Why does it work for ten fingers? It is probably easy to generalize to other bases.How come this works if we didnt have ten fingers and we counted for example in base 8 and had 8 fingers?
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.
Hello, misterchinnery!
Wow! . . . a fascinating trick!
We have two sets of five fingers, each set numbered from 6 to 10.
. .
To multiply :
On the left hand, bend fingers 6 and 7.
On the right hand, bend fingers 6, 7, 8 and 9.
. .
. . .
. . Therefore: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I tried a general proof (for base-ten).
Let and be the two numbers: .
. . .
The product is: .
. . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . .
Hey, it works!