Finding $26$th digit of a number.

Question

I have found an old Regional Olympiad problem (It is so old that you can call it ancient). Which is as follow:

$N$ is a $50$ digit number (in the decimal scale). All digit except the $26$th digit (From the left) are $1$. If N is divisible by $13$ then find the $26$th digit.

My approach:

I went on the basics, I noticed that\ the number $"111111"$ is divisible by $13$, so $111111111....... (24 times)$ will be divisible by $13$. So, we excluded $24$ $1$'s from the left and simlarily $24$ $1$'s from the right. Now, we are only left with the $25$th and $26$th digit, So will be divided by $13$ iff the combo of $25$th and $26$th digit is divisible by $13$.

Since we know that $25$th digit is $1$ (and we have to find $26$th digit). We get $13|10+a$ and immidiatly we get $a=3$.

The problem is that I m not satisfied with this work myself (Though I have done well, I think :) :)). I just want another solution Or indirectly I want to approach this question differently. Any suggestion is heartily welcome.

Thanks.

I really mean it. You've subtracted several multiples of a number divisible by $13$ untill you're left with $1a\cdot 10^{25}$, and we can easily see that $a$ must be $3$ if this is to be divisible by $13$. — Arthur, Dec 07 '16 at 18:44

score 3 · Answer 1 · answered Dec 07 '16 at 18:46

Your solution is already pretty good, although it's a little unclear how easy it is to see that $13\mid 111111$: one way is to recall that $7\cdot 11\cdot 13 = 1001 \mid 111111$. But it's even simpler to see that $13$ divides $11\cdots1$ with $12$ digits: that's just Fermat's little theorem applied to $(10^{12}-1)/9$, and your argument goes through exactly the same.

The dry, mindless approach (since you ask for a different one) is just to write the exact formula for $N$:

$$N = (10^{50}-1)/9 + (a-1)\cdot 10^{24}.$$

By the above, this simplifies mod $13$ to $99/9 + (a-1) = 10+a$, so $a \equiv 3 \pmod{13}$ from which we conclude that $a$ must be $3$.

Thanks and +1. I didn't mentioned that but I was somehow thinking of applying Fermet's little theorem, but I couldn't found my way through it. But now it is pretty clear to me. Thanks. — Vidyanshu Mishra, Dec 07 '16 at 18:49

score 1 · Answer 2 · edited Dec 23 '16 at 14:11

1

Hint:

$$\mathbb{111 111 111 111 111 111 111 111} 1x \mathbb{111 111 111 111 111 111 111 111} =\\ \mathbb{111 111 111 111 111 111 111 111}*(10^{26}+1) + 1x*10^{24}$$

edited Dec 23 '16 at 14:11

answered Dec 07 '16 at 18:45

Jaroslaw Matlak

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Finding $26$th digit of a number.

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