Sure, it's an artifact of base 10 - Edit 2
Before modification by Isaac at 29/02/2012 08:36:56 PM
Same as that trick of multiplying by 9 where the digits add up to 9 or 18 or 27 and so on at very high digits like 9x65802= 592,218... 5+2+9+2+1+8=27=9*3
That phenomenon for 9 will show up in any base for the digit proceed '10', like seven for Octal or 15 for Hexadecimal. It's really no different then multiplying by 5 and always getting the last digit to be 5 or 0, in base 16 any multiple of 8 ends in 8 or 0.
What's going on here is that you can think of 11 more as 10+1, essentially the result of Y x 11 is really 10Y + Y, or Y with a zero following plus itself. If you tihnk of that two digit number as AB then the classic add up way will be:
AB0
+ AB
A(A+B)B
where you just have carry a 1 onto the first A if A+B>9 and this would hold true in other bases besides base 10, with their equivalent of 11, the number one higher than the Base, same as the 9 trick will hold for numbers one less than the Base. For instance in Octal (Base 8 ) 9 is written as 11, and 9x13=117 is written as 11x15=165 ---> A(A+B)B ---> 1 (1+5) 5
There's also a tricks for 6, 6 x Y always ends in Y for any even Y (6 x 8 = 48, 6x 4 =24) and that will carry over into any even-number base system when working with the digit one higher than the halfway point (5 for base 10) as it's just carrying one extra off to x5 (or midpoint of the base) a number of times equal to that number. 6 times any odd number actually just adds that odd number to 5 for the last digit, 6x3= 18 or 15+3, etc.
Edit: seriously we've got to get rid of that 8) makes
drives me nuts
That phenomenon for 9 will show up in any base for the digit proceed '10', like seven for Octal or 15 for Hexadecimal. It's really no different then multiplying by 5 and always getting the last digit to be 5 or 0, in base 16 any multiple of 8 ends in 8 or 0.
What's going on here is that you can think of 11 more as 10+1, essentially the result of Y x 11 is really 10Y + Y, or Y with a zero following plus itself. If you tihnk of that two digit number as AB then the classic add up way will be:
AB0
+ AB
A(A+B)B
where you just have carry a 1 onto the first A if A+B>9 and this would hold true in other bases besides base 10, with their equivalent of 11, the number one higher than the Base, same as the 9 trick will hold for numbers one less than the Base. For instance in Octal (Base 8 ) 9 is written as 11, and 9x13=117 is written as 11x15=165 ---> A(A+B)B ---> 1 (1+5) 5
There's also a tricks for 6, 6 x Y always ends in Y for any even Y (6 x 8 = 48, 6x 4 =24) and that will carry over into any even-number base system when working with the digit one higher than the halfway point (5 for base 10) as it's just carrying one extra off to x5 (or midpoint of the base) a number of times equal to that number. 6 times any odd number actually just adds that odd number to 5 for the last digit, 6x3= 18 or 15+3, etc.
Edit: seriously we've got to get rid of that 8) makes
drives me nuts