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 Ring in 2013 (Posted on 2012-12-31)
Start as you wish had you prove that for any k-digit number M there exists a number n such that the string of first k digits of 2n equals M.

Find a power of 2 whose decimal expansion begins with the 12-digit string "201320132013". It need not be the smallest such number.

Bonus: Find the smallest such number and prove it to be the smallest.

 See The Solution Submitted by Charlie Rating: 4.0000 (1 votes)

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 re: 'Slide Rule' Logic | Comment 2 of 5 |
(In reply to 'Slide Rule' Logic by broll)

Of course 2^1363 (integer power) begins "20131868734285".

It's unfortunate the original formulation mentioned a number n, but the idea is for an integer, otherwise we could use just 2^log2(201320132013) where log2(201320132013) ~= 37.5507004927368408372497476346649

 Posted by Charlie on 2013-01-01 03:30:07

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