# The Cranmer Abacus -- Introduction

The abacus used by individuals who are blind is called the Cranmer Abacus. It is based on the Japanese Soroban abacus with some tactile modifications. The abacus allows students to set up and calculate math problems, without the aid of a calculator. The use of the abacus develops math concepts and skills.

The abacus has 13 vertical rods with 5 beads on each rod.
The column to the farthest right, is the one's column.
The column to the left of that is the ten's. Left of that is the hundred's, then the thousand's, ten thousand's, and continues in place value up to the trillion's column.

A horizontal divider bar separates the single top bead from the lower 4 beads in each column.

On the divider bar, there are 4 vertical lines placed every third column called unit marks. These tactile markers help identify the location of the columns, as unit marks are in the same locations as commas when writing large numbers.

When beads are push toward the divider bar, they are said to be "set".
When all the beads in a column are pushed away from the divider bar, it is said to be "clear".
The beads below the bar each have a value of 1.
The single bead above the bar has a value of 5.

To set the number "1", push one lower bead, in the far right column, up toward the bar. The number "1" is now "set".
To set the number "2", two lower beads are pushed up to the bar.
To set the number "3", three lower beads are push up.
To set the number "4", all four lower beads are pushed up to the bar.
To set the number 5, push the upper bead down to the bar and clear the 4 lower beads.
With the upper bead set, we can continue counting to 6 by setting one more lower bead.
7 has 2 lower beads set
8 has 3 lower beads set
and 9 has all 4 lower beads as well as the upper bead set.
There are no more beads to set in the one's column. To set the number 10, set 1 lower bead in the second column from the right, giving us a 1 in the ten's column. You must then clear the 9 in the one's column. This gives a 1 in the ten's column and a zero in the one's column.

Let's set more numbers. First, clear the abacus by pushing all the beads away from the bar. Numbers on the abacus are set from left to right, in the order that they are spoken. To set the number 47, you first set 4 in the ten's column, and then set 7 in the one's column

To set the number 810 we first clear the abacus, and begin setting the number from left to right. Set 8 in the hundred's column, set 1 in the ten's column, and the one's column will remain clear giving it a value of zero.

Here is another example to set on the abacus. The number to set is 2,508. Locate the thousand's column and set the number 2. Notice that a unit mark on the divider bar is immediately to the right of the thousand's column, where a comma would be placed. Next set the number 5 in the hundred's column. The ten's column will remain clear, giving it a value of zero. Then set the number 8 in the one's column.

You should practice setting more numbers and become comfortable with the process before starting addition.

Addition is done on the abacus using direct and indirect methods.
When we add on the abacus, we will be working from left to right.

First clear your abacus and work the problem 2+2=4 Begin by setting the number 2 in the one's column. To add another 2, simply set 2 more lower beads. The answer is 4. This is direct addition.

Next work the problem 6+3=9. Start by clearing your abacus. Set the number 6 in the one's column, and then add 3 by setting 3 lower beads. The answer is 9 This is another example of direct addition.

Indirect Addition requires the use of logical exchange or memorizing the exchange as "secrets"
Try adding 4+3. Start by clearing your abacus and setting 4 in the one's column. When we try to add 3, we find there are no more lower beads, so we must set the 5 bead. We wanted to add 3 but had to add 5, so we must clear 2. The answer is 7. This problem uses indirect addition.

Here is another problem using indirect addition. Try 8+9=17. First set the 8 in the one's column. There are not enough beads in the one's column to add 9, so you will set one bead in the next column to the left, actually adding 10. You have added 10, but only wanted to add 9, so you must clear 1 beads from the one's column. The answer is 17.

