\[W(n) = \displaystyle \sum_{k=1}^{\lfloor\log n\rfloor + 1}\left( 2^k\left\lfloor\frac{n \pmod{10^k}}{10^{k-1}}\right\rfloor\right).\]

Define the weight of a positive integer \(n\) as above.

Without a calculator (of course), find the last three digits of the largest positive integer \(\tau \in \left(0,10^5\right)\) that satisfies \(16\cdot W(\tau) = \tau\).

**Note.** This is the easy semi-bashy version; if I ever get around to it, I will post my other version.

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