[VA] SRC #008 - 2021 is here !

[Albert Chan]

(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