## April 2010

Our original solution was as follows:

If the digits are x_i, then we have that \sum_{i=0}^N 10^i x_i = \sum_{i=21}^N 16^{i-21} x_i. Defining d_i = 10^{i+21} - 16^i, we get a lattice problem that can easily be solved by backtracking.
15 out of the 16 solutions are divided by 2, 3, or 7, leaving only 228076225315225436100944141061067721086496831646429721832862692146190993521

57014522058098996385254903281545695887189603267201 as a possible candidate. Primality checking verifies that it indeed is a valid solution.

We received the following elegant solution from one of our solvers:

Preliminary estimates show that such a number would have 125 decimal places. Using random seeds, look for fixpoints of the Transformation T: "interpret n_{10}, last 21 digits dropped, as (T(n))_{16}", until you hit a prime.

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