Constant ratio when dividing repeating digit patterns by each other

[androol]

like many people have done with their calculators, I was dividing 987654321 by 123456789. and I realized...

sum(0 to N) 987654321 * 10^(N*j)
divided by
sum(0 to N) 123456789 * 10^(N*j)

where N can be any positive integer and j is in the real domain, there is a constant.


Am I weird and stupid?

 

[CRGreathouse]

sum(0 to N) 987654321 * 10^(N*j) = 987654321 * sum(0 to N) 10^(N*j)
sum(0 to N) 123456789 * 10^(N*j) = 123456789 * sum(0 to N) 10^(N*j)
so their quotient is indeed constant.

I ended up not needing this: For N a positive integer, sum (j=0,N,10^(N*j)) = (10^(N^2+N)-1)/(10^N - 1).

 

[androol]

omg omg I'm like so stup.

something times (thing/thing) = something.

how can I be so stupid...