This article confuses me a bit. Simply because I think it would take longer than 11 seconds to look at all the numbers in a 100 digit number. let alone start doing calculations on it.

This reminds me of a math game that I read about in the hit book: The curious incident of the dog in the night time. In this book the main character, an autistic boy, does cubes in his head to pass the time. (3 cubed = 3*3*3)

so I started doing this during my biweekly drives from New Haven to Manchester. It’s actually pretty fun. You develop all sorts of clever tricks for multiplying and squaring numbers.

Click more for INTENSE geekery

**Example 1:**

Here’s a quick short cut for figuring out squares

13^2 = 169.

12^2 = 144.

144 + (12+1)*2 = 169

so if you know one square, you can quickly figure out the next.

say you knew that 5 squared is 25. But you couldn’t remember what 6 squared was. all you’d have to do is add ((5*2)+1) to 25. 25+11=36!

it works every time and mathmatically it’s really simple. All you’re doing is solving for (X+1)^2 which factors into (x+1)(x+1) which using FOIL comes out to X^2+2X+1. While I’ve known this for years, I’m amazed by the actual application.

**Example 2:**

if the number ends in 5, you can square it by multiplying the first number (x) with the first number plus one (x+1) and throwing that in front of ’25’.

15*15=225.

1*2 =2.

225.

7*8=56

75*75= 5625

mathmatically it’s proved simply as well: 75 = 7*10+5. when squared

(7*10+5)(7*10+5)=(49*100+35*100+35*100+25)

this equals:

(49*100+70*10+25)

or…

4900+700+25

=5625

!

the middle number always ends up being equivilent to the first number times 10 which is where the (+1) thing comes from. Try it with 9.

95*95=(9*10 in front of 25)=9025

or

((9*10)+5)^2 = 8100+900+25 -> 9025!

**Example 3:**

I didn’t figure this one out, but it’s useful.

any number whose digits add up to a multiple of 9 is divisible by 9.

130293 -> 1+3+0+2+9+3 = 18.

18 is divisible by 9

so 130293 is divisible by nine. (it actually turns out to be 14477)

I haven’t figured out a way to prove this except by example.

If you add nine to any number the digit in the tens spot will increase by one, and the digit in the ones spot will decrease by one. If you start with the number 9, you will always have digits that add up to nine. This also occurs from the tens place to the one hundreds place and from the hundreds place to the thousands place etc.

**Example 4:**

when you are multiplying two numbers together who have a difference of 2, you can take the square of the number in between and subtract one for the same result.

16*14= (15*15)-1=224

11*9=(10*10)-1=99

5*7=(6*6)-1=35

This is simple to prove. all you’re doing is multiplying (x-1)(x+1) which solves to (x^2-1) ding! done. it gets cooler though as you seperate the numbers by more than 2. For every step you are further from the middle, you can just subtract the next odd number.

17*13=(15*15)-1-3=221

4*8=(6*6)-1-3=32

18*12=(15*15)-1-3-5=216

this is solved in the exact same manner (x-2)(x+2) = X^2-4

And as I showed in Example 1, the odd numbers happen to add up to the squares of the second number of the foil.

In conclusion, maybe that guy in the article just knows a lot of quick tricks for 100 digit numbers. Either that, or he’s just excessively smart.

If you have any fun math things that you want to share, please do so in the comments. Or, if you have a better proof of the 9 rule, I would love to see it.

Actually #3 is a special case of a more general rule, the digits of any number divisible by 3 sum to be divisible by 3, like 48, 4+8=12, I however can’t tell you a proof for this but I will think about it over the weekend.

Wow. I never knew the three rule. Very interesting. I’ll think about a proof too. With our powers combined we’ll solve this riddle.

Here’s an equation I know! e = i – 1

I think that’s right. But it might not be. All I know is, it’s supposed to be a cool equation.

Ok, I figured it out with the help of the internet. Lets take 1233 which can be rewriten as 1*1000+2*100+3*10+3 which can be rewriten as 1*(999+1)+2*(99+1)+3*(9+1)+3 distributing becomes 1*999+2*99+3*9+(1+2+3+3), so the left most part is divisible by 9 and 3, so if the digits sum to a multiple of either the whole number is divisible by 9 or 3. Incidentally there is a similar trick that can be done to discover multiples of 11, sum the even digits, sum the odd digits and find the difference between them, if that is divisible by 11 the whole number is, since 11-1=10, 11*9+1=100, 91*11-1=1000, 99*11+1=10000, etc.