# multiplying on fingers also work with bases other than ten, why

• Sep 19th 2010, 08:01 AM
misterchinnery
multiplying on fingers also work with bases other than ten, why
At school I was shown how to multiply two intergers, both > 5 and <11 using the fingers on both hands.
Each hand used to represent one of the two numbers.
It works like this e.g 7x8= 56.
On the left hand count from 6 to 7 by folding down two fingers so that two are bend and three are not.
Then on the right hand count from 6 to 8 so that three are bend and two are not.
Count the number of straight fingers on each hand 3x2=6. this represents 6 units.
Count number of bent fingers on both hands and this represent the number of tens, five tens, or fifty.
So the answer is fifty plus six
The question is:
How come this works if we didnt have ten fingers and we counted for example in base 8 and had 8 fingers? If we counted in base 8 we would have 4 fingers on each hand and could multipy numbers from 5 to 8 using this method.
• Sep 19th 2010, 08:31 AM
emakarov
Quote:

How come this works if we didnt have ten fingers and we counted for example in base 8 and had 8 fingers?
Why does it work for ten fingers? It is probably easy to generalize to other bases.
• Sep 19th 2010, 08:35 AM
Soroban
Hello, misterchinnery!

Wow! . . . a fascinating trick!

We have two sets of five fingers, each set numbered from 6 to 10.

. . $\begin{array}{ccccccccccccccccccc}
| & | & | & | & | &&& |&|&|&|&| \\
6&7&8&9&10 &&& 6&7&8&9&10 \end{array}$

To multiply $7 \times 9$:

On the left hand, bend fingers 6 and 7.
On the right hand, bend fingers 6, 7, 8 and 9.

. . $\begin{array}{ccccccccccccccccccc}
\times & \times & | & | & | &&& \times & \times & \times & \times &| \\
6&7&8&9&10 &&& 6&7&8&9&10 \end{array}$

. . . $\begin{array}{c}\text{Left hand} \\ \hline
\text{bent: 2} \\ \text{straight: 3} \end{array} \qquad\quad \begin{array}{c}\text{Right hand} \\ \hline \text{bent: 4}\\ \text{straight: 1}\end{array}$

$\begin{array}{ccccccc}\text{Ten's digit:} & \text{(bent)} + \text{(bent)} &=& 2 + 4 &=& 6 \\

\text{Unit's digit:} & \text{(straight)} \times \text{(straight)} &=& 3 \times 1 &=&3 \end{array}$

. . Therefore: . $7 \times 9 \;=\;63$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I tried a general proof (for base-ten).

Let $\,a$ and $\,b$ be the two numbers: . $6 \,\le\, a,b\,\le\,10$

. . . $\begin{array}{c}\text{Left hand} \\ \hline
\text{bent: }a-5 \\ \text{straight: }10-a \end{array} \qquad\quad \begin{array}{c}\text{Right hand} \\ \hline \text{bent: }b-5\\ \text{straight: }10-b\end{array}$

$\begin{array}{cccccccccc}\text{Ten's digit:} & \text{(bent)} + \text{(bent)} &=&(a-5) +(b-5) &=& a + b - 10 & (T)\\

\text{Unit's digit:} & \text{(straight)} \times \text{(straight)} &=& (10-a)(10-b)&=& 100 - 10a - 10b + ab & (U) \end{array}$

The product is: . $10T + U \;=\;10(a+b-10) + (100-10a - 10b + ab)$

. . . . . . . . . . . . . . . . . . $=\; 10a + 10b - 100 + 100 - 10a - 10b + ab$

. . . . . . . . . . . . . . . . . . $=\quad ab$

Hey, it works!
• Sep 19th 2010, 09:46 AM
misterchinnery
dont know how to edit your solution so that the 10s and the 5s you use will represent the base used. Obviously I can do this by hand but I dont know the notation or type of maths which can do this.
Did you see my other post involving triangles?