Let's try some larger numbers.
The problem is 32+12. First, set the 3 in the ten's columns and the 2 in the one's column. Remember that setting large numbers and performing calculations are done from left to right.
When adding 12, you will begin in the ten's column to add the 1. Then move to the one's column and add the 2 using direct addition. The answer is 44

Next let's try 2,474+5,316
First set 2,474 from left to right in the order that it is spoken. Beginning in the thousand's column, set 2 thousand, 4 hundred, 74. Working from left to right, begin in the thousands column, Add 5 using direct addition. In the hundred's column, add 3, by using indirect addition, set 5 and clear 2. In the ten's column, add 1 using direct addition. Finally, in the one's column, add 6 using indirect addition -- set one bead in the next column to the left and clear 4. The answer is 7,790

Now try 669+333
Begin by setting 669. Add 3 to the hundred's column. Add 3 to the ten's column. When we add 3 in the one's column, we realize that we cannot set one left in the ten's column or in the hundred's column. We must set one in the thousands column. When columns need to be jumped over to set one in the next higher column, you must clear the columns that were jumped over. In this case we must clear the hundred's column and the ten's column. We must then come back to the one's column and clear 7. The answer is 1,002.

# Cranmer Abacus -- Subtraction

Subtraction, like addition, uses direct and indirect methods.

First, work the problem 9-2 using direct subtraction. Start by clearing your abacus and set 9 in the one's column. Subtract 2 by clearing 2 lower beads. The answer is 7.

Next, clear your abacus and try 38-16. Set 3 in the ten's column, and set 8 in the one's column. Remember that setting numbers and calculating are done from left to right. First, locate the ten's column and subtract 1 from it. Then subtract 6 from the one's column. The answer is 22.

Now try some problems using indirect subtraction.
The first problem is 7-3. Begin by clearing your abacus and set 7 in the one's column. To subtract 3, you must subtract 5. You subtracted 5, but only wanted to subtract 3, so you must put 2 beads back. The answer is 4.

The next problem is 26-9. Clear your abacus. Start in the ten's column and set 2. Then set 6 in the one's column. To subtract 9 from the one's column, you find there are not enough beads. You must go to the column to the left and subtract 10 by clearing one bead. You subtracted 10, but only wanted to subtract 9, so you must put one back by setting one bead in the one's column. The answer it 17.

Now try 52-6. Set 52. To subtract 6 from the one's column, you find there are not enough beads, so you must go to the next column to the left and clear one. In this case, to clear one from the ten's column requires indirect subtraction again – clear 5 and set 4. You have subtracted 10, but only needed to subtract 6, so you must put 4 back. Here, you must use indirect addition -- set 5 and clear 1. The answer is 46.

The last subtraction problem to try is 3,002-4. First set 3000 and 2. You find there are not enough bead in the one's column to subtract 4, so you must go to the next column to the left and clear one. This is not possible in the ten's column or the hundred's column. You must go to the thousand's column to clear 1. When you need to clear one from the column to the left, and must jump over a column to do so, that column must be changed to 9. In this problem, we jumped over the ten's column and the hundred's column. We must therefore set a 9 in the hundred's column and a 9 in the ten's column. In the one's column, 10 was subtracted but only 4 needed to be subtracted, so you must put 6 back. The answer is 2,998.

# Cranmer Abacus -- Multiplication

Now that you are comfortable with direct and indirect addition and subtraction on the abacus, we can begin multiplication. It is advisable that your student have studied and memorized the times tables for multiplication before teaching multiplication on the abacus.
Multiplication requires numbers to be correctly located in specific columns.

In the example 7 times 9 = 63, 7 is the multiplicand, 9 is the multiplier and 63 is the product.
On the abacus, the multiplicand, 7, is set at the left side. The multiplier, 9, will be set in a location that is determined by counting the digits in the multiplicand and the multiplier and adding 1. In this problem, there is one digit in the multiplicand and one digit in the multiplier, plus 1 equals 3. Begin counting columns from the right side and set the multiplier, 9, in the third column. Now multiply 7 times 9. In the 2 columns immediately right of the multiplier, set the answer, 63. Now, clear the 9.

