Have I found a new math trick for multiplication?

[dratsab]

This formula is much easier than the standard 10x-x 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/Multiplication-of-9-s-Trick-85433340

 

[rekshaw]

Cool, I still have to read the rest of it, I just wanted to let you know about this:

Take for example 9X25, think of the 25 in a set of 10's starting with 1-10, then 11-20, then, 21-20 which is what 25 is in

EDIT: Ok I read all of it. Uhm...I never heard about the 10x-x method, and I tried doing one calculation with your method and one different calculation with the 10x-x method which you explained at the bottom. Your method MIGHT be easier for 3 digit numbers (as the 10x-x method involves subtracting 3 digit number from a 4 digit number which might prove challenging for some people)...however...there is no discussion that with a 2 digit number...the 10x-x method is 5 times faster.

Your process takes at least 4 steps..and one of those steps is a variation of the 10x-x method using the number minus the approx tens with a 0 dropped.

It is like the Rube Goldberg version of the 10x-x method :) ...[however it might prove simpler and faster with 3 digit numbers]...it all depends how people good are at subtracting big numbers

Props!

 

[dratsab]

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.

 

[JG89]

What's the 10x - x trick?

 

[dilletante]

What's the 10x - x trick?

The 10x-x trick is what it looks like -- 10x-x = 9x, so if you want to multiply 9x25 you can think of it this way: 10 times 25=250, and 250 minus 25 equals 225.

When I multiply a single digit number by a double-digit 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.

 

[MDR123]

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?

 

[uart]

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 "10x-x" 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.

 

[chinyew]

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

 

[dratsab]

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

It's not the same, because it can also break down to 3, 6, or 9. If you can find a way to get the exact answer from it, then it will work, but it isn't narrowed down as simply as 9 is.

 

[dratsab]

I finally uploaded the video I made of me using it:



...and of course in typical dratsab fashion I make a mistake.