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.
We have to look at the multiples of 199, and find which one(s) can end in a 7. We only need to look at the multiples from 09, as we're getting a result in base10.
199 x 1 = 199
199 x 2 = 398
199 x 3 = 597199 x 4 = 796
199 x 5 = 995
199 x 6 = 1194
199 x 7 = 1393
199 x 8 = 1592
199 x 9 = 1791
As 199 x 3 is our only possible choice, we know that our last digit is
3. Now we know we need a remainder of 59 from the second to last 7, thus our last digit of the actual multiple of 199 must be an 8 (...7  8 = ...9). This tells us that we have 199 x 2 + 59 = 457.
Thus, our second to last digit must be a
2, and our remainder from the third to last digit is 45.
To get a remainder of 45, from a number ending in 7, our value must end in a 2. That means our next multiple is 199 x 8 = 1592. We now know the third to last digit is
8, and 199 x 8 + 45 = 1637.
Now, for the final digit, we need a remainder of 163. That tells us the multiple ends in 4, which means we're dealing with 199 x 6 = 1194, 1194 + 163 = 1357.
We now know that the fourth to last digit is a
6. You could continue further, but the problem only asks for the final four digits:
......
6823

Posted by Justin
on 20100205 14:02:04 