tag:blogger.com,1999:blog-8231784566931768362.post3538917991940001138..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: The Largest Odd Factor of 90? Too Easy? How About A Million? A Googol!Dave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-8231784566931768362.post-3983640817882672362008-09-20T22:49:00.000-04:002008-09-20T22:49:00.000-04:00I solved this problem in my head exactly the same ...I solved this problem in my head exactly the same way as Susan.<BR/><BR/>I don't know for sure the best way for students to do it. Experimenting with a calculator sounds good to me for starters. <BR/><BR/>In general I tend to favor a solution that is efficient and "professional". By that I mean that if people who know math well tend to solve a problem a certain way, then that's a "professional" way to solve it.Maria Millerhttps://www.blogger.com/profile/00230743954246449727noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-87176521242946721142008-09-16T20:20:00.000-04:002008-09-16T20:20:00.000-04:00Thanks, Susan. I think these problems were not vie...Thanks, Susan. I think these problems were not viewed as very interesting but I do see their potential for developing prime factorization concepts as well as other related ideas.<BR/><BR/>I agree with your solutions and comments. I can tell you love the reasoning and logical parts of mathematics. You're a problem solver!<BR/><BR/>The idea of starting with a million and having students divide by 2 repeatedly until an odd factor remains is an interesting way to get middle schoolers to analyze a situation. After they try this with a few large even numbers, they should begin to understand why the method works. Of course, we don't have to restrict the numbers to products of powers of 2 and 5.<BR/><BR/>Another interesting variation is play the "million" game on the calculator as early as 4th or 5th grade.<BR/>The teacher (or parent) can play against the child or children can compete against each other. Let's say I go first:<BR/>1000000/2 = 500000<BR/>The the child goes:<BR/>500000/5 = 100000<BR/>Play alternates with each player dividing by either 2 or 5 each time. The player who produces a result of 1 loses!<BR/>Most will figure out after awhile that the player who goes first will always win after 12 divisions. An interesting way to develop the idea of prime factors I think. Students will tire of starting with a million each time so change the the prime factors to say 2,3, and 7 and of course change the starting number to something like (2^4)(3^3)((7^2). Since this number will result in 9 plays, the person who goes first loses!<BR/>What do you think?Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-33745602546842772342008-09-16T19:44:00.000-04:002008-09-16T19:44:00.000-04:00No responses yet?I'm not a math instructor or ...No responses yet?<BR/><BR/>I'm not a math instructor or even a math student, but the method that pops out at me for the million and googol problems is first to express the number as 10^k, then as (2⋅5)^k, and finally as (2^k)⋅(5^k). Then you can say that the largest odd factor is 5^k.<BR/><BR/>The million and googol problems are probably easier than the 90 problem, because they'd actually have to factor 90 (and its factorization is not as easy as (2^k)⋅(5^k)), or repeatedly divide by 2.<BR/><BR/>(Argh, no <sup> allowed!)Anonymousnoreply@blogger.com