ◦ 32 x 12 = 384 ◦ 400-384 = 16  80÷32 = 2 ◦ 32 x 2 = 64 ◦ 80-64 = 16  7÷32 = 0  16+16+7 = 39  39÷32 = 1 ◦ 32 x 1 + 32 ◦ 39-32 = 7  12+2+1 = 15  487÷32 = 15 r7 19 © 2013 Meredith S. Moody

20.  The scaffold method took quite a while, too  Is there a more efficient way to scaffold?  Yes 20 © 2013 Meredith S. Moody

21.  Let’s try another together  Two individuals are to equally share an inheritance of $860. How much should each receive?  To solve the problem, we want to divide 860 by 2  Let’s look at the three ways we could solve (no calculators!) 21 © 2013 Meredith S. Moody

22.  Trying to repeatedly subtract 2 from 860 would take a LONG time  It makes sense to use a faster method 22 © 2013 Meredith S. Moody

23.  Let’s use the extended scaffold division method  First, we break up 860 using place values:  800 + 60 = 860  We can easily divide 800 by 2. 800÷2=400.  Each person would get $400 so far  400+400=800. Since we have ‘used’ $800, we subtract 860-800 = 60. We still have $60 to share. 23 © 2013 Meredith S. Moody

24.  Next, we share the $60.  Dividing $60 by 2 is easy. Each person would get $30.  We need to add another 30 to our quotient.  Notice we place the 30 in the proper place value above the 400. We have ‘used’ the last $60, 60-60 = 0.  We have no money left to share. 24 © 2013 Meredith S. Moody

25.  The last step is to sum the two partial quotients to obtain the final quotient 400+30=$430  Each person would each receive $430 25 © 2013 Meredith S. Moody

26.  Let’s use the standard algorithm  The dividend is 860  The divisor is 2  There are 4 groups of 2 in 8  ‘bring down’ the 6  There are 3 groups of 2 in 6  ‘bring down’ the 0  There are 0 groups of 2 in 0  The quotient is 430 26 © 2013 Meredith S. Moody

27.  What if we had three people and they needed to split $986 evenly among them?  Repeated subtraction would take too long  The extended scaffold division method would take a long time, too  Let’s start with the efficient scaffold division method 27 © 2013 Meredith S. Moody

28.  How much money would each person receive if 3 people had to split $986 evenly?  Each person would receive $328  There would be $2 left over 28 © 2013 Meredith S. Moody

29.  Let’s use the standard algorithm to solve 29 © 2013 Meredith S. Moody

30.  If 14 children had 239 cookies, what is the highest number of cookies each child could receive if each one had to have the same number?  Repeated subtraction would take too long.  The extended scaffold method would take too long  Let’s start with the efficient scaffold method and then try the standard algorithm 30 © 2013 Meredith S. Moody

31. 31 © 2013 Meredith S. Moody 239 ÷ 14 = 17 r 1 Each child would receive 17 cookies There would be 1 cookie left over

32. 32 © 2013 Meredith S. Moody

33.  Use either the traditional scaffold division, efficient scaffold division, or standard algorithm method to solve: 236÷4 33 © 2013 Meredith S. Moody

34.  34 © 2013 Meredith S. Moody

35.  35 © 2013 Meredith S. Moody

36.  36 © 2013 Meredith S. Moody

37.  Use any long division method (repeated subtraction, standard, scaffold, or extended scaffold) to solve 193 ÷ 11  Repeated subtraction solution:  193-11=182-11=171-11=160-11=149-11=138 138-11=127-11=116-11=105-11=94-11=83-11=72 72-11=61-11=50-11=39-11=28-11=17-11=6  17 remainder 6 37 © 2013 Meredith S. Moody

38.  38 © 2013 Meredith S. Moody

39.  39 © 2013 Meredith S. Moody

40.  40 © 2013 Meredith S. Moody