Multiply \[24\times 53\] (or any two 2-digit numbers) using 3 multiplications instead of the usual 4

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Multiply \[24\times 53\] (or any two 2-digit numbers) using 3 multiplications instead of the usual 4

Mathematics
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no takers on this one?
No Takers, Only Under Takers!!
i think this is a computer algorithm problem. cuts down on multiplications

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I can do it using only two multiplications: 20 x 53 plus 4 times 53 oh - ok - misunderstood the question...
yeah that is still 4 right?
use ethiopian multiplication!
ok, what might that be?
rosettacode.org/wiki/Ethiopian_multiplication
http://rosettacode.org/wiki/Ethiopian_multiplication
wow! not exactly what i had in mind though...
method i had was somewhat shorter
here is a list of algorithms for doing multiplication: http://en.wikipedia.org/wiki/Multiplication_algorithm
funny i looked here and didn't see it
One of these is particularly interesting: http://en.wikipedia.org/wiki/Karatsuba_algorithm
hint is \[(a+b)(c+d)=ac+ad+bc+bd\] and you need the numbers \[ac, bd, ad+bc\]
yeah that is it! for example \[53\times 24\] \[5\times 2=10,3\times 4=12,(5+3)(2+4)=8\times 6=48, 48-10-12=26\] 1 2 26 10 __________-- 1 2 7 2
yes - until you brought this up I was wasn't even aware of efficient multiplication methods. thanks for opening the door to new knowledge. :-)
i never knew the name, and only remember it from a discrete algorithms class. in fact it is all i recall from that class
karatsuba! i will try to remember that name
how can we ever forget a name like that! :-)
especially with a bookmark!
lol!
I have just invented a way for me to remember this: rabbits hop around and multiply very fast. so what is their favourite meal? "carrot soup" of course! (hence katatsuba)

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