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There is a trick that is taught to children to help them remember their 9 times tables.

As illustrated below, the digits of the multiplication problem in question can be determined by:

  • Splitting the digits 1 to 0(10) by whatever number you are multiplying 9 by.
  • The resulting sums of the digits on either side of the split make up the the digits of the answer.

(Feel free to edit this description to make it more mathematically correct and/or intuitive)

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How does this work? Why does it only work for the 9 times tables, and not any of the other times tables?

The answer is probably fairly intuitive, but I'd like to hear what math experts have to say about it.

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Two hands have $10$ fingers. Notice that multiples of $9$ up to $9 \cdot 10\,$ follow the pattern where the tens digit and ones digit add up to $9$ $(0+9, 1+8, 2+7$, etc. $)$ which allows for an extra finger to act as a "delimiter" of sorts.

They also follow the pattern where the tens digit increases one at a time, while the ones digit decreases one at a time as you increase by $9$. This is because $9$ is one less than $10$, which you think of as the reason why the tens digit increases by one, yet the ones digit decreases by one. With algebra, you can express a multiple of $9$ as $9x$:

$$9x = (10-1)x \Rightarrow10x-x\Rightarrow 10(x-1)+(10-x)$$

Where the first term represents the tens digit and the latter the ones digit. Thus, the tens digit increases while the ones decreases. And if you add the tens and ones digits:

$$x-1+10-x=9$$

Thus the digits always add up to $9$ up to $9 \cdot 10$.

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It depends on the fact that $9=10-1$, so $9a=(10-1)a=10a-1a=10(a-1)+(10-a)=10(a-1)+(9-(a-1))$. This says the tens digit of the product is one less than the number multiplied by $9$ and the ones digit is the tens digit subtracted from $9$.

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