Very neat math trick.

-Eagle

awsome !

nice!

Anyone tried it with other numbers?

good trick to learn

Cool!

on will say in only one word.it is FANTASTIC

Haha, awesome trick

interessant

i tried it with 8 s and 9 s and you need a large sheet of paper

how about when multiply 102x506?
I do not think this works with zeros...

try to multiply 15x24... I may by dumbass cause I didn't get it right

I tried it and it worked for everything, but I'm not sure what it would do for zeros because I'm not sure how you would go about doing it! Maybe to put 206 you put 20 lines then six? Something to try.

I would have liked to hear the "amazing guitar" player.

Otherwise it was a waste of time.

Paper Calulator:

Have no idea what it was he was trying to do. Was there sound to explain the process.

JP

It basically acts like an abacus. Just draw, for each digit, the lines downwards or across. You then have three parts; a left part, a middle part, and a right part. Take the left part, and count the dots. This is the first digit of the answer, so write it down first. Then the second and third digits for the amount of dots in the other two sections. Sometimes, you will have a multi-digit number for a section. Just take the first digit of that number and add it to the number for the section before it. Just practice, and it gets easy. It also really comes in useful for anyone sitting maths exams where calculators aren't allowed.

this does not work with the zeros!!!!

Very Cooool !!!******

NEAT!!

i tried this out when i saw this it works

Thanks for clipping this Vid. I had seen it before and then missed it. Have to show it to my Son. So thanks for clipping it here.
Cheers. All the best.
Sincerely,
BHARAT.

XD i learned to do this in grade 5 ha. and ya. it DOES work with zeros. when you come up to a zero. you dont draw any lines. and just count the dots on the other corner x.x

123 x 321 = (1x100 + 2x10 + 3x1)x(3x100 + 2x10 + 1x1)
Multiply this applying the distributivity of multiplication concerning addition: (a+b+c)x(d+e+f)=axd + axe + axf + bxd + bxe + bxf + cxd + cxe + cxf
and apply the commutativity of addition as well as its associativity:
((1x100)x(3x100)) + ((1x100)x(2x10) + (2x10)x(3x100)) + ((1x100)x(1x1) + (2x10)x(2x10) + (3x1)x(3x100)) + ((2x10)x(1x1) + (2x10)x(3x1)) + ((3x1)x(1x1))
= 3x10000 + 8x1000 + 14x100 + 8x10 + 3x1
I admit: being drawn, this looks nicer.

I figured out how to make it work for zeros. When a zero comes up do not make a line, make an elipse. when counting up the intersections, any that are made with the elipse to not count. I've tried it quite a few times and it works for me.