xMath :: 6-9 :: Division
xTable of Contents:
  • Introducton
  • Numeration
  • Addition
  • Multiplication
  • Subtraction
  • Division
    • Stamp Game Division - Single Digit
    • Memorization Exercises
      • Division Boards (Bead Boards)
        • Introduction and List of Materials
        • Initial Presentation
          • Preparation for Presentation
          • Presentation
        • Division Booklets
      • Division Charts and Combination Cards
        • Chart I
        • Bingo Game (Chart II)
        • Exercise A:
        • Exercise B.
        • Exercise C.
        • Group Game 1.
        • Group Game 2.
    • Division By More Than One Digit
      • Double Digit Division
        • Intro. to Double Digit Division (on The Change Game)
        • Double Digit Division on The Stamp Game
      • Triple Digit Division
        • Intro. to Triple Digit Division (on The Change Game)
        • Triple Digit Division on The Stamp Game
    • Division Involving Zeros
      • Division with Zero in the Dividend
      • Division with Zero in the Divisor
    • Group Division
      • Single Digit Group Division
      • Double Digit Group Division
    • More Memorization Exercises
      • Special Cases
        • 0- Calculate the quotient
        • 1- Calculate the Divisor
        • 2- Calculate the Dividend
        • 3- Inverse of Case Zero: Calculate the Quotient
        • 4- Inverse of The First Case: Calculate the Divisor
        • 5- Inverse of the Second Case: Calculate the Dividend
        • 6- Calculate the Divisor and the Dividend
      • Search for Quotients
      • Prime Numbers
    • Division With Racks & Tubes
      • Introduction:
      • Materials: racks and tubes:
      • Small Division (1-digit divisor, 4-digit dividend)
        • 1st level
        • 2nd level
        • 3rd Level-Group Division
      • Big Division (2 or more digits in the divisor)
        • 1st level:
        • 2nd level:
        • 3rd level-Group Division with a two digit divisor
      • Long Division
      • Division with Zero in the Divisor
    • Word Problems
      • From Combination Cards
      • Distributive vs. Group Division
    • Divisibility
      • 1. Introduction
      • 2. Div. by 2
      • 3. Div. by 4
      • 4. Div. by 5
      • 5. Div. by 25
      • 6. Div. by 9
    • Synthesis of Multiples & Divisibility
  • Fractions - COMING SOON
  • Decimals - COMING SOON
  • Pre-Algebra - COMING SOON


xStamp Game Division - Single Digit


...wooden stamps of four types: unit stamps printed with the numeral 1, tens stamps printed with the numeral 10, hundred stamps printed with the numeral 100, and thousand stamps printed with the numeral 1000 with three compartments each containing 9 skittles and one counter in each of the hierarchic colors;
...four small plates

The child is given a division problem on a work card. Since this is division, we put the stamps in these little dishes. 'How many must this quantity be distributed to?' Instead of calling children we can use these green skittles, one skittle for each child . Now you begin the division.
The skittles are set out in a row. Beginning with thousands, the child distributes equally to the skittle. When the child runs out of i.e. hundreds before each skittle has received one, the hundreds are returned to the dish, and the child goes on to tens. If he doesn't have enough tens (or if he runs out) one of the remaining hundreds is changed to ten tens. This continues until all of the hundreds have been changed.
When the child has finished distributing, he reads the result-what one skittle received. The problem is recorded in his notebook. Note: If necessary to simplify counting, the stamps of one skittle may be arranged in hierarchic order after the distribution is finished.

Age: 6-7

Aims: to perform the operations at a more abstract level (here there is no visual difference in value according to size)

Note: These materials are used parallel to operations in the decimal system


xMemorization Exercises


a. Introduction and List of Materials

The memorization of division is the synthesis of the four operations. For this reason the child must precede this work with a great deal of work with the other operations, especially multiplication. It is very important that the child know multiplication really well before going on to division.
The child has encountered division before via many materials and regarding many cases: distributive and group division, division with a 1 or 2 digit divisor. In order to go on, the child must memorize certain combinations in division.

