tag:blogger.com,1999:blog-5303307482158922565.post5636269006989996135..comments2024-03-22T13:39:55.941-07:00Comments on Math Mama Writes...: Two Puzzles from Down UnderSue VanHattumhttp://www.blogger.com/profile/10237941346154683902noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-5303307482158922565.post-47213528544084166232009-09-16T09:03:08.392-07:002009-09-16T09:03:08.392-07:00Yep, no logs in problem number 2.
I loved your i...Yep, no logs in problem number 2. <br /><br />I loved your insight that the other problem ends up looking like the handshake problem (a revelation to me). Maybe it'll show up this counting course (= SF math circle) you're leading...<br /><br />Thanks for visiting me here. (I hadn't put two and two together when I saw your comment at Pat's Blog.)Sue VanHattumhttps://www.blogger.com/profile/10237941346154683902noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-7668604900346983182009-09-15T19:30:15.880-07:002009-09-15T19:30:15.880-07:00It was fun working on these problems with you over...It was fun working on these problems with you over on Pat's blog! And it was nice to see you at SF math circle, too.<br /><br />I think the logs are just for the first problem; the second problem needs some other method.Joshua Zuckerhttps://www.blogger.com/profile/04689961247338617418noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-46583477153280574862009-09-13T20:25:16.875-07:002009-09-13T20:25:16.875-07:00Sue: You're right, but there has to be somethi...Sue: You're right, but there has to be something to the fact that the answer comes out the same!<br /><br />I wasn't think about using logs for the second problem, but I can see on Pat's blog that's what they're trying to do. Very interesting. Thanks for the challenge!Tom DeRosahttps://www.blogger.com/profile/06960561773050547167noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-79136369196242562982009-09-12T12:25:27.237-07:002009-09-12T12:25:27.237-07:00But we're not multiplying them, just counting ...But we're not multiplying them, just counting digits in each one. For example, 2*4 = 8, one digit, 2 and 4 both written down is two digits. Could it be like that? I wasn't sure until I did some work with logs. <br /><br />Discussion of the answer to the second problem is at Pat'sBlog.Sue VanHattumhttps://www.blogger.com/profile/10237941346154683902noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-61842893724573289272009-09-12T10:39:25.787-07:002009-09-12T10:39:25.787-07:00Is the answer to Problem 2 = 299?Is the answer to Problem 2 = 299?Tom DeRosahttps://www.blogger.com/profile/06960561773050547167noreply@blogger.comtag:blogger.com,1999:blog-5303307482158922565.post-60545927759733177622009-09-12T10:35:30.782-07:002009-09-12T10:35:30.782-07:00I just spent like 20 minutes working on problem 1 ...I just spent like 20 minutes working on problem 1 with logs, forgetting a simple rule of exponents:<br /><br />(x^a)(y^a) = xy^a<br />so<br /><br />(2^2009)(5^2009) = 10^2009, which would be 2010 digits!<br /><br />(PS I got the same answer using the logs, so at least I know I did that correctly).<br /><br />Now onto problem 2!Tom DeRosahttps://www.blogger.com/profile/06960561773050547167noreply@blogger.com