More Memorization Exercises

Materials: same as for addition snake

Age: from 6 years onwards

This time in constructing the snake, it is not important to use many different colors; rather several bead bars of each of a few colors should be used.
A resulting snake may be: 2 + 4 + 8 + 4 + 2 + 4 +8 + 4 + 2. The snake is counted as in addition.

Control of Error: To one side the ten bars and black and white bars are grouped together. At the other side the original bead bars are grouped according to color. How many times do we have 8? We can say 8×2; the equation is written on a piece of paper. The beads to represent the product 16 are placed below the group of 8 bars. The same is done for 4×4 and 2×3. The three products are added on paper and/or with the second row of bead bars to show that the snake was counted correctly.

Note: The child does this work after completion of the exercises for memorization of multiplication


…box of colored bead bars 1-10
…Chart I

The teacher writes a product on a piece of paper and places it on the table (it will be the heading). Let’s see how many ways we can make 12? Shall we try with one? The child should see that any number can be made using one, therefore it will be omitted for this game.
Beginning with two, the child tries to make 12 by repeating 2, skip counting as he takes out the bars. Stopping at 12, he finds that 2 taken six times equal 12. These bars are placed in a column under the strip. the child writes the combination in his notebook,
12 = 2 x 6, which is read 12 is constructed by taking 2, 6 times.
Go on trying with 3 and 4 following the same steps. When the child tries with 5, he will find that 15 is greater than 12, so the bead bars are put back in the box. Go on with 6. Trying with 7, the child finds that it won’t work, because at the second bar we’ve already exceeded 12. Thus we must stop at 6.


…box of colored bead bars 1-10
…Chart I

Control of error: The child looks on Chart I, finds the combinations he has made, and the absence of 12 in the 5 column, 7 column and so on.

Direct Aim: memorization of multiplication

Indirect Aims:
…preparation for memorization of division
…preparation for decomposition into prime factors
…preparation for the study of multiples

The child is invited to take any number (not to exceed 10) of any color of bead bars. From this pile the child begins to lay them out in a column; saying 4 taken 1 time, putting out a bar and writing either the equation or just the product in his notebook, He continues until the whole pile has been lain out.

Control of error: Chart I. If the child wrote the equation then he has written a table. If only the products were recorded, then he has done progressive numeration (skip counting).

Direct Aim: memorization of multiplication

Indirect Aims:
…preparation for memorization of division
…preparation for decomposition into prime factors
…preparation for the study of multiples


Materials: same as above

Direct Aims: memorization of multiplication understanding of the commutative property of multiplication

The child is asked to take the bar of 7, 6 times. The bars are lain in a column. How do we express this? 7 x 6 = 42 is written down. Now take the bar of 6, 7 times. These are lain next to the others. When the child writes down 6 x 7 =42, he should observe that the product was the same. Are the multiplication problems the same? Notice that the two rectangles are equal, each having 42. even though the order of the factors is not the same. The child should observe in the written form that the factors are the same, just reversed in their positions.
Give several examples of this.


Direct Aim:
…memorization of multiplication
…realization that a number taken by itself n = makes a square.

Indirect Aim: preparation for the powers of numbers

…box of colored bead bars 1-10
…one each of the squares 1-10

Let’s try to multiply all of the numbers by themselves. During this activity the child writes the equations as he goes along. He may also draw this on graph paper as he progresses or when he is finished.
Start with one. One taken one time is one. Put out one bead. Write down 1 x 1 = 1. Two taken two times is four. Place two 2- bars in a column next to one. Write down
2 x 2 = 4. Continue in this manner until 10 x 10, resulting in 9 columns in a row. If the child doesn’t remember a combination he may check Chart I.
We have multiplied all the numbers by themselves. What have we formed? squares. Because these are only bars, we can substitute them with the real squares. With one, there is no square because 1 x 1 is just 1. Replace the 2 bars of 2 with the 2 square and so one up to 10.