...Division Bead Board: the numerals 1-9 across the top on a green background represent the divisors; the numerals 1-9 down
...the left side represent the quotients; the 1 in the green circle in the upper left indicates that the numbers 1-9 below it represent
...units; 81 holes.
...Orange box containing 9 green skittles
...Orange box containing 81 green beads
...Booklet of Combinations (36 pages)
...Box of Multiplication Combinations (same combinations as are found in the booklet)
...Box of orange tiles for bingo game
...Charts I, II (for control)



b. Initial Presentation

Preparation for Presentation
Before-hand, the teacher prepares pads of problems which consist of 81 forms. On each form, in the spaces at the top, is written a number, which is the constant dividend for that form. The forms are arranged on the strip in order according to the dividends, starting with 81, ending with 1. Under the forms on which a combination with remainders appears, the ordinal number of the form is written in red. On each form are written all of the combinations possible to perform on the bead board, beginning always with 9 (except, of course, when the dividend is less than 9) All of the combinations which have a remainder of zero are underlined in red.

In this box are 81 beads which we will distribute to these 9 skittles. The nine skittles are placed along the top green strip of numbers 1-9. These 81 beads are the dividend, so we write 81 under the column 'dividend.' Then write the sign for division. The beads must be divided among 9 skittles; nine is our divisor, so we write 9 under 'divisor.' Now give out the beads in rows until all the beads are given out equally.
Each skittle received nine beads, (note the 9 at the end of the row), so we write 9 under 'quotient.' The last column is for the remainder, but here we have no remainder, so we write zero in that column. Whenever we have a combination that has no remainder, it is very important for our work, so we underline 'it' in red.
Let's try 81 ÷ 8. Remove one skittle and the extra beads to their respective boxes. What is the quotient and the remainder? 81 ÷ 8=9 r 9 In this game there are two rules to be followed:
1) The quotient should never be greater than 9.
2) The remainder may never be equal or greater than the divisor.
Therefore, we cannot have this combination because the remainder is too big.
Go on to 80. Change to a new form. Remove one bead and place it back in a box. Start with 80 ÷ 9=. Write the combination on the form, distribute the beads and count the remainder. 80 ÷ 9 = 8 r 8
Try 80 ÷ 8. Remove one skittle, redistribute the beads, and count the remainder.
80 ÷ 8 = 9 r 8. This cannot be used because the remainder is too big, being equal to the divisor. Erase or cross out this combination.
Go on this way until 72 so that the children see another page on which combinations can be underlined. At this point bring out the prepared roll of forms. On this strip we can see all the combinations that have zero remainder. All of the forms that have at least one combination with remainder zero, have been reproduced into booklet form for your work.

Aim: to understand how the combination booklet was formed



c. Division Booklets

...Division Bead Board with beads of skittles
...Combination booklet
...Chart I

The child chooses a form in the booklet, i.e. 7 Since 7 is the dividend, count out 7 beads into the box cover. The first combination is 7 ÷ 7 =, so 7 skittles are put out; the 7 beads are divided among them. Each skittle receives one; 1 is the quotient. There is no remainder. 7 ÷ 7 = 1 r 0
The next combination is 7 ÷ 6 =. Only six skittles are needed. Each skittle receives one and the remaining bead is placed in the bottom row, or in a box cover. 7 ÷ 6 = 1 r 1. Continue until the form is completed.

Control of error: Chart I. Find the dividend along the top and the divisor in red at the left. Go down and across to find the quotient where the fingers met. If there is a remainder, the box will be empty, thus move along the row to the right until a box is full. There you will find the quotient. To check the remainder subtract the dividend at the top of that column, from your original dividend. The difference is your remainder.

Note: When presenting the chart to the children, we identify the prime numbers as well, since they are shaded in red. 7,5,3,2 and 1 are special numbers because they only have as divisors, themselves and one. 7 can only be divided by 7 and by 1, and so on. These special numbers are called prime numbers.



a. Chart I

Materials: of loose combination cards, only those that have a remainder of zero
...Chart I (as a control)

In this box are only the even division combinations: those having a remainder of zero. The child fishes for a combination, reads it and copies it into his notebook. On the chart he finds the dividend at the top and the divisor on the left side. The place where the two fingers meet is where the answer is found. He writes the quotient to complete the equation.
Later the child can do the combinations in his head, write down the quotient and use Chart 1 only as a control.



b. Bingo Game (Chart II)

Materials: of tiles of combination cards
...Chart II and Chart I (for control)

Note: Much practice should have preceded these exercises.

