FNRttC Christmas Ride and Drinks. Thursday 29th December 2011

[thom]

Happy New Year !

As eny math geek no, every multiple of 9 works - the sum of the digits of a multiple of 9 is always a multiple of 9, so eventually you can get back to 9.

Once upon a time I could prove it.

It boils down to modular arithmetic (and our choice for a base for number representation as 9 + 1).
It is simple to see 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, means that 4878 = ( 4 x 999 + 8 x 99 + 7 x 9 ) + ( 4 + 8 + 7 + 8 ), for example.
The first bracketed bit is obviously divisible by 9 so you only need worry about adding the digits up to get the second bit and seeing if that is divisible by 9.
Since that second number is smaller than the first, you can keep on playing the same trick until you get a single digit number.
9 is the only single digit number divisible by 9 so this is why you always end up with it.
Note if you end up with 3, 6 or 9, you can tell the number is divisible by 3.


A related property concerns division by 11....:
Add and subtract successive digits of a number. If the result is 0, the original number is divisible by 11.
Eg :
121 -> 1 - 2 + 1 = 0, so is divisible by 11.
75691 -> 7 - 5 + 6 - 9 + 1 = 0, so is also divisible by 11.
432 -> 4 - 3 + 2 = 3, so is not divisible by 11.

If we used base 7, we would see similar behaviour for 6 and 11 ( 11 in base 7 is 8 in base 10 ;-) ).

 

[Mice]

Happy New Year !



It boils down to modular arithmetic (and our choice for a base for number representation as 9 + 1).
It is simple to see 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, means that 4878 = ( 4 x 999 + 8 x 99 + 7 x 9 ) + ( 4 + 8 + 7 + 8 ), for example.
The first bracketed bit is obviously divisible by 9 so you only need worry about adding the digits up to get the second bit and seeing if that is divisible by 9.
Since that second number is smaller than the first, you can keep on playing the same trick until you get a single digit number.
9 is the only single digit number divisible by 9 so this is why you always end up with it.
Note if you end up with 3, 6 or 9, you can tell the number is divisible by 3.


A related property concerns division by 11....:
Add and subtract successive digits of a number. If the result is 0, the original number is divisible by 11.
Eg :
121 -> 1 - 2 + 1 = 0, so is divisible by 11.
75691 -> 7 - 5 + 6 - 9 + 1 = 0, so is also divisible by 11.
432 -> 4 - 3 + 2 = 3, so is not divisible by 11.

If we used base 7, we would see similar behaviour for 6 and 11 ( 11 in base 7 is 8 in base 10 ;-) ).

I need to go back to school. I think I'll go by bike...

Thank you though Thom!

M