Trachtenberg system

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The Trachtenberg system is a system of rapid mental calculation. The system consists of a number of readily memorized operations that allow one to perform arithmetic computations very quickly. It was developed by the Russian Jewish engineer Jakow Trachtenberg in order to keep his mind occupied while being held in a Nazi concentration camp. The rest of this article presents some methods devised by Trachtenberg. These are for illustration only. To actually learn the method requires practice. Students begin learning the Trachtenberg system using multiplication algorithms. These initial algorithms are discussed first followed by a more general method for multiplication. Even if you know well how to do arithmetic, the Trachtenberg method can be faster. It is also a method you may use to check work done by more traditional methods.

Beginning to Multiply Using the Trachtenberg Method[edit]

When performing any of these multiplication algorithms the following "steps" should be applied.

Write a zero before the number to be multiplied.

The answer must be found one digit at a time starting at the least significant digit ie the digit to the far right. Move left one digit at the time to work. The last calculation is on the leading zero of the multiplicand. ( In 0128 x 6, the 0128 is the multiplicand and the 6 is the multiplier. In 0382 x 28, the 0382 is the multiplicand and the 28 is the multiplier. The answer may be referred to also as the "product".)

When working with the multiplicand, one digit at a time has your primary focus. The digit immediately to that digit's right is that digit's neighbor. In the Trachtenberg method, never use the digit immediately to the left as the neighbor. The rightmost digit does not have a neighbor as there is no digit to its right. Some algorithms ask for a value of a digit's neighbor. When the focus is on the last digit to the right, the value used for the neighbor is always zero.

The 'halve' operation has a particular meaning to the Trachtenberg system. If a digit is even take half this value. If a digit is odd, mentally subtract one before taking half the value. Alternatively half the odd digit may be taken discarding any decimal, fraction or remainder. For speed reasons people following the Trachtenberg system are encouraged to make this halving process instantaneous. Instead of thinking "half of seven is three and a half, so three" it's suggested that one thinks "seven, three". This speeds up calculation considerably.

The Trachtenberg system also uses something called the tens complement. In this same way the tables for subtracting digits from 10 or 9 are to be memorized. This requires subtracting digits from nine except the last digit to the right of a number which is subtracted from 10. The suggestion is to be able to look at the digits 0-9 and immediately know the result if this digit were subtracted from 9 or subtracted from 10.

Whenever the rule calls for adding half of the neighbor, always add 5 if the current digit is odd.

There will be more discussion on the reasons and details of these methods below.

Multiplying by 10[edit]

It is easier to simply add a final zero to multiply by 10, however it is important to understand also how to multiply by 10 using a Trachtenberg type of technique. This technique will be used when Trachtenberg multiplies by 11, 12, 8, and 9.

We will use the multiplicand ab as an example. In the Trachtenberg method place a leading zero on the multiplicand. We will write the answer just below the multiplicand. Think of 0ab and the answer below it in columns. You will do a procedure to a digit in the multiplicand and this will give the digit in the answer directly below this digit in the multiplicand. The first column considered is that on the far right. In this case this column has a "b" in the m your original multiplicand to your answer.

Example:

        0 3 4 2 5 x 10 = 
        3 4 2 5 0
   (= (0x0)+3) (=(0x3)+4) (= (0x4)+2) (= (0x2)+5) (= (0x5)+0)

Multiplying by 11[edit]

Rule: Add the digit to its neighbor. (By "neighbor" we mean the digit on the right.)

Example:

 03425 x 11 = 
 37675

Each of these digits comes from one times the digit in the multiplicand plus the neighbor ie the digit to the right. Read these from right to left.

(=0+3) (=3+4) (=4+2) (=2+5) (=5+0)

Why does this work? Because 11 = 1+10 . Using the digit itself and the moving the digit to the right over one column to the left (effectively multiplying that digit times ten) accomplishes (1+10 times the multiplier)

Thus,

03425+34250 = 37675

Note: after learning the basic method for the algorithms from right to left, some students practice also from left to right. This entails backing up and correcting for carrying so is mentally more difficult but it is possible if the problems involve higher numbers requiring carrying in the addition.

Multiplying by 12[edit]

Rule: double each digit and add the neighbor. (The "neighbor" is the digit on the right.)

If the answer is greater than a single digit, simply carry over the extra digit (which will be a 1 or 2) to the next operation. The remaining digit is one digit of the final result.

