tag:blogger.com,1999:blog-2433841880619171855.post3435969255102583650..comments2024-03-27T21:09:44.320+00:00Comments on Pat'sBlog: A Nice, but Tough, ProblemUnknownnoreply@blogger.comBlogger1125tag:blogger.com,1999:blog-2433841880619171855.post-87122113979102937192008-10-08T20:13:00.000+01:002008-10-08T20:13:00.000+01:00You don't need to bring e, or ln, into it.All ...You don't need to bring e, or ln, into it.<BR/><BR/>All you need is that 3^2 > 2^3.<BR/><BR/>Then that shows that you want to have zero or one or two 2s and all the rest 3s (with the number of 2s determined by the value of your number mod 3, so with 200 you'll want to have one 2 and 66 3s).<BR/><BR/>Another fun question -- does e have to enter into this one?<BR/>Suppose that instead of partitioning n into integers with the largest product, you take n and choose a value of k, split n into k equal pieces, and multiply those pieces together. What value of k gives the largest result? In other words, for what integer k is (n/k)^k maximized?Joshua Zuckerhttps://www.blogger.com/profile/04689961247338617418noreply@blogger.com