Trachtenberg's method of fast addition
The Trachtenberg addition method is a procedure for adding several numbers quickly and then using another procedure to check that we have counted correctly. The procedure was invented by Russian mathematician Yakov Trachtenberg. We will show the method by an example. Let's try to add these three numbers:
$$9551+7375+9262=?$$
First, we write the numbers underneath each other as follows:
$$\begin{array}{cccc} &9&5&5&1\\ +&7&3&7&5\\ +&9&2&6&2\\ \hline \end{array}$$
Now we start adding the columns, starting from the right. We will be interested in whether the sum is greater than or equal to 11. In the first step, we add the last column, i.e. we add 1 + 5 + 2 = 8. This result is less than 11, so we just write the number below the line:
$$\begin{array}{cccc} &9&5&5&\color{red}{1}\\ +&7&3&7&\color{red}{5}\\ +&9&2&6&\color{red}{2}\\ \hline &&&&\color{red}{8}\\ \end{array}$$
In the second step, add the second column from the right, i.e. add 5 + 7 + 6 = 18. The Trachtenberg method says that whenever the sum is greater than or equal to 11, we subtract 11 from the sum and remember how many times we had to do this. Since 18 is greater than 11, we subtract 18 − 11 = 7. We write the number seven below the line, and since we subtracted 11 once, we note it one line below:
$$\begin{array}{cccc} &9&5&\color{red}{5}&1\\ +&7&3&\color{red}{7}&5\\ +&9&2&\color{red}{6}&2\\ \hline &&&\color{red}{7}&8\\ &&&\color{red}{1}&\\ \end{array}$$
If we go back to the previous column on the far right, we can see that the sum was equal to eight, so we didn't have to subtract the number 11 even once. So let's add the number zero to the table:
$$\begin{array}{cccc} &9&5&5&\color{red}{1}\\ +&7&3&7&\color{red}{5}\\ +&9&2&6&\color{red}{2}\\ \hline &&&7&\color{red}{8}\\ &&&1&\color{red}{0}\\ \end{array}$$
We continue with the third column from the right and add 5 + 3 + 2 = 10. We are not afraid of the number ten, it is a number smaller than 11, so we write it below the line along with zero because we did not have to subtract 11:
$$\begin{array}{cccc} &9&\color{red}{5}&5&1\\ +&7&\color{red}{3}&7&5\\ +&9&\color{red}{2}&6&2\\ \hline &&\color{red}{10}&7&8\\ &&\color{red}{0}&1&0\\ \end{array}$$
And now we add the fourth column from the right, i.e. the first column. We see that 9 + 7 = 16, which is already greater than 11. So we subtract 16 − 11 = 5, remembering that we have already subtracted 11 once, and then we continue with the number 5, which we add with the number in the third row, i.e. we count 5 + 9 = 14. The number fourteen is also greater than 11, so we subtract 14 − 11 = 3. We write this result below the line along with the information that we had to subtract 11 even twice:
$$\begin{array}{cccc} &\color{red}{9}&5&5&1\\ +&\color{red}{7}&3&7&5\\ +&\color{red}{9}&2&6&2\\ \hline &\color{red}{3}&10&7&8\\ &\color{red}{2}&0&1&0\\ \end{array}$$
We'll add one more column with two zeros to the result below the line, it will come in handy later:
$$\begin{array}{cccc} &9&5&5&1\\ +&7&3&7&5\\ +&9&2&6&2\\ \hline \color{red}{0}&3&10&7&8\\ \color{red}{0}&2&0&1&0\\ \end{array}$$
How to get the result of the Trachtenberg method
Let's make another horizontal line and add the last two columns. The first thing we do is add 8 + 0 = 8 and write the result below the line:
$$\begin{array}{cccc} &9&5&5&1\\ +&7&3&7&5\\ +&9&2&6&2\\ \hline 0&3&10&7&\color{red}{8}\\ 0&2&0&1&\color{red}{0}\\ \hline &&&&\color{red}{8}\\ \end{array}$$
From this point on, we have to get more complicated and add the numbers in the shape of the letter "L". So we add the two numbers from the second to last column and add to them the second number from the previous column, thus adding 7 + 1 + 0 = 8:
$$\begin{array}{cccc} &9&5&5&1\\ +&7&3&7&5\\ +&9&2&6&2\\ \hline 0&3&10&\color{red}{7}&8\\ 0&2&0&\color{red}{1}&\color{red}{0}\\ \hline &&&\color{red}{8}&8\\ \end{array}$$
Next, add 10 + 0 + 1 = 11. Now we proceed as in ordinary addition: we write down only the last digit 1 and the ten goes to the next round.
$$\begin{array}{cccc} &9&5&5&1\\ +&7&3&7&5\\ +&9&2&6&2\\ \hline 0&3&\color{red}{10}&7&8\\ 0&2&\color{red}{0}&\color{red}{1}&0\\ \hline &&\color{red}{1}&8&8\\ \end{array}$$
In the next round, we add 3 + 2 + 0 + 1 = 6, where the one at the end is the one that went on in the previous round (i.e. for every ten that went on, we add one here).
$$\begin{array}{cccc} &9&5&5&1\\ +&7&3&7&5\\ +&9&2&6&2\\ \hline 0&\color{red}{3}&10&7&8\\ 0&\color{red}{2}&\color{red}{0}&1&0\\ \hline &\color{red}{6}&1&8&8\\ \end{array}$$
And now we finally use the column with two zeros - it's purely there so we don't forget to do the last sum 0 + 0 + 2 = 2.