Example:

 0316 x12 = 
 3792

Determine neighbors in the multiplicand 0316:

  • digit 6 has no right neighbor
  • digit 1 has neighbor 6
  • digit 3 has neighbor 1
  • digit 0 (the prefixed zero) has neighbor 3
 (6x2)+0= 12   Write the (2 and carry the 1)
 (1x2)+6+1=9   Where did that final 1 come from? This is the 1 you carried from 12 in the column immediately to the right. 
 (3x2)+1=  7 
 (0x2)+3=  3

Multiplying by 5[edit]

  • Rule: to multiply by 5: Take half of the neighbor, then, if the current digit is odd, add 5.

An odd number = (1+an even number). 5 x an odd number = 5 x (1+ an even number ) = (5x1) + (5 x the even number) = 10 x 1/2 x an even number) .

An even number = (0 + an even number). 5 times an even number = 5 x ( 0 + an even number) = (5x0) + (5 x the even number) = 10 x 1/2 x the even number.

Notice that (10 x 1/2) x an even number may equivalently be calculated as 10 x (1/2 x an even number)

Handle the 1 portion of odd numbers when multiplying by 5 by adding 5 to the answer in that column. Handle the even portion of the number by waiting until your focus has moved one column to the left then add half the neighbor to the right. This last bit may also be described as adding half the neighbor, discarding any decimal, fraction or remainder. This fraction is not being thrown away. This additional portion was handled one column to the right from your current focus when you added 5 if the number was odd. You do not wish to count this portion twice.

Examples:

 042×5=
 210
Half of 2's neighbor, the trailing zero, is 0.
Half of 4's neighbor is 1.
Half of the leading zero's neighbor is 2.
 043×5 = 
 215
Half of 3's neighbor is 0. Remember also to add 5 because 3 is odd
Half of 4's neighbor is 1. Remember to discard any fraction when you divide by 2 for this method.
Half of the leading zero's neighbor is 2.
 093×5=
 465
Half of 3's neighbor is 0, plus 5 because 3 is odd, is 5.
Half of 9's neighbor is 1, plus 5 because 9 is odd, is 6. Remember to discard any remainder when dividing by 2.
Half of the leading zero's neighbor is 4. Remember to discard any remainder or fraction when dividing by 2.

Multiplying by 6[edit]

6= 1 + 5

One times the digit, and add in the multiplication by 5 rule.

  • Rule: to multiply by 6: Take the digit, add 5 if the digit is odd, then add half of the neighbor to each digit, .

Example:

 0357 × 6 = 
 2142

Working right to left,

7 + 5 (since 7 is odd)+0 since 7 has no neighbor to the right = 12. Write 2, carry the 1.
5 + 5 (since the starting digit 5 is odd) +half of 7 (3) + 1 (carried) = 14. Write 4, carry the 1.
3 + 5 (since 3 is odd) + half of 5 (2) + 1 (carried) = 11. Write 1, carry 1.
0 + 0 (since 0 is even) + half of 3 (1) + 1 (carried) = 2. Write 2.

Note: remember when taking half of a number in the Trachtenberg method to discard any remainder or fraction.

Multiplying by 7[edit]

7 = 2 + 5

Double each digit and also add in the multiplication by 5 rule.

Rule: to multiply by 7:

  1. Double each digit.
  2. Add half of its neighbor.
  3. If the digit is odd, add 5.

Example:

 0523 × 7 = 
 3661
Double the 3, add half the neighbor (zero as there is no neighbor, and add 5 as 3 is odd. 11 so write 1 and carry 1.
    (2x3) +(1/2 x 0) + 5     = 11
Double the 2, add half of 3 discarding any fraction. Two is even. Add in the one carried from the 11 total in the prior column.
    (2x2) + (1/2 x 3) +(0) +(1) = 6
Double the 5, add half of 2, add 5 since the 5 is odd. 16 so record 6 and carry 1.
    (2x5)+ (1/2 x 2)+(5) = 16  
Double the 0, add half of 5 discarding any fraction. Zero is even. Add the carried 1 from the 16 total in the prior column.
    (2x0) + (1/2 x 5)+ (0) + (1)=3

Multiplying by 9[edit]

Multiplying by 9 is equivalent to multiplying by (-1 + 10).

This method also relies on the fact that adding a number to a problem then subtracting that same number results in no net change to the problem.