Exercise A:
Spread out the tiles face up. The child fishes for a combination, writes it down including the quotient, and finds the corresponding tile. The tile is placed on Chart II appropriately.

Exercise B.
With all of the tiles in the box, the child fishes for a tile. He thinks of a combination that would yield that quotient and writes the equation in his book, i.e. 8 = 56 ÷ 7. The tile is placed on Chart II appropriately.

Exercise C.
All the tiles are stacked as usual (this time forming a parallelopiped) The child chooses a stack, and one at a time he thinks of all the possible combinations that will yield that quotient, writes them down and places the tiles on Chart II appropriately. This continues until all the tiles of the stack which was chosen, are used. The child uses Chart I to check if he found all of the possible combinations and if they were placed correctly.

Note: For many children the aim of this work can be to fill up the entire board.

Group Game 1.
The teacher or a child leading a group of children draws a combination and reads it. One child responds.

Group Game 2.
The teacher draws a tile and reads the quotient. One child may try to give all of the possible combinations, or each child in turn may give one until all of the possibilities are exhausted.


xDivision By More Than One Digit


a. Intro. to Double Digit Division (on The Change Game)

...golden bead materials and numeral cards
...ribbons: green, blue, (also red for later)

The teacher prepares the numeral cards and asks a child to bring the corresponding quantity. Now I would like to distribute this quantity among twelve children. Twelve children are called, but this creates such confusion. How can we solve this problem? Twelve is made up of ten and two units. Two children can represent the two units and can be given green ribbons. These ten children must choose one who will represent ten. What color ribbon do we give the representative?
Now we are ready to distribute this quantity. If we give one thousand to the child who represents ten children, what will each of the other children get? 100 (because 1000 is 10 hundreds) also each of these children receives 100another thousand for the group of ten, another hundred for this one, and another hundred for this one. This continues, making all of the necessary changes, until all has been distributed (perhaps leaving a remainder)
What is our result? The result is what one unit receives. This child who represents ten children has enough on his tray so that each of these ten children will receive what one child received. (This quantity may be distributed if necessary.)

Control of error: The quantity may be added together again, making the necessary changes to form the original number.

Age: 6-7

Aim: to learn the concept of two-digit division: that if the ten receives a certain quantity, the unit receives 1/10 of that quantity.



b. Double Digit Division on The Stamp Game

Materials: stamp game work cards

The first quantity for this problem is formed with the stamps and placed in little dishes. We need to divide this among twelve. This blue skittle can be used to represent 10, and these green skittles will represent the two units. Now we must give out this quantity. One thousand is given to the ten, so how much does each unit receive? 100. The child distributes and changes as necessary, 'one hundred to the ten, ten to the units, another hundred to the tenand so on.' What did one receive? The child records the problem in his notebook.

Control of error: The quantity may combine the quantities distributed, count them, change them, to obtain the original number.

Age: 6-7

Aim: to practice two-digit division



a. Intro. to Triple Digit Division (on The Change Game)

...golden bead materials and numeral cards
...ribbons: green, blue, (also red for later)

Given a division problem, the child can see that it would be impractical to distribute one by one to over a hundred people. The child representing 100 wears a red ribbon. When red (100) receives 1000, blue (ten) will receive 100, and units (green) will receive 10.

Control of error: The quantity may be added together again, making the necessary changes to form the original number.

Age: 6-7

Aim: to learn the concept of three-digit division.



b. Triple Digit Division on The Stamp Game

As in the previous presentation, skittles are used. Here one red skittle represents 100.


xDivision Involving Zeroes


Materials: stamp game work cards

When the child places the stamps in the dishes to show the dividend, the hierarchy that has zero is indicated by an empty dish. When the child needs to change to that hierarchy, the procedure is the same.



