I'm sorry there's an algebra error below, but fortunately the proof
still works. It should be
>From this, algebra provides
10^e - 1 == 0 (mod 9Y) and
10^e == 1 (mod 9Y).
But 9Y and 10^e are still coprime, so we're good.
Here is code that seems to be working fine.
#include
int main(int argc, c
I'm sorry there's an algebra error here, but fortunately the proof
still works. It should be
>From this, algebra provides
10^e - 1 == 0 (mod 9Y) and
10^e == 1 (mod 9Y).
But 9Y and 10^e are still coprime, so we're good.
Here is code that seems to be working fine.
#include
int main(int argc, ch
This is very beautiful. Here is a less elegant proof, though it leads
to an efficient algorithm.
In a different problem some time ago, there was a statement that every
number with a 3 in the units position has a multiple that consists of
all 1s. The proof needs a little number theory. Any numbe
@dave: +1 dude...:-)
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Theorem: In any set of (n+1) distinct integers there exists two whose
difference is divisible by n.
Proof: Each of these integers leaves a remainder between 0 and n-1
inclusive upon division by n. Since there are n possible remainders
and (n+1) integers that means there exist two which have the sa
Theorem: In any set of (n+1) distinct integers there exists two who
difference is divisible by n.
Proof: Each of these integers leaves a remainder between 0 and n-1
inclusive upon division by n. Since there are n possible remainders
and (n+1) integers that means there exist two which have the same