Lets multiply qrstx9 where q,r,s,and t are each digits in a 4 digit number. A leading zero is placed in front of qrst giving us 0qrst. Always remember to place a leading zero in front of the multiplicand (the number you are going to multiply).

0qrstx9 = (-1 x 0qrst)+ (10 x 0qrst)


How will you work with -1 x 0qrst? Try adding a multiple of 10 that is just larger than qrst ie a one followed by as many zeros as there are digits in your multiplicand excluding its leading zero. In this case this means we will use 10000 as qrst is a 4 digit number. If you add in 10000 this will change the problem unless you subtract this back off later.

        10000     
-  1 x  0qrst    
+ 10 x  0qrst 
       -10000
      _______
        9 x 0qrst

In the initial step, you can see why each digit in qrst is subtracted from 9 except the right most digit which is subtracted from 10. This process completes 10000 - (1xqrst). This step is called finding the "tens complement". This procedure is not unique to the Trachtenberg method. See "Method of Complements" in Wikipedia.

Adding the digit to the right in the multiplicand 0qrst is equivalent to multiplying by ten and adding this to your answer.

Finally when working under the leading zero of the factor 0qrst, DO NOT try to create some sort of tens complement for this zero. Remember this is just a place holder and not part of your original multiplicand. Add the neighbor ie q of qrst minus one. You must remember to use q as a neighbor, otherwise you havent completely multiplied qrst by 10. Subtracting 1 in this final step is the equivalent of subtracting 10000. You added 10000 to perform the problem. It must be removed or the problem will be incorrect.

Your digits in your answer should be (and please read these right to left):

0qrst x 9 = q-1___(9-q)+r___(9-r)+s____(9-s)+t____(10-t)

Editors note: Read the underlines as separating the different digits.

Rule:

  1. Subtract the right-most digit from 10.
    1. Subtract the remaining digits from 9.
  2. Add the neighbor.
  3. For the leading zero, subtract 1 from the neighbor.

For rules 9, 8, 4, and 3 only the first digit ( the digit to the far right ) is subtracted from 10. After that each digit is subtracted from nine instead.

Example: 02,130 × 9 = 19,170

Working from right to left:

  • (10 − 0) + 0 = 10. Write 0, carry 1.
  • (9 − 3) + 0 + 1 (carried) = 7. Write 7.
  • (9 − 1) + 3 = 11. Write 1, carry 1.
  • (9 − 2) + 1 + 1 (carried) = 9. Write 9.
  • 2 − 1 = 1. Write 1.

Multiplying by 8[edit]

Rule:

  1. Subtract right-most digit from 10.
    1. Subtract the remaining digits from 9.
  2. Double the result.
  3. Add the neighbor.
  4. For the leading zero, subtract 2 from the neighbor.

To understand what is going on here, understand that 8 = -2+10.

Again let's use qrst as an example 4 digit number.

 -  1 x  0qrst           
 -  1 x  0qrst    
 + 10 x  0qrst 
    __________
     8 x 0qrst

How do you work with - 0qrst? You work with the tens complement. This means adding in a multiple of 10 with as many zeros as digits in your original multiplicand i. e. 4 for 0qrst. Since your working with -0qrst twice you must add this multiple of 10 twice. This is only to make the working easier. You must remove these later if your answer is to remain correct.

This is equivalent to:

 +       10000
 -  1 x  0qrst
 +       10000    
 -  1 x  0qrst    
 + 10 x  0qrst
        -20000           
       __________
        8 x 0qrst 

This simplifies to:

 2 x ( 10000 -  0qrst )
 + 10 x    0qrst
 - 20000
       __________
        8 x 0qrst

So you find the 10s complement of a digit and double it. Add the neighbor to the right i. e. use the multiply by 10 procedure and add that. When working under the leading zero of the multiplicand, get the neighboring digit to the right in the multiplicand and subtract 2. This step subtracts the 20000 you added to work the problem. Use this neighbor to the right minus 2 as your left most digit in your answer.

Your digits in your answer should be (and please read these right to left):

0qrst x 8 = q-2 ____ {2 x (9-q)}+r ____ {2 x (9-r)}+s ____ {2 x (9-s)}+t ____ {2x(10-t)}

Editors note: Read the underlines as separating the different digits.

