Year Today, Gone Tomorrow?
Consider the set {1,11,111, …, ((10^2007) – 1)/9}.
Prove at least one of these numbers is divisible by 2007.
Is the same true for 2008 (replacing 10^2007 with 10^2008, of course)?
Communicated by Michael Varnava.
Consider the set {1,11,111, …, ((10^2007) – 1)/9}.
Prove at least one of these numbers is divisible by 2007.
Is the same true for 2008 (replacing 10^2007 with 10^2008, of course)?
Communicated by Michael Varnava.
jake: email me at sjm1 AT williams.edu
let me know if you want a hint/soln (sjm1 AT williams.edu)
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.
you can do this without actually dividing –ie, you can prove one exists without knowing which one works!
Euclidean Algorithm for divisibility helps greatly.
sure //s
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?
I don’t believe this logic is correct.
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
i think we can use Congruences to prove this easily ????????