New method of multiplying

The guy is pretty smart if he found this by himself.

i am bored to ckeck it but:

I wonder if this works for all multyplying equations. His examples use numbers that consist of the 1,2 and 3.
 
Not sure if it was posted before, but I found it very interesting and quite unexpected :)
http://www.blogz.ru/2006/11/21/novyjj_sposob_bystrogo_umnozhenija.html
He hasn't done anything clever - more like stupid.

All he has done is to replace learning a few multiplication tables and with a lot of tedious counting/addition! :rolleyes:

For example, try seeing how painfully slow it would be to do 99*99!

If you want something slightly more useful for doing multiplication by hand, try looking up the Karatsuba-Ofman algorithm.
 
I'm smarter than he is:
23*54

split into := 2|3 * 5|4
2*5 = 10, 3*4 = 12, 3*5 = 15 + 1 = 16

When multiplying the middle part you need to always add 1 on the result.

Now I have:
10, 16, 12
L|R, L|R, L|R
1|0, 1|6, 1|2

Multiply out the first set by the last.
1*1 = 1
1*2 = 2
0*1 = 0
0*2 = 0

I have my first 2 digits of the answer: 12

Do the same but to middle, but leave the 1 out:
6*1 = 6 -take the 6, multiply out to the left
6*0 = 0 -take the 6, multiply out to the left
6*1 = 6 -take the 6, multiply out to the right
6*2 = 12 -take the 6, multiply out to the right

Add the result: 6+6+12 = 24, swap the digits = 42

Add the digits 12 and 42:
1242.

So what's the answer to 23*15?
23*54 = 1242

What did I get? 1242. :)
 
Surprising no one has noticed before.

Not surprising that it works: each place in each number is multiplied by each place in the other... The diagonal grouping is the unit places again.
 
I always just convert numbers to binomials (sometimes recursively). in my head and multiply them, e.g.

123 x 321 = (100 + (20 + 3)) (300 + (20 + 1) = 30000 + 6000 + 900 + 2000 + 100 + 400 + 80 + 3 = 39483

3-digits * 3-digits being a big example. For most 2 digital multplication, it's really trivial. Just round up or down to nearest multiple of 5 or 10 and then add in an error correction term.
 
For most 2 digital multplication, it's really trivial. Just round up or down to nearest multiple of 5 or 10 and then add in an error correction term.

This sounds interesting, how would i find the error term? I tried some by myself but couldn't come up with something that is simple enough to satisfy me.
 
Surprising no one has noticed before.
Not surprising to me.

He's basically replacing adding and multiplying small numbers (how hard is 3x3?) with counting. Why the heck would anyone think this is easier?

rendezvous, DemoCoder is just saying make one number a multiple of 5. For KILER's example of 23x54, do 23*55 = 115+1150 = 1265 and subtract 1x23. Or do 23*50 = 1150 and add 23*4 = 92.


EDIT:
There are a few other tricks that are nice, like 42*48 = 2016 (4x(4+1) gives 20, 2*8 give 16), which works when the first digits are the same, and last add to 10. Another nice one is multiplying "around" an easy square, e.g. 42*58 = 50*50 - 8*8 = 2436 (both numbers are 8 away from 50). Flipping that trick the other way around, you can figure out squares more easily too. 38*38 = 40*36 + 2*2 = 1444.

But other than that, what DemoCoder described (which is really identical to junior school multiplication) is about as fast of a general method as there is.
 
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Thanks, I tried to round both number to the closest evenly dividable by five. having two error terms was not easier than doing it the normal way; writing one number below the other and multiply one number at the time and finally add the partial products.
The other tricks were neat too, but I have to admit that I need to look at the first one again tomorrow, after a good nights sleep.
 
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