A number N consisting of a string of sevens
i.e. N=77777...7777, is divisible by 199.
Find the last 4 digits of the quotient.
No programming, please.
I found the solution in essentially the same way as Justin without a program.
I wanted to make sure such a number exists. I was pretty sure it did but I couldnt figure out how to prove it without programming. So I wrote a little program to find the remainders of 7/199, 77/199, 777/1999, etc.
I used my calculator so I wont share the program itself except that the crux is if R is a remainder, the next remainder is
10R+7  199*INT((10R+7)/199)
This does eventually give a remainder of zero when N is 99 digits long. (And the last few agree with Justin as well)
Incidentally N could just as well have been 11111...1111 since 199 is not divisible by 7.

Posted by Jer
on 20100205 16:30:09 