…box of colored bead bars 1-10
…box of signs for the operations
…Chart I
…pieces of white paper

Direct Aim:
memorization of multiplication
understanding of the distributive property of multiplication over addition

Indirect Aim: preparation for the square of the polynomial

Binomial Presentation:
The teacher writes on a strip ( 5 + 2 ) x 3 =, which is read, take 5 plus 2, 3 times. This is a multiplication problem. The problem is prepared as before with bead bars for the addends, signs, parentheses and the multiplier, written on a little card. Recall that the operation inside the parentheses is done first. The child places the 7-bar for the sum under the parentheses, places signs, multiplier and the product, represented in bead bars. The work is recorded.

On a different day:
When you find a problem of this kind, you can also multiply one term at a time by the multiplier. The other way will be put aside for now. (The equation in beads and cards:
7 x 3 = 21 is placed off to the side, leaving the slip of paper and the original layout of beads)
First take 5, 3 times, 5 x 3 is written on a strip and 3 bars of 5 are placed below the original 5 barplus (put out the sign)2 taken 3 times, 2 x 3 is also written on a strip and 3 bars of 2 are lain out. Now we must find these products. The products are placed below the group in a perpendicular position. Add 15 + 6 and put out the result, The result is the same as the equation we put aside. The child writes in his notebook:

Trinomial Presentation: On yet another day
The teacher writes a problem on a strip such as: ( 5 + 2 + 3 ) x 4 =. The child lays out the corresponding beads for 5, 2, and 3, the signs, parentheses and a little card for 4. As before we must multiply each term by the multiplier. Then, for control, the child may add the addends within parentheses and multiply the sum by 4. When his work is written in his book it should be:

Note: After the child has learned to multiply such a problem term by term, he should not go back to the first way of adding first, then multiplying. In this way the following aims will be achieved.


…box of colored bead bars 1-10
…one each of the squares of 1-10
…rubber bands

The teacher presents the hundred square, and the child identifies it as the square of 10, or the hundred square. Observe that it has 10 beads on one side and ten on the other. Write 10 x 10 = 100 on a strip of paper.
Because we know this square so well, we are going to perform a small division of the square. The child is asked to count 6 beads along one side. A rubber band is placed after the 6th bead around the square, (Note: the result will be two perpendicular rubber bands.)
How many parts has the square been divided into? Let’s see how the 4 parts are composed: 6 on one side, 6 on the other, 6×6; 4 by 6 or 4 taken 6 times , 6×4, 4×4. Let’s write this down and while writing, we can reconstruct the square with colored bead bars, As each combination is written, the product is recorded, as the appropriate beads are lain out.

Add the products. Push the bead bars together and place the 100 square over it to verify sensorially that the decomposition was done correctly.

Note: Later, after the passing of a year and much work with the decomposition of a square and the powers of numbers, the child will learn the exact way of writing this:

10² = (6+4)² = (6+4) x (6+4) = (6×6) + (6×4) + (4×6) + (4×4)


…box of colored bead bars 1-10
…one each of the squares of 1-10
…rubber bands

Control of Error: Compute the sums of the three columns and add them together. Then slide all the bead bars toward the center, and place the 100-square on top. These are the two ways to prove that this equals one hundred.

This time we’ll decompose the square in a different way. Count 3 along one side, and place the rubber band around the square. Count 3 along the other side and put on a rubber band. Now continue counting 5 more beads, and put a rubber band around. Do the same on the other side.
Into how many parts have we decomposed the square? Observe how each of the 9 parts is composed. Notice the three squares which all lie on the diagonal, and the various rectangles formed.
As before, we’ll construct the squares as we write it all down.