...stamp game cards

The child lays out the first quantity in the dishes. This quantity must be divided among 104. I don't need any tens skittles, but so as not to forget the tens, we place a blue counter in its place. The child begins distributing: 1000 to the hundred, the ten would get 100, but I give it nothing, and the unit received 10and so on.
By this time the child should realize that any remainder must be less that the divisor.


xGroup Division


...stamp game cards (divisor up to 9)

The division problem says 24 divided by 4 =____. How many groups of four are we able to make with this number. The four skittles are placed in a bunch and groups of four units are in rows in front of this group. Changes are made as necessary. When the distribution is complete, how many groups of four were we able to make from 24? The rows are counted. Six is the result of this group division.
The child then does the same problem using distributive division to see that the result is the same.

Age: 6-7

Aim: to understand a different aspect of division

Note: This is presented parallel to abstract division. Memorization has begun.



...stamp game cards


1. Given a division problem, the child prepares the stamps and the skittles. Since we want to do a group division, we put the skittles together in a group. How many groups of 26 can be made from this number? The child places two tens and six units in a row, continuing his distribution by making all horizontal rows of 26 in a column. (The stamps are always placed in hierarchic order) Here, the skittles serve only as a reminder of the number in the group.

2. This time we will first make groups only of tens. Groups are made of two tens and lain out in rows. How many groups of ten did I make? So that each group has 26, I must make the same number of groups of units. Groups of six units are made in rows that line up with the groups of ten, yet in a separate column. When the child finds that more units are needed, one group of tens is returned to the dish so that they may be changed. How many groups of tens do I have ? 4 How many groups of units? 4 is our answer.

Age: 7

Aims: learn the concept of group division continue towards further abstraction in division


xMore Memorization Exercises


As for the other operations, we examine the special cases using as a starting point the combination that is most familiar. A chart will be constructed as follows:

0- Calculate the quotient
72 ÷ 9 = ? ( 72 divided by 9 gives me what number?)

1- Calculate the Divisor
72 ÷ ? = 8 ( 72 divided by what number gives me 8?)

2- Calculate the Dividend
? ÷ 9 = 8 ( What number when divided by 9 gives me 8?)

3- Inverse of Case Zero---Calculate the Quotient
? = 72 ÷ 9 ( what number will we obtain by dividing 72 by 9)?

4- Inverse of The First---Case, Calculate the Divisor
8 = 72 ÷ ? ( We obtain 8 as a quotient when dividing 72 by what number?)

5- Inverse of the Second Case ---Calculate the Dividend
8 = ? ÷ 9 ( We obtain 8 as a quotient when dividing what number by 9?)

6- Calculate the Divisor and the Dividend
8 = ? ÷ ? ( We obtain 8 by dividing a certain number by another number. What is the first number and the second number?)

Note: Here we also see the relationship between multiplication and division. In cases 2 and 5 the child must multiply to find the dividend.



...Chart II tiles for multiplication

Have the child find one bingo tile to match all the dividends along the top of Chart II. These are placed in a box cover or something.
The child fishes for a tile, i.e., 24. Let's try to find all the quotients with zero remainders that can be made with this dividend. Start with 24 ÷ 9 =. It won't work so leave it blank, and go on24 ÷ 8 = 3 and so on with the child giving the correct quotients. At the end erase those that would not yield zero remainders, thus leaving space to correspond with Chart I. Notice how the column of quotients matches the column under 24 on the chart.
These are the combinations I wanted, because now we can do 3 x 8 = 24. Write this to the right of 24 ÷ 8 = 3. 24 was my dividend: now it is my product. Go on in the same way for the other combinations making a second column.
The child will realize that if 24 ÷ 8 = 3, then 24 ÷ 3 = 8. It is a sort of game where the numbers change positions.