Now try the problem 3 times 21.
Set the 3 in the first column at the left.
Count the number of digits in the problem and add 1. The result for this problem is 4.
Beginning at the right side, count over to the fourth column where you will begin setting the number 21
First, multiply 3 times 1 and set the answer in the two columns immediately to the right of the multiplier. This answer has a leading zero before the 3. It is important to say the leading zero to maintain correct column positioning in multiplication. Now, clear the 1.
Next multiply 3 X 2 . Set this two digit answer immediately to the right of the 2. The answer is 06.
Now, clear the 2. The answer is 63.

The next problem 8 X 76
Set the 8 in the first column from the left.
Count the number of digits in the problem and add 1. The result is 4.
Beginning at the right side over to the fourth column, and set 76.
First, multiply 8 X 6. In the 2 columns immediately to the right of the 6, set the 48.
Now, clear the 6.
Next multiply: 8 X 7. In the two columns immediately right of the 7, set the answer, 56You will need to add the 6 of 56 to the column where the 4 of 48 was set. To do this, you must set 1 left, and clear 4.
Now, clear the 7.

The next problem is 26 X 73
Beginning in the left most column, set 26.
Count the number of digits in the problem and add 1. The result is 5.
Beginning from the right side, count to the fifth column and set 73.
First multiply 2 X 3. In the 2 columns immediately to the right of the 3, set 06. Remember that when a partial product is a single digit, it must have a leading zero. Keep the index finger of your right hand on the 6 in the second position.
Next multiply 6 X 3. Beginning in the column where your finger is placed, set 18.
Now clear the 3 from the multiplier.
Next multiply 2 X 7. In the 2 columns immediately to the right of the 7, set 14. Keep the index finger of your right hand on the 6 in the second position.
Next multiply 6 X 7. Beginning in the column where your finger is placed, set 42.
Now clear the 7 form the multiplier.
The product is 1,898

The next problem is 67 X 50
Beginning in the left most column, set 67.
Count the number of digits in the problem and add 1. The result is 5.
Beginning from the right side, count to the fifth column and set 50.
There is nothing to multiply for the 0, however the zero must be counted to set the multiplier in the correct columns.
Multiply 6 X 5. In the 2 columns immediately to the right of the 5, set the 30. Keep the index finger of your right hand on the zero in the second position.
Next multiply 7 X 5. Beginning in the column where your finger is placed, set 35.
Now clear the 5 from the multiplier.
The product is 3,350

The last problem is 27 X 902
Beginning in the first column from the left, set 27.
Count the number of digits in the problem and add 1. The result is 6
Beginning from the right side, count to the sixth column and set 902.
Multiply 2 in the multiplicand times the 2 in the multiplier. In the two columns immediately to the right of the multiplier, set the answer, 04. It is important to keep your right index finger on the 4.
Next multiply 7 in the multiplicand times 2 in the multiplier. Beginning in the column that your right index finger is on, set 14, adding the 1 to the column that contains the 4 by using indirect addition to set 5 and clear 4. Then set 4 in the next column to the right.
Now, clear the 2 from the end of the multiplier.
There is nothing to multiply for the zero, so next multiply 2 in the multiplicand times the 9 in the multiplier. In the 2 columns immediately to the right of the 9, set the answer, 18. Keep your index finger of your right hand on the second column where the 8 is set.
Next multiply 7 in the multiplicand time the 9 in the multiplier. Beginning in the column where your right index finger is placed, set the answer 63. In the first position, you find you must add the 6 of 63 to the 8 of the 18 that was set. You must set 1 left and clear 4; to clear 4, you must clear 5 and set 1. You now go to the second position and set the 3.
Now clear the 9 from the multiplier.

# Cranmer Abacus -- Short Division

When performing division on the abacus, the divisor is set on the left side, and the dividend is set at the right side. The quotient is set in the middle with the number of columns after it equal to the sum of the digits in the divisor plus 1.