Aim: indirect preparation for the square of a binomial

…box of colored bead bars 1-10
…one each of the squares of 1-10
…rubber bands

The child identifies the square chosen by the teacher, i.e. the square of 4. Let’s use this square to build a square of 5. Allow the child to suggest and discover ways of doing this. Guide the work, giving guidelines such as: ‘The sides must be built upon.’
Four bars are placed to the right and the bottom of the square. A one-bead fills in the hole left in the lower right hand corner. Superimpose the square of five to verify successful completion. What was added to the square of 4 to make the square of 5? two bars of 4 and one unit.


…box of colored bead bars 1-10
…one each of the squares of 1-10
…rubber bands

The teacher volunteers to assist in constructing a square of 9 from a square of 5. What is the difference when you take 5 from 9? So, how many more bars of 5 must be added to this side to make 9? Four bars of 5 are added to the right side. In the same way 4 5-bars are added to the bottom. Again, there is a hole. Count the number of beads on the sides of the hole…4 by 4. Four 4-bars can be replaced by the square, Use the square of 9 to control:


a. List of Materials

…Board of powers (though it is not named as such at this point)
…Cubes, long chains, squares, short chains
…Two Boxes:
…arrows for short chains, i.e. for 5 we have 1, 2, 3, 4, 5, 10, 15, 20, 25
…arrows for long chains, i.e. for 3 we have 1, 2, 3, 6, 9, 12, 15, 18,21, 24, 27


b. Construction of Geometric Figures

Direct Aim: reinforcement of law: the smallest possible polygon must have three sides

Indirect Aim: preparation for perimeters of polygons: preparation for multiples and divisibility

Materials: short chains

Using the short chains the child tries to construct regular polygons. With the short chain of 2, he cannot make anything, though with 3, he can make a triangle; with 4, a square; with 5, a pentagon, and so on.


c. Decanomial (a polynomial having ten terms) & The Construction of Chart I

List of Materials

Note: This material is presented parallel to memorization of multiplication, in three different presentations.

…box of bead bars, 1-10, 55 of each
…square of the numbers 2-10
…Multiplication (memorization) Charts I and III


c. Decanomial (a polynomial having ten terms) & The Construction of Chart I

Vertical Presentation

Beginning with the ones table, reconstruct Multiplication Chart I, 1 x 1 =…2 The child states the product and puts out one bead, 1 x 2 = …2 The child takes a unit bead (1) two times and places them in a column, 1 x 3 = …3 The child puts out three unit beads in a column and so on to 1 x 10 = 10.
Construct the twos table 2 x 1 =…2 Place the two bar at the beginning of a new column, so that it lines up with 1 x 1. Go on making a column for the table of 2…2 x 2, 2 x 3, 2 x 4, 2 x 5, 2 x 6…making new columns for each multiplicand, ending the last column with 10 x 10. Use Chart I for control, as you go along, if the child needs it for recalling products. The result will be a square, made up of 10 vertical strips each strip being the width and color of that bead bar.


c. Decanomial (a polynomial having ten terms) & The Construction of Chart I

Horizontal Presentation

This time, the multiplier will remain constant, as we progress along the rows of Chart I. Begin with 1 x 1=…1 Place the unit bead 2 x 1 = 2 Place the two bar next to it forming a row. Go on to 10 x 1 =…10. Beginning the second row with 1 x 2, each of the bars must be taken twice. Place the unit beads in a column forming the beginning of the second row. Continue in this way up to the end of the last row-10 x 10
The result of this work is the same square as before. We can’t tell by looking at the finished product whether it was made the first way or the second way. This square was formed by 10 horizontal strips, varying in width as before, but now with each having the same multi-colored pattern.