Aims: find quotients with zero remainders realize the relationship between multiplication and division memorization of division

Indirect Aim: indirect preparation for the divisibility of numbers



Materials: Chart I

Recall the child's attention to the numbers in pink on the chart, which were called prime numbers, and which can be divided only by themselves and 1. These are very important numbers because they form all of the other numbers. We can see that this is true. (refer to the chart) 1, 2, and 3 are prime numbers. 4 is not a prime number, but is made of 2 x 2, and 2 is a prime number. Go on to 6 which is not prime. It is made up of 3 x 2; 3 and 2 are prime numbers.
If you try to decompose any number, you will find that it is made up of prime numbers. Invite the child to choose one of the dividends, and think of one combination: 24 = 3 x 8. 3 is a prime number, but not 8, 8 is made up of 2 x 4. 2 is a prime number, but not 4, 4 is made up of 2 x 2. 2 is a prime number.

Try another combination of 24 to check.
24 = 6 x 4 neither 6 nor four is prime
6 = 2 x 3 both are prime
4 = 2 x 2 both are prime

Direct Aim: to realize the importance of prime numbers

Indirect Aim: prepare for divisibility of numbers, LCM-least common multiple GCD-greatest common divisor and
...reduction of fractions to lowest terms

...(for all of division) to memorize the combinations necessary for division. stimulate an interest that will help him to use the experiences acquired previously.


xDivision With Racks & Tubes


The operations of addition, subtraction and multiplication can be performed with the large bead frame. But since the beads are connected to the wires, the beads frame cannot be used for division. The hierarchic material for division consists of loose beads.
Up to this point division has been done with the decimal system material to give the concept, including division with a 2-or 3-digit divisor, and group division. These concepts were reinforced with the stamp game. Following a research of the combinations necessary for memorization, where the quotient was limited to a maximum of 9. Division was dealt with indirectly in many of the multiplication activities. Using this material the dividend may have up to 7 digits and a divisor of 1-, 2- or 3-digits may be used.
Maria Montessori referred to this material "as an arithmetical pastime for the child." This work clarifies the analytic procedure for the development of the operation. The fundamental difficulty of division is obtaining the digits of the quotient, recognizing their values and placing them in their proper hierarchical position.
At this level more importance is given to the quotient, that is, what each unit receives, and not so much to the quantity to be divided.



...7 test tube racks: 3 white, 3 gray and 1 black
...each rack contains 10 test tubes

deposit each tube contains 10 loose beads
[These are the "deposit" from which quantities are drawn. The racks are white for the simple class, gray racks for thousands and 1 black rack of green beads for units of millions.]

7 bowls - 1 to correspond to each rack:
dividend the exterior of the bowl corresponds to the color of the rack the interior of the bowl corresponds to the color of the beads The dividend is formed in these bowls, just as was done with the stamp game.

3 bead boards: in hierarchical colors for units, tens, hundreds
For a 1-digit divisor the green board is used
divisor For a 2-digit divisor the green and blue boards are used
For a 3-digit divisor the green, blue and red boards are used.
[As in memorization, the distribution is done on the boards.]
Box with three compartments containing nine skittles of each of the three hierarchic colors that represent the divisor.

Also: A large tray to hold the racks while they are not in use.



1st level

9764 ÷ 4 =



Isolate the racks that are needed to form the dividend. Place the other racks on the tray. Pour the quantities into the respective bowls. Place the green skittles on the green board for the divisor.
Begin distributing "bringing down" the units of thousands, that is moving the rack and bowl closer to the board. After distributing the units of thousands, record the first digit of the quotient in its hierarchical color, reading the number at the left side of the board.
Remove the beads from the board and place them back into the tubes. There is one units of thousand bead remaining in the bowl, which can't be distributed as is. Change it for 10 hundreds (pour the hundreds into the hundreds bowl). Having finished with the units of thousands, place the rack and the bowl out of the way on the tray.
Continue in the same way for hundreds, tens and units.

9764 ÷ 4 = 2441



Note: Here also the operation is reduced to the level of memorization.

Recall the problem 81 ÷ 8 = which couldn't be done before. This does not mean that it couldn't be done, just that it couldn't be done with that material. Try it using this new material.
It is important to emphasize that every time a hierarchy is considered, a digit must be placed in the quotient. If there are not enough beads to distribute, we must still record a zero. This is where the child could easily make a mistake.

2nd level

9216 ÷ 3 =



Set up the problem as before and begin distributing the units of thousands. Record the first digit of the quotient using the hierarchical color green. There are no beads in the bowl, so the remainder is zero. Write the remainder under the 9. Put the thousands away.

