A number of 50 digits has all its digits equal to 1 except the 26th digit. If the number is divisible by 13, then find the digit in the 26th place.
(In reply to
re(3): solution by Richard)
I don't want to knock you Richard....but you typed this:
we have a=m*(a/m) + a%m=m*q+r. It is quite easy to show that (a+b)%m= (a%m+b%m)%m and (ab)%m= ((a%m)*(b%m))%m, which means we can remainder first before adding or multiplying without affecting the remainder of the result. It is this that makes casting out nines work as a checksum. Since (10^n)%9=1, 176923%9 =(1+7+6+9+2+3)%9=1. The same idea can be applied to digit problems using any value for m, although m=9 is the nicest for decimals, and m=11 is second nicest (since (10^n)%11=(1)^n).
Now...read that over. That does not make *any* sense at all!! You're telling me, when you typed it, you thought it would make sense to a normal 17 year old!? E.g. the first equation... a=m*(a/m)...I can understand that, as the 'm's cancel and so you are left with 'a', but wait...you've mysteriously added on a%m...and that is equal to m*q + r. What the heck is 'q' and 'r'?
Also...you said "It is this that makes casting out nines work as a checksum". It is *what*? Here is an example...sdfffggggg...it is that that makes eggs lay placebo generating coefficients of friction. It makes no sense!!! PLUS....what is "casting" out nines??? What is a "checksum"! If you're going to explain yourself, Richard, please do it properly!

Posted by Kirk
on 20031122 18:49:19 