tag:blogger.com,1999:blog-5244458674117363159.post626468951712146915..comments2023-09-24T02:39:05.162-07:00Comments on Daily GRE: Math GRE - #16Paul Liuhttp://www.blogger.com/profile/16809371907394009052noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-5244458674117363159.post-14572364104262583282013-05-22T16:57:22.462-07:002013-05-22T16:57:22.462-07:00Hey, I solved this problem and all the other GRE m...Hey, I solved this problem and all the other GRE math subject test questions. You can view my solutions at http://rambotutoring.com/GRE-math-subject-test-68-solutions.pdfAnonymoushttps://www.blogger.com/profile/10917551019935272317noreply@blogger.comtag:blogger.com,1999:blog-5244458674117363159.post-16002954020588228542011-08-17T19:59:48.123-07:002011-08-17T19:59:48.123-07:00@3314509773579284909.0
Yes, but spotting i = 80 i...@3314509773579284909.0<br /><br />Yes, but spotting i = 80 is a bit difficult for those who are not good with mental arithmetic (myself included).<br /><br />I haven't found any quick tests of determining of a specific number of zeros exist at <a href="http://oeis.org/A027868" rel="nofollow">OEIS</a> either, though there are a few computation intensive formulae that give the exact number of trailing zeros in k! for any k. My guess is that there is no easy way (i.e. without much computation) to find if 99 zeros are possible.Paul Liuhttps://www.blogger.com/profile/16809371907394009052noreply@blogger.comtag:blogger.com,1999:blog-5244458674117363159.post-33145097735792849092011-08-17T19:45:49.778-07:002011-08-17T19:45:49.778-07:00You're right we can ignore the 2s and just foc...You're right we can ignore the 2s and just focus on the 5s. <br /><br />Let's say div(i,j) is the integer div operation.<br /><br />For a given k, let's say that i=div(k,5). Then k!, when prime-factorised, will contain 5 to the power of <br /><br />i + div(i,5) + div(i,25) + div(i,125) + ...<br /><br />All we need to show is that there exists an i such that this sum equals 99. As it turns out, i=80 satisfies this, as<br /><br />99 = 80 + div(80,5) + div(80,25) = 80 + 16 + 3<br /><br />Therefore, if i=80 and i=div(k,5), then k! will figure 5 to the power of 99 in its prime-factorisation. Solving div(k,5)=80 gives k=400,401,402,403,404.Raoul Golanhttps://www.blogger.com/profile/06197787742625805513noreply@blogger.com