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.
(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 20040124 17:51:29 