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Topic: Functional Equation (Read 338 times) |
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ThudnBlunder
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Functional Equation
« on: Oct 15th, 2004, 6:41am » |
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Consider a function which, for a given integer n, returns the number of 1s required to write out all of the integers between
0 and n.
For example, f(13) = 6
After F(1) = 1, what is the next n such that f(n) = n?
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| « Last Edit: Oct 15th, 2004, 6:42am by ThudnBlunder » |
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Fresno_Bob
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Re: Functional Equation
« Reply #1 on: Oct 15th, 2004, 11:10am » |
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Here's a shot at it
f(999,999,999) = 900,000,000
f(1,099,999,999) = f(999,999,999) + 100,000,000 + 80,000,000 = 1,080,000,000
f(1,109,999,999) = f(1,099,999,999) + 20,000,000 + 7,000,000 = 1,107,000,000
f(1,110,999,999) = f(1,109,999,999) + 3,000,000 + 600,000 = 1,110,600,000
f(1,111,099,999) = f(1,110,999,999) + 400,000 + 50,000 = 1,111,050,000
f(1,111,109,999) = f(1,111,099,999) + 50,000 + 4,000 = 1,111,104,000
f(1,111,110,999) = f(1,111,109,999) + 6,000 + 300 = 1,111,110,300
f(1,111,111,099) = f(1,111,110,999) + 700 + 20 = 1,111,111,020
f(1,111,111,109) = f(1,111,111,099) + 80 + 1 = 1,111,111,101
f(1,111,111,110) = f(1,111,111,109) + 9 = 1,111,111,110
1,111,111,110
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