tag:blogger.com,1999:blog-20067416.post8900098649877924116..comments2021-09-24T17:58:19.157-04:00Comments on Recursivity: The 2010 Bernoulli TrialsUnknownnoreply@blogger.comBlogger7125tag:blogger.com,1999:blog-20067416.post-2434922704856136612010-03-30T17:07:39.694-04:002010-03-30T17:07:39.694-04:00ivanm - I take it back, I see the error you point ...ivanm - I take it back, I see the error you point out.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-20067416.post-79753838405351558022010-03-30T17:04:38.411-04:002010-03-30T17:04:38.411-04:00ivanm: double check that. There's a second ne...ivanm: double check that. There's a second negative in the numerator of the last term which you missed.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-20067416.post-46031400783399357472010-03-23T02:13:38.317-04:002010-03-23T02:13:38.317-04:00I just discovered an error in the 2006 Bernoulli T...I just discovered an error in the 2006 Bernoulli Trials solutions, in the last problem, #16. The limit there is -1/30, not 2/15, and thus the correct answer is FALSE, not TRUE. (It looks like the only mistake in the solution is that b_3 - a_3 is given as -1/2 when it's actually +1/2.)<br /><br />Does this mean that the winner of that contest was awarded wrongly? Yikes!IvanMnoreply@blogger.comtag:blogger.com,1999:blog-20067416.post-34582188349586449412010-03-18T14:03:52.541-04:002010-03-18T14:03:52.541-04:00Harrison: That's not quite right for #11. You ...Harrison: That's not quite right for #11. You only need the common difference to be divisible by a primorial (i.e., 2*3*5*7*11*13*... in general). Also, which primorial is needed depends on the starting term.<br /><br />Since we don't have spoiler tags here, I'll give my answers in sorta-cryptic form:<br /><br />The problem numbers whose answers are true are exactly those which are congruent modulo ten to either the additive identity or one of the first two primes.<br /><br />Let me know if I made any embarrassing mistakes.IvanMnoreply@blogger.comtag:blogger.com,1999:blog-20067416.post-15812986353952176642010-03-17T23:27:18.601-04:002010-03-17T23:27:18.601-04:00The last question: "A rectangular box with si...The last question: "A rectangular box with sides of length 1, 2 and 3 hovers above the <br />flat ground. The<br />maximum possible area of its shadow on the ground is equal to 7"<br /><br />Nope, it's infinity.<br />The light source isn't directly overhead.<br /><br />Heh, just kidding. I hope to work on this one.Mirandanoreply@blogger.comtag:blogger.com,1999:blog-20067416.post-11689125320156647552010-03-17T14:37:33.945-04:002010-03-17T14:37:33.945-04:00For #4 it seems to be false as if you take the set...For #4 it seems to be false as if you take the set of p_n for n such that n=(2*10^k)-1 you get a sequence 1/1,11/19,111/199 which is decreasing but cannot go below 5/9. When n=10^k-1, the sequence equals exactly 1/9. Since after each of the tens powers the probability gets larger, it will oscillate between going above 5/9 and at 1/9 so a limit will not exist.<br /><br />For #11 the common difference must be divisible by 2,3,4,5,6,7, and the lowest common multiple is 2*5*6*7 which is 420, and thus the progression cannot exist with all primes less than 900.<br /><br />I'm feeling rusty right now so I'm not sure if these are right but I'm just throwing them out here.hpgrosshttps://www.blogger.com/profile/10518506913832606439noreply@blogger.comtag:blogger.com,1999:blog-20067416.post-17385255858019505832010-03-17T09:03:42.171-04:002010-03-17T09:03:42.171-04:00Hint for number 11: seven primes in arithmetic pr...Hint for number 11: seven primes in arithmetic progression must have a common difference divisible by what?Jeffrey Shallithttps://www.blogger.com/profile/12763971505497961430noreply@blogger.com