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 2010-02-05 16:30:09