Re: Russian Peasant Multiplication / was question on how to do this

*To*: mathgroup at smc.vnet.net*Subject*: [mg131349] Re: Russian Peasant Multiplication / was question on how to do this*From*: d.a.paxton at gmail.com*Date*: Sun, 30 Jun 2013 03:29:22 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <kqm6u7$hp5$1@smc.vnet.net>

On Saturday, June 29, 2013 2:47:03 AM UTC-6, Richard Fateman wrote: > Dave -- > > 1. You should try to come up with a useful subject line > > in the future. > > 2. It is called Russian Peasant Multiplication (which you may > > find on Google). > > 3. There is no reason to believe that a procedural algorithm > > has a formula, but in this case I think the inverse is known > > as division :) I am sorry for posting a question that is improper. It is a matter of rounding as a floor or ceiling question I guess wondering nomenclature. If the lowest count on the numbers in a multiply are let's say 7 and 5. We multiple these and get 35. How does on drop the 5 into the lowest of the answer and carry the 3? Since 7 and 5 and also 35 are know known we do this.. In base ten we multiply by ten. Divide out the seven and get fifty. Divide that by ten and get the 5 drop out number. Then put the the thirty five minus the drop out number and then divide by ten again. Gives three. The carry number. So I guess in using this for any base number system we do have a divide as an inverse function for any variables in the course of a multiply.