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Let k be the smallest digest then a_{n+1} = a_n + 2^k

Source: AIMO 3/1, German Pre-TST 2010

July 16, 2011
algebra unsolvedalgebra

Problem Statement

A sequence (an)\left(a_n\right) with a1=1a_1 = 1 satisfies the following recursion: In the decimal expansion of ana_n (without trailing zeros) let kk be the smallest digest then an+1=an+2k.a_{n+1} = a_n + 2^k. How many digits does a9⋅102010a_{9 \cdot 10^{2010}} have in the decimal expansion?
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