$$\begin{array}{cccc} &9&5&5&1\\ +&7&3&7&5\\ +&9&2&6&2\\ \hline \color{red}{0}&3&10&7&8\\ \color{red}{0}&\color{red}{2}&0&1&0\\ \hline \color{red}{2}&6&1&8&8\\ \end{array}$$
This is our result, which we obtained using the Trachtenberg method. That is,
$$9551+7375+9262=26188.$$
Verification of intermediate results
The advantage of the Trachtenberg method is that we can relatively easily verify that we have calculated correctly without doing the whole calculation again. Trachtenberg found a way to verify intermediate results and the final result by a completely different procedure. So if you calculate the sum using Trachtenberg's method and run the test at the same time, you can be fairly confident that you have calculated correctly. So let's try to verify that our result is correct.
We can divide our whole calculation into three parts. We have the addends, the intermediate results, and the result:
$$\begin{array}{cccc} &\color{red}{9}&\color{red}{5}&\color{red}{5}&\color{red}{1}\\ +&\color{red}{7}&\color{red}{3}&\color{red}{7}&\color{red}{5}\\ +&\color{red}{9}&\color{red}{2}&\color{red}{6}&\color{red}{2}\\ \hline \color{green}{0}&\color{green}{3}&\color{green}{10}&\color{green}{7}&\color{green}{8}\\ \color{green}{0}&\color{green}{2}&\color{green}{0}&\color{green}{1}&\color{green}{0}\\ \hline \color{blue}{2}&\color{blue}{6}&\color{blue}{1}&\color{blue}{8}&\color{blue}{8}\\ \end{array}$$
In the first stage, we check that we have calculated the intermediate results correctly. To do this, we need to perform column numbering. We get the digitization of a number by adding all the digits of the number. For example, we get the digit of 456 by adding 4 + 5 + 6 = 15. If the result is greater than 9, we perform another digitization, i.e. we still calculate 1 + 5 = 6. The digit of 456 is thus equal to 6.
Now we return to the addends and compute the digit of the last column, i.e., compute the digit of 152, compute 1 + 5 + 2 = 8.
$$\begin{array}{cccc} &9&5&5&\color{red}{1}\\ +&7&3&7&\color{red}{5}\\ +&9&2&6&\color{red}{2}\\ \hline &&&&\color{red}{8} \end{array}$$
Next we calculate the digit of the penultimate column, i.e. the number 576, which is 5 + 7 + 6 = 18, next we get 1 + 8 = 9.
$$\begin{array}{cccc} &9&5&\color{red}{5}&1\\ +&7&3&\color{red}{7}&5\\ +&9&2&\color{red}{6}&2\\ \hline &&&\color{red}{9}&8 \end{array}$$
For the remaining two columns, we get 5 + 3 + 2 = 10, since the number has two digits, we get 1 + 0 = 1:
$$\begin{array}{cccc} &9&\color{red}{5}&5&1\\ +&7&\color{red}{3}&7&5\\ +&9&\color{red}{2}&6&2\\ \hline &&\color{red}{1}&9&8 \end{array}$$
and for the first column we get 9 + 7 + 9 = 25 and then 2 + 5 = 7:
$$\begin{array}{cccc} &\color{red}{9}&5&5&1\\ +&\color{red}{7}&3&7&5\\ +&\color{red}{9}&2&6&2\\ \hline &\color{red}{7}&1&9&8 \end{array}$$
The resulting digits are equal to : 7198. Now we look at the intermediate results:
$$\begin{array}{cccc} \color{green}{0}&\color{green}{3}&\color{green}{10}&\color{green}{7}&\color{green}{8}\\ \color{green}{0}&\color{green}{2}&\color{green}{0}&\color{green}{1}&\color{green}{0}\\ \end{array}$$
We copy the second row and paste it at the end, so the second row will be there twice:
$$\begin{array}{cccc} 0&3&10&7&8\\ 0&2&0&1&0\\ 0&2&0&1&0\\ \end{array}$$
Now we do the same as with the addends. We create a number from each column and calculate its digits. Again we can start from the end, so we get 8 + 0 + 0 = 8, the next column will give us 7 + 1 + 1 = 9, the next 10 + 0 + 0 = 10 and then 1 + 0 = 1 and finally 3 + 2 + 2 = 7:
$$\begin{array}{cccc} 0&3&10&7&8\\ 0&2&0&1&0\\ 0&2&0&1&0\\ \hline 0&\color{red}{7}&\color{red}{1}&\color{red}{9}&\color{red}{8} \end{array}$$
At this point we compare the results of the digitization. The resulting digitizations of the adders gave us the number 7198, as did the digitizations of the intermediate results - so we have calculated correctly. If these results were different, we made a mistake somewhere. Either during the test or during the count itself.
Verification of the result
Verifying that we have calculated the result correctly is easy at this point. We calculate the digitization of the result and the digitization of the digits and it should come out the same. So, we take our result 26 188 and calculate its digitization:
$$\begin{eqnarray} 2+6+1+8+8&=&25\\ 2+5&=&7 \end{eqnarray}$$
Now we calculate the digitization of our calculated digits, i.e. the digitization of 7198:
$$\begin{eqnarray} 7+1+9+8&=&25\\ 2+5&=&7 \end{eqnarray}$$
We can see that both digitizations are equal to seven, so the test went well and we calculated correctly!