Example: 456 × 8 = 3648

Working from right to left:

  • (10 − 6) × 2 + 0 = 8. Write 8.
  • (9 − 5) × 2 + 6 = 14, Write 4, carry 1.
  • (9 − 4) × 2 + 5 + 1 (carried) = 16. Write 6, carry 1.
  • 4 − 2 + 1 (carried) = 3. Write 3.

Multiplying by 4[edit]

4 = -1 + 5

-1 times a number means in the Trachtenberg method a tens complement is used.  
5 times the number should suggest to you the multiply by 5s rule.
 -1 X 0qrst
 +5 x 0qrst
 ___________
 4 x 0qrst

Your working this as

      10000
 -    0qrst
 +5 x 0qrst
 -    10000
 _______________
  4 x  0qrst

The 10000 - 0qrst is the tens complement. Subtract each digit from nine unless it is the digit to the far right. For the digit to the far right subtract it from 10. The leading zero in 0qrst is not used to find the tens complement. In other words, this is the same tens complement procedure you have been using.

Add to your answer i. e. to your tens complement, the multiplication by 5 procedure. Remember you add 5 if the number is odd and zero if even. Add half the number to the right discarding any remainder or fraction.

While working under the leading zero of your factor decrease your answer for half the neighbor by 1. You added 10000 to make the work easier. You must subtract that back out at the end or you will have changed the problem and end up with a wrong answer.

Your digits in your answer should be (and please read these right to left):

0qrst x 4 = (q-1) __(9-q) +5 if q is odd + (r/2) __ (9-r) +5 if r is odd + (s/2) __ (9-s) +5 if s is odd + (t/2) __ (10-t)+ 5 if t is odd

Remember in set of equations for the digits in this example when a number divided by 2 is written, discard any remainder or fraction.

Rule:

  1. Subtract the right-most digit from 10.
    1. Subtract the remaining digits from 9.
  2. Add half of the neighbor, plus 5 if the digit is odd.
  3. For the leading 0, subtract 1 from half of the neighbor.

Example: 346 * 4 = 1384

Working from right to left:

  • (10 − 6) + Half of 0 (0) = 4. Write 4.
  • (9 − 4) + Half of 6 (3) = 8. Write 8.
  • (9 − 3) + Half of 4 (2) + 5 (since 3 is odd) = 13. Write 3, carry 1.
  • Half of 3 (1) − 1 + 1 (carried) = 1. Write 1.

Multiplying by 3[edit]

The general idea for multiplying by 3 is that 3 = -2 + 5.

 - 2 x   0qrst 
 + 5 x   0qrst 
 __________________
  3  x   0qrst

Work this as follows:

 2 x ( 10000
     - 0qrst)
+ 5 x 0qrst
-     20000
__________________
  3  x   0qrst

Find the tens complement and double it. Then follow the 5s rule. Remember that doubling the tens complement means you added 10000 twice to make the math easier. You must subtract that when calculating from the leading zero in 0qrst. This means you must subtract 2 from that leading number in your answer before you record it. This is usually expressed as taking 2 from half the neighbor to the right when calculating that last digit.

Your digits in your answer should be (and please read these right to left):

0qrst x 3 = (q-2) __2x(9-q)+5 if q is odd +(r/2) __ 2x(9-r) +5 if r is odd+(s/2) __ 2x(9-s)+5 if s is odd + (t/2) __ 2x(10-t)+ 5 if t is odd

Remember in set of equations for the digits in this example when a number divided by 2 is written, discard any remainder or fraction.

Rule:

  1. Subtract the rightmost digit from 10.
    1. Subtract the remaining digits from 9.
  2. Double the result.
  3. Add half of the neighbor, plus 5 if the digit is odd.
  4. For the leading zero, subtract 2 from half of the neighbor.

Example: 492 × 3 = 1476

Working from right to left:

  • (10 − 2) × 2 + Half of 0 (0) = 16. Write 6, carry 1.
  • (9 − 9) × 2 + Half of 2 (1) + 5 (since 9 is odd) + 1 (carried) = 7. Write 7.
  • (9 − 4) × 2 + Half of 9 (4) = 14. Write 4, carry 1.
  • Half of 4 (2) − 2 + 1 (carried) = 1. Write 1.

Multiplying by 2[edit]

  • Rule: to multiply by 2, double each digit.

Further Note: If you really understand why you are doing the steps, you can create new procedures for yourself such as for multiplying by 15 by using the 10 rule plus the 5 rule ie add 5 if the number is odd and add one and a half times the neighbor discarding any fraction or remainder.