c. Decanomial (a polynomial having ten terms) & The Construction of Chart I

Angular Presentation

As always we begin with 1 x 1 = 1. Go on to 2 x 2 = 2; put out a two bar, making a row, 2 x 1 gives us the same as 1 x 2; put out two unit beads making a column 2 x 2 =…4; put out two 2-bars. Outline the formation with a finger to help the child to observe the square that was formed.
Continue with 3 x 1, then 1 x 3…3 x 2, then 2 x 3 and finally 3 x 3 to fill in the arrangement to make a larger square. Continue in this way with the child stating the products for each combination and stopping to observe each square that is formed.
The result of this work is the same square as before. Notice the various geometric forms; point, lines, rectangles, and squares. Substitute real squares for the bars: 1 x 1 is still 1 so we leave the bead. Replace 2 x 2, 3 x 3 and so on up to 10 x 10. We can see that the squares are placed on the diagonal just as they are on Chart III. (outlined in bold black lines)

Control of Error: visible arrangement; number of bead bars in the box

Direct Aim:
…memorization of multiplication
…development of mental flexibility

Indirect Aim: preparation for the Decanomial


c. Decanomial (a polynomial having ten terms) & The Construction of Chart I

Commuted Decanomial

…Two boxes of bead bars 1-10, 55 of each
(actually 1 box of 55 each, and nine 8-bars, twenty six 9-bars, and forty five 10-bars)
…squares and cubes of the numbers 2-10
…Multiplication Charts I and III

Part One 
The Decanomial is lain out already with the squares substituted along the diagonal.
“We’re going to change its beautiful colors. As usual we start with 1×1. It stays the same. Here we have 2×1 which is a green bar. 1×2, which is the same thing as 2×1, is a poor imitation of a bar. We’ll exchange this fake bar for a real 2-bar. Then we have 2×2, which is a square; it remains the same.”
Continue with 3 x 1, and 1 x 3 which are the same thing. Therefore, we’ll make them look the same by placing a pink 3-bar in place of these unit beads. We have 3 x 2 and 2 x 3 which are both equal to 6, because changing the order of the factor doesn’t affect the product. Change the 2-bars for two 3-bars. Then there is 3 x 3, which is the square -9; we have it there.
Continue changing the beads following the same order as the angular presentation that preceded. After we have reached 4 x 4 (see diagram) the changing pattern of colors begins to appear. The strip of yellow outlines two sides of the square, which as before becomes bigger and bigger. After four we’re going to make the right angle blue. Continue with 5 up to 10, where the angle is gold.

Part Two-Building the Tower
The square has changed its colors, but we haven’t finished here. We can see that the diagonal is made up of several squares. The diagonal is like the spinal cord of the square, because, like your spinal cord, it supports everything and keeps it straight.

(Note: present one of these methods to the children)

1st Method: 1 x 1 = 1 It remains the same. Look at the green angle. We have 2 x 2 and
1 x 2. When we place these 2-bars together we make a square of 2. Exchange these 2 bars for a square of 2, since there is no room for it in the places vacated by the 2-bars.
Starting at the column of three, slide the bars of 33 x 3 towards the bar of 3+ ( 3 x 1 ) to see that when combined they form a square of three. Substitute the bars for a square which is placed in the vacant column. Going to the row of three, do the same3 x 2 + 3 x 1 = a square of three.
Continue with the column of four. Push one bar from 4 x 2 down to meet 4 x 3:
4 x 3 + 4 x 1 forms a square. Replace these bars with a square. On the row repeat the procedure: 4 x 3…+…4 x 1…forms a square. Replace the bars with a square. At the head of the row and the column combine the single bars to form 4 x 2 and 4 x 2. Combine these two groups to make a square which is placed on top of the one on the diagonal.
Continue with the column of five: 5 x 4 + 5 x 1 a square. Repeat on the row. The on the column: 5 x 2 + 5 x 25 x 1 = a square. Repeat on the row.
As a result the square will be transformed into an uneven arrangement of squares on the diagonal, with 2 squares stacked at each even number. Superimpose all of the squares to make an oblique line of stacks. These are then transformed into cubes and stacked to make a tower.
Our spinal column has been transformed into a tower, just like the pink tower. Everything that was spread out on this table has been used to construct the tower. Our Pythagorean table has been transformed into a beautiful multi-colored tower just like the ugly duckling became a swan.

