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70 arithmetic/digits/power.two.p |
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Articles / TULARC / Self Growth / Puzzles / | |
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70 arithmetic/digits/power.two.p |
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This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.
Prove that for any 9-digit number (base 10) there is an integral power
of 2 whose first 9 digits are that number.
arithmetic/digits/power.two.s
Let v = log to base 10 of 2.
Then v is irrational.
Let w = log to base 10 of these 9 digits.
Since v is irrational, given epsilon > 0, there exists some natural number
n such that
{w} < {nv} < {w} + epsilon
({x} is the fractional part of x.) Let us pick n for when
epsilon = log 1.00000000000000000000001.
Then 2^n does the job.
Continue to:
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