General multiplication[edit]

The method for general multiplication is a method to achieve multiplications  a\times b with low space complexity, i.e. as few temporary results as possible to be kept in memory. This is achieved by noting that the final digit is completely determined by multiplying the last digit of the multiplicands. This is held as a temporary result. To find the next to last digit, we need everything that influences this digit: The temporary result, the last digit of a times the next-to-last digit of b, as well as the next-to-last digit of a times the last digit of b. This calculation is performed, and we have a temporary result that is correct in the final two digits.

In general, for each position n in the final result, we sum for all i:

a \text{ (digit at } i\text{ )} \times b \text{ (digit at } (n-i)\text{)}.

People can learn this algorithm and thus multiply four digit numbers in their head – writing down only the final result. They would write it out starting with the rightmost digit and finishing with the leftmost.

Trachtenberg defined this algorithm with a kind of pairwise multiplication where two digits are multiplied by one digit, essentially only keeping the middle digit of the result. By performing the above algorithm with this pairwise multiplication, even fewer temporary results need to be held.

Example: 123456 \times 789

To find the first digit of the answer:

The units digit of 9 \times 6 = 4.

To find the second digit of the answer, start at the second digit of the multiplicand:
The units digit of 9 \times 5 plus the tens digit of 9 \times 6 plus

The units digit of 8 \times 6.
5 + 5 + 8 = 18.

The second digit of the answer is 8 and carry 1 to the third digit.

To find the fourth digit of the answer, start at the fourth digit of the multiplicand:
The units digit of 9 \times 3 plus the tens digit of 9 \times 4 plus

The units digit of 8 \times 4 plus the tens digit of 8 \times 5 plus
The units digit of 7 \times 5 plus the tens digit of 7 \times 6.
7 + 3 + 2 + 4 + 5 + 4 =  25 + 1 carried from the third digit.
The fourth digit of the answer is 6 and carry 2 to the next digit.
Two headed arrows drawn from each digit of the multiplier to two digits of the multiplicand
2 Finger method

Trachtenberg called this the 2 Finger Method. The calculations for finding the fourth digit from the example above are illustrated at right. The arrow from the nine will always point to the digit of the multiplicand directly above the digit of the answer you wish to find, with the other arrows each pointing one digit to the right. Each arrow head points to a UT Pair, or Product Pair. The vertical arrow points to the product where we will get the Units digit, and the sloping arrow points to the product where we will get the Tens digits of the Product Pair. If an arrow points to a space with no digit there is no calculation for that arrow. As you solve for each digit you will move each of the arrows over the multiplicand one digit to the left until all of the arrows point to prefixed zeros.

Addition[edit]

A method of adding columns of numbers is presented in the book by Trachtenberg. His addition method uses intermediate totals. These are used in an L-shaped algorithm to check for accuracy without repeating the initial procedure. This method allows the precise column in which an error occurs to be identified. For the procedure to be effective, the different operations used in each stages must be kept distinct, otherwise there is a risk of interference.

Division[edit]

Setup for division using Trachtenberg Method
Setting up for Division

Division in the Trachtenberg System is done much the same as in multiplication but with subtraction instead of addition. Splitting the dividend into smaller Partial Dividends, then dividing this Partial Dividend by only the left-most digit of the divisor will provide the answer one digit at a time. As you solve each digit of the answer you then subtract Product Pairs (UT pairs) and also NT pairs (Number-Tens) from the Partial Dividend to find the next Partial Dividend. The Product Pairs are found between the digits of the answer so far and the divisor. If a subtraction results in a negative number you have to back up one digit and reduce that digit of the answer by one. With enough practice this method can be done in your head.

Publications[edit]

  • Rushan Ziatdinov, Sajid Musa. Rapid mental computation system as a tool for algorithmic thinking of elementary school students development. European Researcher 25(7): 1105-1110, 2012 [1].
  • The Trachtenberg Speed System of Basic Mathematics by Jakow Trachtenberg, A. Cutler (Translator), R. McShane (Translator), was published by Doubleday and Company, Inc. Garden City, New York in 1960.[1]

References[edit]

  1. ^ Trachtenberg, Jakow (1960). The Trachtenberg Speed System of Basic Mathematics. Translated by A. Cutler, R. McShane. Doubleday and Company, Inc. p. 270.