tag:blogger.com,1999:blog-5159574672638793285.post8463624531702398505..comments2017-06-26T04:16:09.160-04:00Comments on The Center of Math Blog: Problem of the Week: 5-16-17Center of Mathematicshttp://www.blogger.com/profile/06584574038886207370noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-5159574672638793285.post-37146058276929347242017-05-16T15:48:26.945-04:002017-05-16T15:48:26.945-04:00I think the given solution is misleading, you shou...I think the given solution is misleading, you should rather have used "mod 3" in the first instance and "mod 5" in the second instance. As it stands, it looks as if the first is divisible by 3 because 12 is, and the second is divisible by 5 because 10 is. But for example, 12 is also divisible by 6 but 14^n+11 is never divisible by 6 for n>0. The key point is that 12 *and* 15 are divisible by 3, and idem, mutatis mutandis for the second case.MFHhttp://www.blogger.com/profile/04360855081789112747noreply@blogger.com