9216 ÷ 3 =






Bring down the hundreds, that is, move the bowl closer to the board and write the 2 next to the 0. Since we can't distribute these two beads, write the next digit in the quotient and write the remainder. Remove the beads.

9216 ÷ 3 =







The two hundreds must be changed for tens. Put away the hundreds. Bring down the tens. Now we must distribute 21 tens. Continue in this way.

9216 ÷ 3 =











Note: It is important to work through a problem such as 1275 ÷ 3 = to demonstrate the grouping of the first two digits. In a case such as this we do not record a zero in the quotient for the first hierarchy



3rd Level-Group Division

Note: The child will never reach abstraction using the distributive division technique.

Recall the concept of group division. Take out 15 loose golden beads. Invite the child to find out how many groups of three can be formed.
Relate word problems to demonstrate the difference between distributive division and group division:

1. I have 12 pencils which I must give to 6 children. How many pencils will each child get? 2. What kind of division did we do? distributive division.

2. I have 25¢. I want to buy pencils costing 5¢ each. How many pencils may I buy? 5. This time I had to think of how many groups of 5 are in 25. This is group division.

Note: The Difference here is mostly in language, for this is an important step in the development toward abstraction. At this level the child incorporates the other operations in a conscious way. The quotient is no longer written in the hierarchical colors, because by this time, the concept should be firm in the child's mind.

7687 ÷ 5 =



Set up the materials as before. We must see how many times this group of 5 (indicate the skittles) is contained in this 7 (the units of thousands). Distribute the beads. Record the quotient. One group of 5, that is 5 x 1 = 5. Write 5 under the 7 and subtract. This is our remainder. Check to see if the number of beads in the bowl matches the difference.

Change the remaining two units of thousands to hundreds and bring down the hundreds.
Now we must find how many groups of 5 are contained in 26.

Age: from 6 - 7 years ( at 7 years old, the child should reach abstraction)

Note: When the child has reached abstraction of division, he has actually reachedabstraction for all the operations, since division involves all of the operations. Before progressing to Big Division (having a divisor of more than one digit) the child should have reached abstraction with small division, that is, without the materials.



Recall with the child the presentation of the concept of division with a two-digit divisor, using the decimal system materials and the arm ribbons. Introduce the bead board for tens. Recall that each blue skittle represents 10 units. If I give 10 to the blue skittle, what must I give to the green skittle? After this concept is already recalled, begin division.

1st level:

37,464 ÷ 24 =



Set up the material as before. Bring down the tens and units of the thousands, one for each board. Distribute the beads: one 10,000 for the tens; one thousand for the units. The first digit of the quotient is 1, but 1 what? The result is what one unit receives, so it is one thousand. Record the digit in color.

37,464 ÷ 24 = 1



Remove the beads from the board. Change the 10,000 to ten units of thousands and out away the ten thousands. Move the rack and bowl of units of thousands to the left, to the tens board. Bring down the hundreds. Distribute.
When the bowl of the lesser of the two hierarchies being considered is emptied, continue changing and distributing. However, when the bowl of the greater hierarchy is emptied, we must stop, record the digit of the quotient, and move on to another hierarchy.

37,464 ÷ 24 = 1561



2nd level:

7886 ÷ 35 =



As for the second level of the small division, record the remainder and bring down the digits of the dividend.

3rd level-Group Division with a two digit divisor

Recall the meaning of group division in the same way as before.

8847 ÷ 24 =



Set up the problem as usual. Bring down the first two hierarchies. How many times is 24 contained in 88? First, we must find how many times 2 is contained in 8. Distribute the beads 4 times. Now we must see if 4 is contained in 8, 4 times also. Distribute the beads. It doesn't work. Take off one row of beads from the board and place them in the bowl. Change a thousand bead to 10 hundreds and distribute.
Since we want to make groups of 24, there must be the same number of groups of 2, as there are groups of 4. Thus 3 groups of 24 were made. 3 what? Refer to what one unit received - 3 hundreds.
Multiply 3 x 24, carrying mentally and recording the product beneath 88 in the dividend. (The number that we had to carry in the small multiplication, corresponds to the number of changes that were made while distributing.) Subtract; the difference should match the quantity that remained in the bowls. Remove the beads. Change. Bring down a new hierarchy and continue as before.



