All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
50 - Digit Number (Posted on 2003-11-15) Difficulty: 3 of 5
A number of 50 digits has all its digits equal to 1 except the 26th digit. If the number is divisible by 13, then find the digit in the 26th place.

See The Solution Submitted by Ravi Raja    
Rating: 3.3333 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution re(2): a little more straightforward | Comment 8 of 39 |
(In reply to re: a little more straightforward by Tristan)

10^24 is not divisible by 13 by unique factorization into primes as 10^24 factors uniquely as 2^24 * 5^24 and the prime 13 does not appear. The 50-digit number written out in normal number (arabic) notation looks like
111111 111111 111111 111111 x1 111111 111111 111111 111111
where I have put in spaces to make the blocks of 6 ones stand out. Thus the number is the sum of multiples of 111111, which is divisble by 13, and x1*10^24. Thus to make the whole thing divisible by 13, x1*10^24 must also be divisible by 13. Since 10^24 certainly is not divisible by 13, x1 must be. The only two-digit number ending in 1 that is divisble by 13 is 91=13*7, so x=9 is the unique solution.
  Posted by Richard on 2003-11-19 01:59:52

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information