tag:blogger.com,1999:blog-5730391.post1532688444135770876..comments2024-04-12T03:59:51.417-07:00Comments on Blaine's Puzzle Blog: This Number is a Two-Timer...Blainehttp://www.blogger.com/profile/06379274325110866036noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-5730391.post-57225435339346198182008-02-19T17:52:00.000-08:002008-02-19T17:52:00.000-08:00Don and Pantheist are exactly correct. The last d...Don and Pantheist are exactly correct. The last digit is 2 which will get moved to the front. Therefore the next digit has to be double this:<BR/>...42<BR/>Then double the 4:<BR/>...842<BR/>Then double the 8 (remember the carry):<BR/>...(1)6842<BR/>Then double the 6 and add the carry (13), that gets you another carry:<BR/>...(1)36842<BR/>Double the 3 and add the carry (7)<BR/>...736842<BR/>Double the 7 (remember the carry)<BR/>...(1)4736842<BR/>Double the 4 and add the carry (9)<BR/>...94736842<BR/>Double the 9 and remember the carry<BR/>...(1)894736842<BR/>Double the 8 and add the carry (17), remember the new carry:<BR/>...(1)7894736842<BR/>etc.<BR/>...(1)57894736842<BR/>...(1)157894736842<BR/>...3157894736842<BR/>...63157894736842<BR/>..(1)263157894736842<BR/>.5263157894736842<BR/>(1)05263157894736842<BR/>Now when you double you get back to the digit 2 so you are done.<BR/><BR/>105,263,157,894,736,842<BR/><BR/>P.S. This is the shortest number. You could actually repeat this sequence of digits as much as you like.Blainehttps://www.blogger.com/profile/06379274325110866036noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-12441022175790585712008-02-19T09:28:00.000-08:002008-02-19T09:28:00.000-08:00105263157894736842105263157894736842Don Hodunhttps://www.blogger.com/profile/10629317119387671011noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-44309675488218081012008-02-19T09:06:00.000-08:002008-02-19T09:06:00.000-08:00I did it in 18 places. Any takers??I did it in 18 places. Any takers??Don Hodunhttps://www.blogger.com/profile/10629317119387671011noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-84996269629249071992008-02-18T19:47:00.000-08:002008-02-18T19:47:00.000-08:00You're on the right track when you start with "the...You're on the right track when you start with "the second to last digit must be 4". You know that because you know the last digit is 2, and doubling a number that ends in 2 gives a number that ends in 4.<BR/><BR/>Now continue from there. If you know it ends in 42, then the next digit on the left of 4 must be 8. This is because doubling any number that ends in 42 gives a number that ends in 84. Hence the last 3 digits are 842<BR/><BR/>Keep going with this reasoning. The trick is not to give up!Pantheisthttps://www.blogger.com/profile/01072012869065860952noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-62322587503056325602008-02-18T08:28:00.000-08:002008-02-18T08:28:00.000-08:00Well, the first digit is a 1. The next-to-last di...Well, the first digit is a 1. The next-to-last digit is a 4. That fulfills the basic requirements given by the problem statement.Don Hodunhttps://www.blogger.com/profile/10629317119387671011noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-86177534412771518552008-02-17T00:17:00.000-08:002008-02-17T00:17:00.000-08:00You tried 3 or 4 variables but have you gone highe...You tried 3 or 4 variables but have you gone higher?Blainehttps://www.blogger.com/profile/06379274325110866036noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-9419708254844971452008-02-17T00:03:00.000-08:002008-02-17T00:03:00.000-08:00Ben,It has an unknown number of digits which is wh...Ben,<BR/><BR/>It has an <I>unknown</I> number of digits which is why I included an ellipsis ...Blainehttps://www.blogger.com/profile/06379274325110866036noreply@blogger.comtag:blogger.com,1999:blog-5730391.post-11427838601600677302008-02-16T06:20:00.000-08:002008-02-16T06:20:00.000-08:00I played around with it but I keep ending up with ...I played around with it but I keep ending up with equations that look something like this:<BR/><BR/>2000A+200B+20C+2=2000+100A+10B+C<BR/>1900A + 190B + 19C = 1998<BR/>19(100A + 10B + C) = 1998<BR/><BR/>1998 not being evenly divisible by 19, there doesn't seem to be any integer combination, even if you use a zero, that will get the job done. The same holds true for four variables, etc.<BR/><BR/>Guess I'm on the wrong track?Benhttps://www.blogger.com/profile/04754156636762779545noreply@blogger.com