Note: If the child has reached abstraction in division with a 2-digit divisor, he will encounter very little difficulty here because the mechanics of the operation are the same. Thus, the material will be used less by the child. The material is used for the presentation to be certain that the child has understood the concept. At this level, group division is used immediately.

Recall the activity done with the decimal system material and arm ribbons. Note the difference that the centurion received over the decurion and the unit. Present the materials.

56,438 ÷ 234 = 241 r.44


r. 44



The procedure follows the pattern set down previously, now using 3 bead boards, and bringing down 3 hierarchies at a time. Remember that the first digit tells what all of the others must receive: How many times is 2 contained in 5? 2 We must see if 3 is contained in 6 2 times and if 4 is contained in 4 2 times.



51,252 ÷ 207 =

Recall the similar case in the stamp game where a counter took the place of zero for the skittles. Here the board without any skittles reminds us of the zero. Move down the three hierarchies- one for each board. The hierarchy by the empty board reminds us of what the tens would receive if there were any. Distribute as before using group division.

19,293 ÷ 370 = 676 ÷ 300 =

Make the child conscious of what the units would have received, if there were any , in order to determine the value of the digit of the quotient. Hierarchic colors can be used for recording the quotient.

Dividend: 70,569 ÷ 229 =

Place the dividend in the bowls, leaving one empty. Do not put it back on the tray since it will be needed for making changes. Bring down the hierarchies as usual, ignoring the fact that one bowl is empty; that hierarchy corresponds to one of the digits in the divisor.

Age: 9 years (by 9 1/2 the child should reach abstraction)

...mastery of long division
...knowledge of the reasons for every aspect of the procedure


xWord Problems


The teacher prepares seven special combination cards and mixes them with the other 21. The child fishes for one, solves it, writing it down substituting the ? for the answer in red.
Also, word problems are prepared and mixed with the others. The child copies the text of the word problem, writes an equation with the answer in red, and writes a conclusion, that is, a complete sentence which answers the question stated in the word problem.

Aim: further understanding of the concept of multiplication



These word problems are to aid the child's understanding of the difference between distributive division and group division:

Example: (distributive) Mother has 24 cookies. She wants to give these out to her three children equally. How many cookies will she give to each child?

Example: (group) Mother has 24 cookies. She wants to make up packages with three cookies in each package. How many packages can she make?




...decimal system materials (wooden)
...blackboard or blank chart

Write a number and invite the child to bring that quantity with the materials. Try to make two equal groups from this quantity, making changes as necessary .If it is possible to make two equal groups, write "yes" next to the number and underline the last digit. If not, write "no". Add or subtract one unit, and repeat the process. Examine many numbers in the same way. At a certain point the child will realize the rule: When the last digit is an even number or zero, the number is divisible by 2.
If the child does not reach this point of consciousness on her own, ask questions to call her attention to the relationship between the yes or no and the oddness or evenness of the last digit.

1126 yes
1125 no
1124 yes
1123 no
1122 yes
78 yes
79 no
80 yes
12 yes



Materials: Same as above

The procedure is the same as before, except that the last two digits are underlined. Now it is no longer a matter of oddness or eveness.
Rule: When the last two digits are divisible by 4, or they are both zeros, the number is divisible by 4.

816 yes
817 no
818 no
819 no
820 yes



Materials: Same as above

The procedure is the same as before, only the last digit is underlined.
Rule: When the last digit is 5 or 0, the number is divisible by 5.

125 yes
126 no
127 no
130 yes
45 yes
100 yes



...ten bars and 40 or more golden unit beads
...small white square pieces of paper,

Put out one group of 25 with a little card over it. Place another group next to it in an inverse position to make it easy to visualize the group as 50. Place a little card over it ("50"). Continue placing groups of 25 (7 tens, 5 units for 75) to the left of the previous group, the respective card is placed over the top. Continue up to 8 groups of 25, substituting 100 squares after 100.

