10 Comments

1. Steven Miller on September 14, 2012 at 3:08 pm

   jake: email me at sjm1 AT williams.edu

   
   
   
2. Steven Miller on May 7, 2012 at 5:03 pm

   let me know if you want a hint/soln (sjm1 AT williams.edu)

   
   
   
3. Mohamed Ilyas A on May 7, 2012 at 4:58 pm

   No idea about 1st question. But I’m trying.
   2008 is not possible I think. Since, the set contain only odd number, cannot be divided using 2008.

   
   
   
4. Steven Miller on April 9, 2012 at 12:46 am

   you can do this without actually dividing –ie, you can prove one exists without knowing which one works!

   
   
   
5. Andy Cai on April 8, 2012 at 6:45 pm

   Euclidean Algorithm for divisibility helps greatly.

   
   
   
6. Steven Miller on December 24, 2011 at 4:52 pm

   sure //s

   
   
   
7. Mike Varnava on December 24, 2011 at 4:07 pm

   Steve, long time, no see! I hope you are well!

   Can we hopefully tip our fellow riddle solvers, by saying that I used Fermat’s little theorem in solving this myself?

   
   
   
8. Steven Miller on December 5, 2011 at 1:54 pm

   I don’t believe this logic is correct.

   
   
   
9. Thilina on December 5, 2011 at 6:20 am

   got it in an easy way! if you notice the set of numbers, the number : 111111111 is devisible by 9 and 2007 is also devisible by 9. So that answers will be respectively, 12345679 and 223. as the set contains numbers more than 9*223, we can find a number which has 9*223, 1s: that means 2007th number. So this number is devisable, so that set contains atleast one number which is divisible by 2007.

   second part : no it’s not true. it’s not divisible. if you consider the ending of 2008, is 8. when 8 is multiplied by any number from 1 to 9, the ending number is again even. but no number in the given set it even. So it’s not divisible

   
   
   
10. Thilina on December 5, 2011 at 5:36 am

    i think we can use Congruences to prove this easily ????????