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Have I found a new math trick for multiplication? 
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#1
Mar3110, 07:47 AM

P: 40

This formula is much easier than the standard 10xx to do in your head. I have been able to do 3 digit numbers with this trick easily, and I'm generally not that great at doing math in my head. Anyway, he it is:
http://dratsab.deviantart.com/art/Mu...Trick85433340 


#2
Mar3110, 09:58 AM

P: 15

Cool, I still have to read the rest of it, I just wanted to let you know about this:
Your process takes at least 4 steps..and one of those steps is a variation of the 10xx method using the number minus the approx tens with a 0 dropped. It is like the Rube Goldberg version of the 10xx method :) ...[however it might prove simpler and faster with 3 digit numbers]...it all depends how people good are at subtracting big numbers Props! 


#3
Apr110, 01:59 PM

P: 40

Ugg, that's the 2nd mistake I made, thanks for pointing it out. I definitely think for 3 digits my way is easier, I tried both ways and I find it hard for me to calculate arithmetic that high, but I'm pretty average in math ability.



#4
Apr210, 07:15 PM

P: 726

Have I found a new math trick for multiplication?
What's the 10x  x trick?



#5
Apr410, 09:45 AM

P: 96

When I multiply a single digit number by a doubledigit number I usually do it backwards. For example, to multiply 7 times 67 I would think: 7 times 60 is 420 and then add 7 times 7 to get 469. Likewise 9 times 25 is 180 plus 45=225. 


#6
Apr710, 02:08 AM

P: 3

What if we apply the methodology to two examples, 100 and 101?
I believe that for the case of 100, we should consider: 100  10 = 90 and then 9 + 0 + 0 = 9 and therefore the answer is 900 As for the case of 101, we should consider: 101  11 = 90 and then 90 + 0 = 9 and therefore the answer is 900? 


#7
Apr710, 11:06 PM

Sci Advisor
P: 2,751

Essentially this method involves an approximation step (that is based on "10x x" but with an easier subtraction) followed by a correction step (that is based on the modulo 9 technique of "casting out nines").
Lets looks in detail at how your method works and why it might be easier for some people.  The first step is to factor out the "10" from the "10xx" method to give a numerically smaller subtraction. 9x = 10x  x = 10(x  x/10). This is exact but it requires subtracting a decimal, "x/10", and so holds no advantage.  Replacing "x/10" with the approximation [itex]\lceil x/10 \rceil[/itex] (ceiling function) gives the required simplification to the subtraction but results in an approximate answer, having a maximum error of 9.  Finally the correction step uses the fact that the approx answer is always "under" by 0..9 to allow a correction by the "digit sum" modulo 9 method. See http://en.wikipedia.org/wiki/Casting_out_nines The only problem (as pointed out with MDR123's example above) is that the error is 0..9 and 0 and 9 are congruent modulo 9. In my opinion easiest way to get around this is just to stipulate that : if any rounding up is performed by the ceiling function then the correction step must also add a strictly positive "modulo 9 correction". For example, in the "101*9" case given above we would not be able to stop at the preliminary "900" answer just because it's modulo 9 correct. Since [itex]\lceil 101/10 \rceil[/itex] did require a finite upward rounding then the preliminary answer of 900 must be corrected upward to the next modulo 9 consistent value of 909. 


#8
Apr2410, 07:21 PM

P: 2

this method can be used by multiplication of 3 too..
for example: 3 is a factor of 27, 2+7=9 which can be divided by 3. same case with multiplication of 9 


#9
Apr2510, 08:50 PM

P: 40




#10
May1910, 08:56 AM

P: 40

I finally uploaded the video I made of me using it:
http://www.youtube.com/watch?v=QPjH0mzUL4c ...and of course in typical dratsab fashion I make a mistake. 


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