How to Divide any number By 9, 90, 900 and so on in just 5 seconds

http://mathema-tricks.blogspot.com/2012/01/dividing-by-9-90-900-and-so-on-into.html

by mquaes • 14 years ago

69 29 14 years ago

mathema-tricks.blogspot.com

lignuist• 14 years ago

Cool Trick. Reminds me a bit of vedic math:

http://en.wikipedia.org/wiki/Vedic_math

swatkat• 14 years ago

Yes! And, this reminds of amazing "Mathemagic" series of videos by Khurshed Batliwala. Here's one: (a+b)²=a²+2ab+b² - But Why? http://www.youtube.com/watch?v=49_TJymgXgM

finnw• 14 years ago

The same technique for dividing by 9 in base 10, also works for dividing by 255 in base 256, so you can quickly calculate N % 255 by adding the bytes of N together.

Useful for algorithms like this: http://stackoverflow.com/q/295579/12048

Sembiance• 14 years ago

Uhm, 5 seconds? I can type X/9 into Google or my smart phone in 1 to 2 seconds.

thewisedude• 14 years ago

I think I can reason why you are down voted. May be it was the delivery style of the message than the message itself. But I think it was unfair to you as you have a valid point. If you strictly want to asses the fastest way to arrive at this result in today's day and age. Your solution is probably one of the fastest ways to do that. I think the author's mathematical tricks would have been very useful in a world where personal computing is not where it is today... say prior to the advent of calculators.

why-el• 14 years ago

Unrelated, but relates to 9: if you take any two digit number,say 59, and you deduct the sum of the two digits 5 and 9, it yields 45, a multiple of 9. Interestingly, this is true for any two digit number. I used this trick in a game where I labeled all multiples of 9 with a sign, asked my sister to perform the operation above, and claim that I can guess the sign, because all multiples of 9 share the same sign. :)

rimantas• 14 years ago

  > Interestingly, this is true for any two digit number

Not only for two digits numbers. Also there is a trick based on that: ask someone to think of a number (two-digits, or greater). Ask to subtract the sum of digits from that number. Ask to multiply result by any number. Ask to cross-out any digit, except 0 and then say remaining digits to you in any order. Then you say what digit was crossed out. The secret is simply: you add the digits told to you till you get one digit number, and then subtract it from nine.

E.g.:

  Original number => 42
  Subtract the sum => 42 - 6 = 36
  Multiply by any number => 36 * 666 = 23 976
  Cross out 3 => 2976
  Name the digits in any order => 2679

So you got 2679. Calculate digital root: 2 + 6 + 7 + 9 = 24, 2 + 4 = 6.

9 - 6 = 3 <= there is your answer.

http://en.wikipedia.org/wiki/Casting_out_nines

muyuu• 14 years ago

Actually this generalizes for all number of digits (down to 1) and to all multiples of 3, 9 or any number only divisible by 3. And it's not just an if, it's an iff.

132144 => 1+3+2+1+4+4 = 15 => 1+5 = 6 (divisible by 6 - obviously you can stop at 15 too)

enkrs• 14 years ago

Unrelated note, you can do simple multiplication by 9 with your fingers: place your palms on a table and imagine your fingers are numbered 1-10 from left to right.

4 * 9? Lift up your fourth finger - you have 3 to the left and 6 to the right - answer is 36.

8 * 9? Lift up your eight finger - 7 to the left and 2 to the right - answer is 72.

I hope it's not totally common knowledge thing.

neovive• 14 years ago

Never knew this one. Thanks for sharing!

emehrkay• 14 years ago

My son is going to flip. Thanks

lkozma2• 14 years ago

Unrelated, you can square quickly a (reasonably small) number ending in 5: multiply the number without the 5 to the number one larger and append 25 to it. Example:

  85*85 = (8*9).25 = 7225

onemoreact3• 14 years ago

Did you mean it's only fast when it's a small number, or does it fail at some point?

EX: works for 10005^2 and 12345^2.

lkozma2• 14 years ago

It works for any number, I meant "reasonably small", because for those it can be done mentally.

obtino• 14 years ago

Interesting fact: If the sum of all digits in the dividend is a multiple of 9, then it can be assumed that the dividend is a multiple of 9.

no-espam2• 14 years ago

Professor Trachtenberg (early 1900) comes to mind for doing speed math.

no-espam2• 14 years ago

Wow, downvoted for posting info related to a professor who was held in nazi concentration camp and kept his mind lucid with math?

http://en.wikipedia.org/wiki/Trachtenberg_system

downx3• 14 years ago

+1 for the Trachtenberg system. I only wish we were taught these methods at school, in addition to the 'normal' way...

jsvaughan• 14 years ago

An alternative way (but more intuitive) to find approx a / 9 in your head: result = (a x 0.1) + (a x 0.01) + ...

so e.g. take a random number, 22103 answer is about 2210 + 221 +22 + 2 = 2455

lotharbot• 14 years ago

To be more complete: .1a + .01a + ... + (digit sum / 9) will give it to you. That's actually what this method is doing, it's just poorly explained.

22103/9 = 2210 + 221 + 22 + 2 + (2+2+1+0+3)/9 = 2455 + 8/9

So your ones digit ends up being 2+2+1+0+(any whole part left over from the digit sum), your tens digit is 2+2+1, your hundreds digit is 2+2, and your thousands digit is just 2.

rytis• 14 years ago

This is obviously cool and all that, but I see one problem with these kind of shortcuts - children (who this site targets) have no idea why this works and how. They memorise it and go about their lives using it. They know it works, but have no idea why.

I'm ranting about it because I have 2 kids who learn all sorts of similar stuff at school, but teachers fail to explain why these things work. (and in most cases I suspect they don't actually grasp the idea themselves)

budgi3• 14 years ago

The trick seems to be a combination of both. Learning by rote in combination with an effective method seems to be key to strong number sense in kids. Take a look at the Triple Code Model (Dehaene & Cohen) for some more detail.

Jach2• 14 years ago

I'd correct you by saying that they'll memorize it for as long as it's useful for passing a test because the teacher wants them to solve something a particular way. They'll soon forget it. I learned long division in 3rd grade, relearned it in 6th grade, learned the algebraic polynomial version in 10th grade (then was shown/told synthetic division which is cooler)... and at this point I know I'd struggle doing any of those. I'd rather use a calculator and get back to doing whatever it was that required the answer. In these crazy times where even most elementary kids have phones with calculator capabilities there's no excuse to be parading these mental tricks as "math".

I do agree though that a proof of this would be much more interesting than the technique itself. If I have kids I intend to teach them math myself. I don't think I was introduced to the idea of formal proof until 8th grade geometry, and everyone hated those proofs since it was a bookkeeping exercise of writing down every algebraic step and what theorem/axiom allowed you to perform it.

pg_bot• 14 years ago

Why don't you just divide by three two times? This would be a lot easier to remember and most likely will be faster than the other method.

Jach2• 14 years ago

Or get an approximate answer by dividing by the appropriate power of 10, which is pretty much the fastest method. I think the "5 seconds" editorialization is BS.

jgrahamc2• 14 years ago

Also fun is squaring two digit numbers in your head: http://blog.jgc.org/2010/03/squaring-two-digit-numbers-in-yo...

jgrahamc2• 14 years ago

Why this works: http://blog.jgc.org/2012/03/how-to-divide-by-9-really-really...

How to Divide any number By 9, 90, 900 and so on in just 5 seconds

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