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Friday, January 30, 2009

There is a technique for multiplying numbers between 6 and 10 on your fingers called peasant multiplication (graphic illustration).  I explained how to do it two years ago, and I mentioned an easy proof of it.  Since I was subsequently asked for the proof, I present it now with an improvement.  (If you don't already know how to perform the multiplication, please check the links above, first)

Why Peasant Finger Multiplication works for any Bilaterally Symmetric Species

As long as your fingers (or tentacles) on each hand (or appendage) are bilaterally symmetric and you count in a base defined by your total number of fingers, this method will work.  First, let's start with definitions:

Since "digits" has different meanings in physiology and math, I will say "digits" when I mean mathematical place-value digits, and "fingers" when I mean dangly meat appendages with which you are counting/multiplying.

let n = factor #1 to be multiplied
m = factor #2 to be multiplied
p = product n * m

d = the number of fingers on each hand (total fingers on both hands = 2*d)

Since we (humans) count each number (n or m) from 6 to 10 on a hand with 5 fingers, the actual number of fingers used on a hand would be n-5, or more generically for any species,

a = n - d    (number of fingers used on one hand)
b = m - d   (number of fingers used on other hand)

If the product p for which we are searching is defined as

p = n * m


p = n * m = (a + d)*(b + d) = ab + d*(a+b) + d2

With peasant math, the tens digit (base 10) is represented by adding the number of fingers on each hand representing digits.  Here I multiply it by 10 to represent that it will be placed in the 10s digit:

p10  = 10*(a + b)

or more generally for any species, the "d's" digit is,

pd  = 2*d*(a + b)

Similarly, for the value of the ones digit, you add the remaining (unused) fingers on each hand:

pones  = (d - a)*(d - b)

which expands to

pones  = d2 - d*(a + b) + ab

In peasant math, the product p is acheived by combining the 10s digit with the ones digit:

p = pd + pones = 2*d*(a + b) + d2 - d*(a + b) + ab
= ab + d*(a+b) + d2 =  (a + d)*(b + d)
= n * m

which is the original definition of the product you are seeking to find.  

Peasant math works, regardless of your species, as long as you're bilaterally symmetric and your numeric base is that of your total number of fingers, tentacles, or tough scaly claws. 

Burton MacKenZie


slyght said...

if i recall, you posted about 1=2 once, and mentioned that the exercise was flawed by a division by 0. i found a similar math conundrum you might interested in, and possibly enlighten me as to the error. it was posted on 5-feb-09, so it might be a few posts down.


burton mackenzie said...

@slyght: I think you mean this one. It just so happens I reviewed that proof about six months ago. The deception is that when taking the square root of (3-4)^2 and (5-4)^2, there are two answers for each side of the equals sign.

For instance, 2*2 = 4, but (-2)*(-2) = 4, as well. So, the square root of 4 is both 2 and -2.

So, there are four intersections, and two unique ones. Just as there are four solutions to the more generic sqrt(x^2) = sqrt(y^2)

+x = +y
+x = -y
-x = -y
-x = +y

The correct square roots for a valid solution would be:


The other two solutions were spurious roots of the equation.

slyght said...

i thought that was the answer, but wasn't completely sure. thanks man, you rock!


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