How to Divide any number By 9, 90, 900 and so on in just 5 seconds(mathema-tricks.blogspot.com)
mathema-tricks.blogspot.com
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
13 comments
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)
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)
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.
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.
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.
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
Useful for algorithms like this: http://stackoverflow.com/q/295579/12048
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 = 7225Did 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.
EX: works for 10005^2 and 12345^2.
It works for any number, I meant "reasonably small", because for those it can be done mentally.
Cool, thanks I was sitting there trying to think why it would fail for large numbers. But, (x * 10 + 5)^2 = x * x * 10 * 10 + 2 * x * 50 + 25 = x * (x + 1) * 100 + 2 .
edit '+ 2 .' should be '+ 25.'
so: (x * 10 + 5)^2 = x * x * 10 * 10 + 2 * x * 50 + 25 = x * (x + 1) * 100 + 25.
so: (x * 10 + 5)^2 = x * x * 10 * 10 + 2 * x * 50 + 25 = x * (x + 1) * 100 + 25.
Also fun is squaring two digit numbers in your head: http://blog.jgc.org/2010/03/squaring-two-digit-numbers-in-yo...
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
so e.g. take a random number, 22103 answer is about 2210 + 221 +22 + 2 = 2455
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.
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.
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. :)
> Interestingly, this is true for any two digit number
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
9 - 6 = 3 <= there is your answer.
See also: http://en.wikipedia.org/wiki/Digital_root
http://en.wikipedia.org/wiki/Casting_out_nines
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)
132144 => 1+3+2+1+4+4 = 15 => 1+5 = 6 (divisible by 6 - obviously you can stop at 15 too)
Cool Trick. Reminds me a bit of vedic math:
http://en.wikipedia.org/wiki/Vedic_math
http://en.wikipedia.org/wiki/Vedic_math
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
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.
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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.
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.
Professor Trachtenberg (early 1900) comes to mind for doing speed math.
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
http://en.wikipedia.org/wiki/Trachtenberg_system
+1 for the Trachtenberg system. I only wish we were taught these methods at school, in addition to the 'normal' way...
Uhm, 5 seconds? I can type X/9 into Google or my smart phone in 1 to 2 seconds.
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.
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.