Variation on the 1st Method
Since the newly obtained square of two will not fit in the vacated places, we can move the original square and place one square in the column and one in the row. When the problem arises at each even number, re-arrange the portions of the square slightly so that half are in the column and half are in the row. The result will be a broken diagonal, having a vacant space at each even number.

2nd Method
As before, 1 x 1 is left alone, and 2 x 1 is combined with 2 x 1 to form a square which is stacked on top of the other square. The combination of groups of bead bars will always be with reference to the existing square.
For three, choose the bar on the column which is farthest from the square and combine it with the group of bars on the row which are nearest to the square: 3 x 1 + 3 x 2 = a square. Place the new square on top. Continue combining: 3 x 2+3 x 1
gives a square. The third square completes the stack.
For four, start with the bar on the column furthest from the square, combining it with the group of bars on the row nearest the square4 x 1 + 4 x 3 = a square. Then 4 x 2…+…4 x 2… = a square. Continue until all the squares are stacked forming an oblique line. Proceed as before.

Part Three-Decomposition of the Tower
This collective activity is similar to the bank game for the decimal system. One child acts as the banker, while the others change the quantities.
Begin by dismantling the tower of cubes to make a diagonal of cubes. Starting with two, the child recalls that the cube is made from 2 squares of two. The cube is replaced by a stack of squares. Continue until all of the cubes are transformed into stacks.
Take one square of two and ask, ‘What is the square made of?’ two 2-bars. Exchange the square. Where will we put them? They correspond to 2×1 and 1×2; therefore place them accordingly.
Continue to break down squares in a way that is the opposite of the way you chose to construct them. Notice again the angles of color being formed, with the square always serving as the vertex. The result will be the original Pythagorean table from which we began.
This activity is complicated only in the sense that the tasks must be divided among the numbers of the group.
Note: ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 )2 = 552 = 3025
13 + 23 +33 +43 +53 +63 +73 +83 +93 +103 = 3025


c. Decanomial (a polynomial having ten terms) & The Construction of Chart I

Numeric Decanomial

Age: between 6 1/2 and 81/2 years

…to incarnate the geometric figures formed by multiplication
…an indirect preparation for square roots and the square of polynomials

…a control chart
…envelopes numbered 0-9: 0 contains 10 blue squares, 1-9: rectangular pieces on which products are written (i.e. 5 has products of
5 x 6, 5 x 7, 5 x 8, 5 x 9, 5 x 10, 6 x 5, 7 x 5, 8 x 5, 9 x 5, 10 x 5)

Take out the contents of envelope 0 and lay them out in order on the diagonal. If the child hasn’t had powers, simply explain that 102 is one way we can write 10 x 10.
Examine the contents of envelope 1, forming pairs of like rectangles. Mix them up and fish for one. The child reads the product, thinks of the combination and places the rectangle in the formation, just as in the Bingo Game. Allow the child to continue choosing envelopes in any order. At a later stage, eliminate the envelopes 1-9 and mix the pieces in a basket from which the child fishes.
Later the child will realize that this puzzle like the table of Pythagorean is symmetrical, having equal products on opposite sides of the diagonal.


…multiplication combination booklet
…black and red pens
…special combination cards

The procedure is as for special cases in addition and subtraction, resulting in a chart as below: (The words in parentheses do not appear on the chart, but are given orally)

0- Calculate the product
2 x 3 = ? ( two taken three times; what does it give me?)

1- Calculate the Multiplier
2 x ? = 6 ( two taken how many times will give me six?)

2- Calculate the Multiplicand
? x 3 = 6 ( what number taken three times will give me six?)

3- Inverse of Case Zero—Calculate the Product
? = 2 x 3 ( what number do I obtain when I take two, three times?)

