[VA] SRC #008 - 2021 is here !
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01-14-2021, 10:49 PM
(This post was last modified: 01-15-2021 09:16 AM by Albert Chan.)
Post: #38
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RE: [VA] SRC #008 - 2021 is here !
(01-14-2021 10:18 PM)Vincent Weber Wrote: Well if x is this number, then 2021/x is each equal number contributing to the sum, and the product is (2021/x)^x ... Instead of letting x = number of products, it may be better letting x = base. P = x ^ (2021/x) ln(P) = (2021/x) * ln(x) ln(P)/2021 = ln(x)/x We like to maximize P, so find the local extremum. 0 = (x*(1/x) - ln(x)*1) / x^2 → 1 = ln(x) → x = e Quote:e=2.718... the closest integer that comes to mind is obviously 3 The safer way is not going for closest integer, but actually check value of ln(x)/x 3^2 > 2^3 2 ln(3) > 3 ln(2) ln(3)/3 > ln(2)/2 = ln(4)/4 This showed ln(x)/x maximized when x=e, and we should pack as many 3's as possible. (with the exception of N mod 3 = 1, since 3+1 > 3×1) 2021 = 3*673 + 2 → P = 3^673 * 2 |
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