# Can you prove why Popsicle Stick Multiplication works?

This is a unique way of multiplying numbers by using sticks. Let's call it "Popsicle Stick Multiplication". Or maybe "Linear Algebra" quite literally. Take a look at both images that I've drawn below. Can someone explain why this works out? Notice how you must carry the 10's digit when the sum of the intersections is greater than 9, and if you have more than 99, then carry the hundreds digit just like any other multiplication problem. I'm not sure if it's faster to do it the long way or to do it this way. Perhaps this is the way the Egyptians did it while building the great pyramids? This works for multiplying any amount of sticks.

• I'm glad you realized that complex-multiplication does not mean complicated-multiplication. :)
– user856
Jan 15 '13 at 6:08
• Just for your reference, when we say "infinite" sticks in mathematics, it literally means there are no end to the sticks, not just that we can put as many sticks as we want. In overly dramatic fancy prose, we'd write "an arbitrarily large finite number of" sticks. To be less pretentious, we might just write "any amount of" sticks. Jan 15 '13 at 6:10
• As Maverick's father says, "I believe ya". :-) Is an integral, the area of an arbitrarily large finite count of two perpendicular sets of sticks? Jan 15 '13 at 6:12
• There is a typo in the second example: it computes $412 \times 1231$, not $412 \times 4231$. Unfortunately since it is an image, I cannot edit it. Jun 16 '14 at 21:29
• I fixed the image. Oct 26 '14 at 2:16

$$\begin{matrix} & 1 & 2 \\ \times & 3 & 4 \\ \hline & 4 & 8 \\ 3 & 6 \\ \hline 3 & 10 & 8 \\ \hline 4 & 0 & 8 \end{matrix}$$
$$\begin{matrix} & & & 4 & 1 & 2 \\ \times & & 1 & 2 & 3 & 1 \\ \hline & & & 4 & 1 & 2 \\ & & 12 & 3 & 6 \\ & 8 & 2 & 4 \\ 4 & 1 & 2 \\ \hline 4 & 9 & 16 & 11 & 7 & 2 \\ \hline 5 & 0 & 7 & 1 & 7 & 2 \end{matrix}$$
Consider the first image: $$12*34 = (10 + 2)(30 + 4) = (1*3)*10^2 + (2*3 + 1*4)*10^1 + (2*4)*10^0$$ This is just a visual form of the distributive law.