In the example 56 divided by 7 = 8
Set the divisor, 7 at the left and the dividend, 56 at the right.
The location of the quotient will be determined by the calculation.
First, see if the divisor 7 will go into the first digit of the dividend, 5. It will not, so calculate 7 into 56. The answer is 8. Because the division was done with two digits of the dividend, the answer will go in the column immediately left of the 56.
Now we multiply the divisor 7, times the quotient 8, to get 56. This product, 56, is subtracted from the two digit dividend, 56, clearing the last two columns at the right.
The quotient is 8. We know it is 8 (and not 80 or 800) because there will be 2 columns after the quotient. The number of columns after the quotient is equal to the sum of the digits in the divisor plus 1.

The next problem is 75 divided by 5 = 15
Set the divisor, 5 at the left and the dividend, 75 at the right.
The location of the quotient will be determined by the calculation.
First, see if the divisor 5 will go into the first digit of the dividend, 7. It will, so calculate 5 into 7. The answer is 1. Because the division was done with one digit of the dividend, the answer will be set two columns left of the 7.
Now we multiply the divisor 5, by the quotient 1, to get 05. This product, 05, is subtracted from the two columns immediately to the right. Next, the divisor 5 goes into 25. The answer is 5 and will be set in the column immediately to the left because the division was done with two digits of the dividend.
Multiply the dividend 5 by the answer 5 to get 25. This product is then subtracted from the 25, and the last two columns are cleared.
The quotient is 15. We know it is 15 (and not 150 or 1500) because there will be 2 columns after the quotient. The number of the columns after the quotient is equal to the sum of the digits in the divisor plus 1.

The next problem is 374 divided by 6 = 62 r2
Set the divisor, 6 at the left and the dividend, 374 at the right.
The location of the quotient will be determined by the calculation
First, see if the divisor 6 will go into the first digit of the dividend, 3. It will not, so calculate 6 into 37. The answer is 6. Because the division was done with two digits of the dividend, the answer will be set immediately to the left of the 37.
Now we multiply the divisor 6, times the quotient 6, to get 36. This product, is subtracted from the 37, leaving a 1 in the second column from the right.
Next, 6 goes into 14. The answer is 2. Because the division was done with two digits of the dividend, the answer will be set immediately to the left of the 14.
Multiply the divisor 6 by the answer 2 to get 12. This product is then subtracted from the 14, leaving a 2 in the last column as a remainder.
The quotient is 62 with a remainder of 2. We know it is 62 (and not 620 or 6202) because there will be 2 columns after the quotient. The number of columns after the quotient is equal to the sum of the digits in the divisor plus 1.

The last problem to try is 7283 divided by 8 = 910 r3
Set the divisor, 8 at the left and the dividend, 7283 at the right.
The location of the quotient will be determined by the calculation
First, see if the divisor 8 will go into the first digit of the dividend, 7. It will not, so calculate 8 into 72. The answer is 9. Because the division was done with two digits of the dividend, the answer will be set immediately to the left of 72.
Now multiply the divisor 8, times the answer 9, to get 72. This product, 72, is subtracted from the two digits of the dividend, 72, clearing the two columns.
Next, see if the divisor 8 will go into the first digit of remaining dividend 8. 8 will go into 8, and the answer 1 will be set two columns to the left of the 8 because the division was done with one digit of the dividend.
Now multiply the dividend 8 by the answer 1 to get 08. This product is subtracted from the 8, leaving the column clear. Next see if the divisor 8 will go into the remaining dividend 3. It will not, so this number is the remainder.
The answer must be checked for how many columns should follow the quotient. Remember that the number of columns will equal the number of digits in the divisor plus 1. In this problem, the 1 digit in the divisor plus 1 determines that there will be 2 columns after the quotient. Therefore, the quotient has a zero at the end. The final answer for this problem is 910 with a remainder of 3.