tag:blogger.com,1999:blog-2400513859305780710.post3534956154457758480..comments2023-04-04T07:53:53.789-07:00Comments on Mathematical Food For Thought: Rounding Time! Topic: Number Theory. Level: AIME.Jeffrey Wanghttp://www.blogger.com/profile/11114458640271201663noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-2400513859305780710.post-13971974731637764462005-12-24T11:41:43.000-08:002005-12-24T11:41:43.000-08:00Merry Christmas! =) Miss you!Merry Christmas! =) Miss you!Liz & Stanleyhttp://www.freewebs.com/blue_giraffenoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-18567145097688855652005-12-25T20:47:11.000-08:002005-12-25T20:47:11.000-08:001. We want to take \sum 4k^2 + 1 until it becomes...1. We want to take \sum 4k^2 + 1 until it becomes very close to 600, then keep adding terms until it exceeds 600. We want 4(n)(n+1)(2n+1)/6 + n > 600. To find an approximate answer, we will just write 4n^3/3 ~ 600, 4n^3 ~ 1800, n^3 ~ 450, which gives us either 7 or 8. 7 gets us 4(7)(8)(15)/6 + 7 = 567, which is as close to 600 as we can risk. This means we are using \sum 4k^3 + k = 4( 8(9)/2 )^2 + 8(9)/2 = 5220 terms.<br><br>The terms after that are all going to have f(k) = 8, and we still have to make up for a sum of at least 23, so we need 8 * 23 = 184 more terms for a total of m = 5404 terms. I think.QCnoreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-31573849226732062352005-12-28T00:31:43.000-08:002005-12-28T00:31:43.000-08:00Everything's right, except the sum of 4k^3+k f...Everything's right, except the sum of 4k^3+k from k=1 to 7 is actually<br><br>4[(7*8)/2]^2+(7*8)/2 = 3164.<br><br>So the answer should be 3164+184 = 3348.paladin8noreply@blogger.comtag:blogger.com,1999:blog-2400513859305780710.post-14379753391716792712009-12-29T10:44:57.000-08:002009-12-29T10:44:57.000-08:00600-567=33 :P...but shouldn't there be an extr...600-567=33 :P<br>...but shouldn't there be an extra term added so that the sum [i]exceeds[/i] 600?morinhuurnoreply@blogger.com