Invite the child to speculate on the next few multiples of 25.

Rule: When the last two digits of a number are 23, 50, 75 or two zeros, the number is divisible by 25.



...Peg board
...Box of pegs in hierarchic colors
...Small white square pegs

9 is a very important number, Why? It is the last digit that can remain loose in our system. It is the square of 3, and 3 is a perfect number. Therefore in this work we will consider 9 as the square of 3.
On the peg board use 9 green pegs to construct a square of 3 by 3. This is a square of 3. Dissolve the square into a column at the top left corner of the board, labeling it 9.
Form two more squares of 3 side by side. Remove one unit from one square and add it to the other. Now one group has ten, change the ten green unit pegs for one blue ten peg. Dissolve the pegs into two columns next to the previous column. Label 18.
Repeat the procedure with 3 squares. Add one peg to each of the first two squares, taking the pegs from the last square. Change each group of 10 to a blue peg and dissolve. Continue in this way up to 10 squares = 90.
Observe that as the units decrease, the tens increase. At one extreme there are 9 units, at the other, 9 tens. This special pattern exists only in the table of 9. The sum of each pair of digits is 9. i.e. 27, 2 = 7 9.
Rule: A number is divisible by 9 when the sum of its digits equals 9 or a multiple of 9. If a number is divisible by 9, it is also divisible by 3.

Note: This characteristic may have been noticed in the multiplication booklet for memorization.



Materials: Decimal system materials

Take the thousand cube and try to make 9 equal groups. 1000 is not divisible by 9, but take away one unit and try again. 999 is divisible by 9. Write 1000 ­ 1 = 999. Repeat the procedure for 100 and 10. Write 100 ­ 1 = 99, 10 ­ 1 = 9. Choose a number: 5643. Convert it to expanded notation:

40 = 4 x 10
............10 ­ 1 = 9...........40 - 4 = 36..............r.4
600 = 6 x 100
.......100 ­ 1 = 99 .......600 - 6 = 594...........r.6
5000 = 5 x 1000
...1000 ­ 1 = 999 ....5000 - 5 = 4995....+ r.5
.................................................................................... 18

10 ­ 1 = 9, but we have 4 tens. We must multiply the whole equation by 4, which gives us 36. 36 from our original 40 leaves us 4 as a remainder. Continue for the others. Add all the remainders in the end. Since 18 is divisible by 9, the whole number is divisible by 9. (cont.)
Show an example of multiplication: 643 x 1527 = 981,861. Make sums of 9 in the digits - 643 x 1527 = 981,861. Total the remaining digits (4 from the multiplicand, 1 and 5 from the multiplicand) multiply the sums ( 6 x 4 = 24 ), and add the digits of the product
( 2 + 4 = 6 ). This number should correspond to the sum of the remaining digits in the original product - 6. 6 is also the remainder when the original product ( 981,861 ÷ 9 = 109,095 r. 6) is divided by 9. The "remaining sums" from above multiplicand and multiplier (4 and 6) will also be the remainder when divided by 9: ( 643 ÷ 9 = 71 r. 4 and 1527 ÷ 9 = 169 r. 6).

Age: Ten years


xSynthesis of Multiples & Divisibility


Refer back to the multiples work and the charts constructed to find the multiples of numbers up to 10 (circling the numbers in different colors). Repeat this work making new observations-i.e. A number is a multiple of (is divisible by) 6 if it is also a multiple of 2 and 3. This is noticed when the charts for 2, 3, and 6 are done simultaneously. A multiple of 6 intersects the lines of multiples of 2 and 3.
This work really shows the close relationship of multiples and divisibility. Knowledge of one reinforces the other.

Take Table C with the prime factors. Here also we can find, for example, by what numbers 18 is divisible, by making all possible combinations of the prime factors:
18 = 2 x 3 x 3.
18 is divisible by 2, 3, 6, and 9.
18 is even, thus it is divisible by 2.
1 + 8 = 9, thus 18 is divisible by 9, which means it is also divisible by 3.
Since 18 is divisible by 2 and 3, it is also divisible by 6.


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