Have I found a new math trick for multiplication?

In summary, the conversation discusses a simpler formula for mental multiplication, specifically for multiplying by 9. The formula involves approximating the tens digit and then using the modulo 9 technique to correct the answer. While it may be easier for some people with 3 digit numbers, the 10x-x method is still faster for 2 digit numbers. This method can also be applied to multiplication by 3, but it is not as precise as the 9x trick.
  • #1
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:

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  • #2
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

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  • #3
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
What's the 10x - x trick?
  • #5
JG89 said:
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.
  • #6
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
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.
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  • #8
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
chinyew said:
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.
  • #10
I finally uploaded the video I made of me using it:

...and of course in typical dratsab fashion I make a mistake.
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1. How do I know if I have found a new math trick for multiplication?

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5. Can I publish my new math trick for multiplication?

Yes, you can publish your new math trick for multiplication in academic journals, books, or online platforms. It is important to properly cite any references and give credit to previous work that may have influenced your discovery.

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