- #1

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http://dratsab.deviantart.com/art/Multiplication-of-9-s-Trick-85433340

- Thread starter dratsab
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- #1

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http://dratsab.deviantart.com/art/Multiplication-of-9-s-Trick-85433340

- #2

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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 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!

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

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!

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- #3

- 36

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- #4

- 726

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What's the 10x - x trick?

- #5

- 94

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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.What's the 10x - x trick?

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

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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

uart

Science Advisor

- 2,776

- 9

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.

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 :

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- #8

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for example: 3 is a factor of 27, 2+7=9 which can be divided by 3.

same case with multiplication of 9

- #9

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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.

for example: 3 is a factor of 27, 2+7=9 which can be divided by 3.

same case with multiplication of 9

- #10

- 36

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

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

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

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