IBM Research | Ponder This | August 2014 solutions
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August 2014

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If the costs were equal, the best way to solve this problem would be a binary search. However, since erring on one side costs more, we should skew the search.
Using the following dynamic programming:
f(1) = 0
f(n) = min_{0<i<n} [i/n*(f(i)+10) + (n-i)/n*(f(n-i)+1)]
f(100) = ?
we can find the best strategy, which costs, on average, 26.5 cents.

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