Thread: Find the Value View Single Post
 2007-06-26, 22:44 #2 Orgasmic Troll Cranksta Rap Ayatollah     Jul 2003 64110 Posts so here's what I do know.. if n = 999, then we have 99 0s from the ones digit and 90 0s from the tens digit if n = 9999, we have 999 0s from the ones digit, 990 0s from the tens digit and 900 0s from the hundreds digit if $n = 10^{k+1}$ then there are $10^{k-1}-1$ from the ones digit, $10^{k-1}-10$ from the tens digit, and so on, thus if $n = 10^{k+1}$, then we have accumulated $\sum_{i=0}^{k-1} 10^{k-1} - 10^i$ 0s, this equals $k*10^{k-1} - \sum_{i=0}^{k-1} 10^i = k*10^{k-1} - \frac{10^k-1}{9}$ now, somewhere in between 10^11-1 and 10^12-1, we have more 0s than n, in fact, we have 88,888,888,889 more 0s, so, assuming that the difference is monotone increasing (I believe it is, although I haven't checked formally) it has to happen for some 11 digit number (or it doesn't happen at all)