tag:blogger.com,1999:blog-2237906900456925233.post6450469896745076552..comments2023-10-18T23:36:58.396+08:00Comments on 數學資料庫手記: Elementary number theoryMathDBhttp://www.blogger.com/profile/18353093974880979999noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-2237906900456925233.post-36268636649299641602009-12-25T03:20:38.267+08:002009-12-25T03:20:38.267+08:001989k = 9*13*17*k =r(111...1), where r between 1-9...1989k = 9*13*17*k =r(111...1), where r between 1-9<br /><br />Numbers divisible by 13<br />alternating sum of three-digit, if it is divisible by 13, then the number is divisible by 13.<br /><br />so 111,111 is divisible by 13 as (111)-(111)=0<br /><br />Numbers divisible by 17<br />alternating sums of 8-digit groups!<br /><br />so 1...16-digit...1 is divisible by 17.<br /><br />so 1....48-digit...1 is the smallest integer of this form which can be divisible by both 13 and 17.<br /><br />and so the smallest positive integer k s.t. all digit of 1989k are the same = 9*13*17*k = (r...48 digit...r)<br /><br />if r=1, 48 != 0 mod 9, so not the answer.<br />if r=2, 96 !=0 mod 9, so not the answer.<br />if r=3, 144 = 0 mod 9, 3...48digit...3 is the required answer!<br /><br />the given k = 167588402882520529579353108764873470755823697 have 45 digits,<br />16758<br />8402882520<br />5295793531<br />0876487347<br />0755823697<br /><br />so this k times 1989 will have 48 digits.<br /><br />If the question guarantee this k times 1989 would produces a number with all digit are the same, then i can 100% agree.Pophttps://www.blogger.com/profile/17406316084081487132noreply@blogger.com