4- Inverse of The First—Case, Calculate the Multiplier
6 = 2 x ? ( six is equal to two taken how many times?)

5- Inverse of the Second Case —Calculate the Multiplicand
6 = ? x 3 ( six is equal to what number taken three times?)

6- Calculate the Multiplier and the Multiplicand
6 = ? x ? ( six is the number I obtain when I take a certain number, a certain number of times; what are these numbers?)

Note: In cases 1, 4 and 7, the child performs multiplication, but in all others division is indirectly involved

The seven special combination cards are left at the disposition of the children, combining these with the cards given previously for special cases of addition and subtraction.

Aim: further understanding of the concept of multiplication


Age: 7 years

Aim: development of mental flexibility to prepare for mental calculating

This activity is the culmination of the many skills the child has developed: addition, multiplication at the level of memorization, multiplication and division by powers of 10 and changing from one hierarchy to another. This work is parallel to the large bead frame.

…box containing 9 series of white cards – product
…4 series of colored cards – multiplication
…3 series of gray cards – multiplier
…box of signs for the operations

Invite the children to lay the cards out in columns, naming them as they go. The child should recognize the numerals up to 9 million. If necessary identify for him 10 million and 100 million.
One child must go to the bank. (For the first demonstration use a one-digit multiplier. In all other work a two digit multiplier will be used) Write the problem on a piece of paper. The child takes this to the bank and sets up the multiplication, using colored cards for the multiplicand, signs for the operation and a gray card for the multiplier.

4876 x 6 =

We then decompose the multiplicand, just as was done on the bead frame for multiplication, and begins multiplying:

6 x 6 = 36. He asks the banker for 36. The two cards are placed separately at the near edge of the table. Move the cards for the operation and the multiplier to align them with the next digit – 70. 7 x 6 = 42, 42 tens. He asks the banker for 420. There are placed with the previous product cards so that columns are being formed.

The operation continues in this way. When all of the digits have been multiplied the children-assembles the multiplicand. To find the product, he begins with the lowest hierarchy, combining and making changes. The product cards are assembled and placed by the equal sign. The children record the equation.

Let’s try a different one. Write the problem on a piece of paper:
6835 x 48 =. The child sets up the problem with the colored cards and the gray cards. We don’t have a gray card for forty, so we place a 4 next to a 0 to make 40, then place the 8 on top of the 0 to make 48.
As before, the child decomposes the multiplicand. We only want to multiply by one digit of the multiplier at a time. Remove the 40 and set it aside, yet together for later retrieval
Begin multiplying as before: 5 x 8 = 40. The banker gives 40, etc. After the multiplicand has been multiplied by the units of the multiplier, we can begin multiplying by the tens. ( 8 and 40 switch places, 8 is turned over) However, just as with the bead frame we have the rule: the multiplier must be units. Transfer the zero from the multiplier to the multiplicand.
Continue multiplying. The child will realize that 8 x 4 = 32 and 3 zeros after it- makes 32,000. In the end make changes in the product, carrying mentally, The product is assembled, the equation is recorded.

Notes: Since there is only one set of cards for the product much changing will be involved, calling for quick mental addition and subtraction at the level of memorization, on the part of the banker.
In multiplication with a two-digit multiplier, the child never stops in this game to record partial products, because the aim here is to develop agility in changing from one hierarchy to another. With only one set of cards it would be difficult to set aside partial products.
It is interesting to perform the same problem using the bank game, the checkerboard and the large bead frame to recognize the similarities and differences in the work. Be aware of the limits of the bead frame.


As for addition and subtraction, word problems are prepared on cards which deal with each of the seven special cases. These are mixed in with those given for addition and subtraction.

Steve has 48 stamps in his collection at school. Each day he brought a certain number of stamps, for a certain number of days. How many stamps did he bring each day? and how many days did he bring stamps?

The child writes his answer:

48 = 6 x 8
48 = 8 x 6

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