The Trachtenberg speed method

TODO — replace with Arosh’s own write-up. The Maths below is correct; the words are a starting point.

Jakow Trachtenberg worked out a system of mental arithmetic while imprisoned in a Nazi concentration camp, with no paper and nothing else to do. The system replaces the times tables with a short list of rules, one per multiplier, each of which you apply digit by digit — so you never hold more than one small sum in your head at a time.

Two of the rules are worth learning even if you ignore the rest.

Multiplying by 11: add the neighbour

Write the number with a zero at each end. Then each digit of the answer is that digit plus the digit on its right.

Take $3254 \times 11$. Writing it out as $0\,3\,2\,5\,4\,0$ and working from the right:

Digit	Neighbour on its right	Answer digit
4	0	4
5	4	9
2	5	7
3	2	5
0	3	3

Reading the answer digits back: $3254 \times 11 = 35\,794$.

If a column comes to 10 or more, write the units digit and carry the 1 into the next column, exactly as in ordinary addition.

Multiplying by 12: double, then add the neighbour

Same idea, one extra step: each answer digit is twice the digit, plus the digit on its right.

For $3254 \times 12$, working from the right:

• $2 \times 4 + 0 = 8$
• $2 \times 5 + 4 = 14$ → write 4, carry 1
• $2 \times 2 + 5 + 1 = 10$ → write 0, carry 1
• $2 \times 3 + 2 + 1 = 9$
• $0 + 3 = 3$

Giving $3254 \times 12 = 39\,048$.

Why the rules work

This is the part worth understanding, because once you see it you can derive the rule for any small multiplier yourself rather than memorising a list.

Write the number in terms of its digits:

$$N = \sum_{i} d_i \cdot 10^{i},$$

where $d_i$ is the digit in the $10^i$ column, and $d_i = 0$ for any column past the end of the number.

Now split the multiplier. Since $11 = 10 + 1$,

$$11N = 10N + N.$$

Look at what lands in column $i$ of the answer. The $N$ term contributes $d_i$, its own digit. The $10N$ term is the whole number shifted one place left, so what it contributes to column $i$ is $d_{i-1}$ — the digit that was one place to the right. Adding them:

$$\text{column } i = d_i + d_{i-1}.$$

That is exactly “add the neighbour”. The zeros at the ends are just a reminder that $d_{-1} = 0$ and that the leading digit still has a column of its own to spill into.

The rule for 12 falls out the same way, since $12 = 10 + 2$:

$$12N = 10N + 2N \quad\Longrightarrow\quad \text{column } i = 2d_i + d_{i-1}.$$

“Double the digit and add the neighbour.”

Making your own rules

The same argument gives a rule for any multiplier you can write as a small combination of powers of ten. For $13 = 10 + 3$ the rule is triple the digit and add the neighbour; for $9 = 10 - 1$ it is subtract the digit from its neighbour, with borrowing.

Trachtenberg’s full system has cleverer rules for the harder multipliers — 6, 7 and 8 involve halving as well — but they are all the same trick underneath: break the multiplier into pieces that only ever move digits between neighbouring columns, so no step needs more than single-digit arithmetic.