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.)
re: Digit number | Comment 31 of 39 |
(In reply to Digit number by Purna)

You write "For a given number to be divisible by 13, the sum of all the digits of the given number must be divisible by 13." This is false, as 13 itself testifies. The sum of the digits is divisible by 9 if and only if the number is divisible by 9, but with 13 in place of 9, this doesn't work. The digits must be weighted by numbers congruent to the power of 10 of their place, modulo the divisor in question. With the divisor 9, things work out so nicely because 10^n is always congruent to 1 modulo 9. Modulo 13, the powers of 10 are not all congruent to 1, however.

  Posted by Richard on 2004-01-24 17:51:29

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

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information