Thread: Find the Value
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Old 2007-06-26, 22:44   #2
Orgasmic Troll
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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)
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