N is a 50digit nomber(in the decimal scale). All digits except the 26th digit(from the left) are 1. If N is divisible by 13, find the 26th digit.
we can use the facts that
so
so
also
then combining this with our other idendity
so
so our number is
which we can write as..
using our idendities from before, we know that for instance
since 48 = 7*6 so
which simplifies to
so if then
the only value of k where this is true is
thus this is the answer.
What I wanted to say(although I have not tried) is that we could substitute the 26th digit as 1 and perform divison(which will be easy due to repeating 1's) and then find find the value of k by guessing the remainder that the 26th digit should give while performing division.Originally Posted by malaygoel
I think I have found it.Originally Posted by malaygoel
By performing simple division, I found that 111111 is divisible by 13
Now,
there are 25 1's before k and 24 1's after k.
24 1's before k and 24 1's after k are divisible by 13
We are left with and hence the answer 3.
Hello, ThePerfectHacker!
Good call!
These numbers are called "repunits".
We can however show: if a repunit is prime, then the number of ones must also be prime.
"Repunits" is an abbreviation for repeated units.
Even if the numbers of 1's is a prime number, there is no dependable rule.
Let represent a number composed of n 1's.
I would be happy if you post here the proof of the statement in the quoted work.But the question still remains:Is there any criteria by which we can say that " is prime or composite depends on the value of k"?Originally Posted by ThePerfectHacker
Similarly, what can you say about the numbers .
Keep Smiling
Malay