2 Strange mathematical facts .

[thom]

Whats the odd prime number.

2



(odd because its even...)

Edit: probably the most wondrous proper math relationship would be :

e ^ ( i x pi ) + 1 = 0

where :
e is the natural number ~ 2.71...
i is the imaginary number
pi is 3.142.... , ie. pi is pi...

 

[marinyork]

111,111,111 x 111,111,111 = 12345678987654321

As described elsewhere that's a property of 11 in base 10. However there are other similar ideas. If you look up your sloane's you will find two sequences in there that are called automorphic numbers (can hear the echo of my old supervisor cursing at the name in anger).

An automorphic is a number where you square it and on the end the same number appears. 1 and 0 are obviously automorphic. If you count digits of any length then for each digit there are two automorphic numbers (if you include 0 infront as a digit or otherwise there are at most 2 automorphic numbers per digit).

So 5x5=25
25x25 = 125
625x625= 390625
90625x90625 = 8,212,890,625
and so on

6x6=36
76x76 = 5776
376x376 = 141,376

You may have noticed that 5+6 is 11, 25+76 = 101 and 625+376 = 1001. As with many other of these ones one can show it with modular arithmetic.

 

[marinyork]

If bet you can't give me an example that can't be done in 3 though ;-)
(see Goldbach conjecture )

That's the problem with trying to do story arcs in threads. It might sound ridiculous to someone else, but it is current even. That and I couldn't think of a short and snappy way of incorporating the Jones polynomial instead.

 

[thom]

As described elsewhere that's a property of 11 in base 10.

There's a little more to it... and to be a little less elliptical, similar is true if the number chosen, 111,111,111 has fewer digits ( nine ) than the base (ten).
It remains trues in base eleven, twelve, thirteen etc...
Eg, 11,111 ^ 2 is similarly behaved for base six and above

 

[ohnovino]

111,111,111 x 111,111,111 = 12345678987654321

Looks particularly satisfying when done as long multiplication:

          111111111
        x 111111111
        -----------
          111111111
         1111111110
        11111111100
       111111111000
      1111111110000
     11111111100000
    111111111000000
   1111111110000000
+ 11111111100000000
-------------------
  12345678987654321

 

[Fnaar]

666?

 

[marinyork]

There's a little more to it... and to be a little less elliptical, similar is true if the number chosen, 111,111,111 has fewer digits ( nine ) than the base (ten).
It remains trues in base eleven, twelve, thirteen etc...
Eg, 11,111 ^ 2 is similarly behaved for base six and above

Similarly there's a bit more to automorphic numbers, but hey. There are plenty of other automorphic numbers in other bases where the equivalent will repeat itself (given how you can divide it) in other bases. You can get problems dividing things by 4 with smallers numbers than 4 and take out primes so base 6 upwards will get some (3,4) and so on. Anyway all of this stuff of both sorts of problem and much more besides is in sloane's as you should already know and someone else may want to know.

 

[thom]

666?

I expected you to go for 59009

 

[Maz]

When I was at junior school, we had a method of working out the square root of any given number, to n decimal places. It looked like a long multiplication in layout, but I'm buggered if I can remember how we did it!

 

[Davidc]

When I was at junior school, we had a method of working out the square root of any given number, to n decimal places. It looked like a long multiplication in layout, but I'm buggered if I can remember how we did it!

There's an algorithm for doing it that looks like that. I learnt it when about 8 and understood it when I returned to it 10 years later.

Much more useful at the same time was learning the method of successive approximations.

Edit: Just had a look and found the method you remember on the web HERE

 

[Maz]

There's an algorithm for doing it that looks like that. I learnt it when about 8 and understood it when I returned to it 10 years later.

Much more useful at the same time was learning the method of successive approximations.

Edit: Just had a look and found the method you remember on the web HERE

Good find that, Davidc!
The method we used was as per the example given on that link ("Find √645 to one decimal place").

 

[02GF74]

any number ending with 0, 2, 4, 6 or 8 is divisiable by 2
any number ending in 5 or 0 is divisible by 5
any number whose digits added up are divisible by 3 is itself divisible by 3
any number ending in 0 is divisible by 10

 

[thom]

any number ending with 0, 2, 4, 6 or 8 is divisiable by 2
any number ending in 5 or 0 is divisible by 5
any number whose digits added up are divisible by 3 is itself divisible by 3
any number ending in 0 is divisible by 10

There's one about eleven too:
Take a number and add together the odd position digits, the first, third, fifth etc, then subtract all the even position digits, ie. the second, fourth, sixth etc. Your original number is divisible by 11 if and only if the total is 0.
Eg,
14683046 : 1 + 6 + 3 + 4 - ( 4 + 8 + 0 + 6 ) = -4 so not divisible by 11
146830464 : 1 + 6 + 3 + 4 + 4 - ( 4 + 8 + 0 + 6 ) = 0 so is divisible by 11

 

[Dayvo]

666?

999 - you're effing nicked, me old mate!

 

[Maz]

Why was 6 afraid of 7?