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If I have the equation-

$$ 2^b \bmod 11 = 7 $$

How can I solve this to find out what $b$ is?

I know $b$ is $7$ but I'd like to know how this is done mathematically rather than guessing.

Soham
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TheRapture87
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  • Look at this article – rogerl May 07 '16 at 18:42
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    There is nothing unmathematical about systematically trying $b=1,2,3,\dots$. There are only 9 possibilities and that is by far the fastest method! – almagest May 07 '16 at 18:51
  • $b=7$ is not the only solution (try $b=77$). – Dietrich Burde May 07 '16 at 20:31
  • The general problem is called discrete logarithm. This article provides some good approaches: https://math.dartmouth.edu/~carlp/dltalk09.pdf – Fred Yang May 08 '16 at 15:08
  • I agree with the comment by @almagest; that's how I did the problem. You can slightly speed up that approach by noticing, while you're trying successive powers of $2$ mod $11$, that $2^2=4$ and $2^5\equiv-1$, so $2^{2+5}\equiv-4\equiv7$. (If you know some theoretical results, then you'd compute $2^2$ and $2^5$ before any other powers of $2$, to verify that $2$ is a primitive root modulo $11$.) – Andreas Blass Jul 18 '16 at 04:51

2 Answers2

1

Since $2$ are $11$ are relatively prime, using Fermat's little theorem, we get that $2^{10}=1 (mod 11)$. Hence you only need to check for only $b=1,2,...,10$

joy
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I am giving you a genaralised solution for this-

You need this theorem-

$\color{red}{a\equiv b\pmod c\implies na\equiv nb\pmod c}$

Proof:-

$a\equiv b\pmod c\implies c|(a-b)$ and $na\equiv nb\pmod c\implies c|(na-nb)\implies c|n(a-b)$ Now,we have already proved $c|(a-b)$

Coming back to the problem-

Notice that $2^n$ for $n<3$ is less than $11$ hence remainder is $0$.

So,start working from $n=4$.

By small and simple calculation $2^4\equiv5\pmod{11}....[1]$

By our above theorem,multiplying by $2$ we have,

$2^5\equiv10\pmod {11}...[2]$($2^4\times2\equiv5\times2\pmod{11}$)

Similarly by repeated multiplication by $2$ have the following congruences-

$$2^6\equiv9\pmod {11}...[3]$$ $$2^7\equiv7\pmod {11}...[4]$$ $$2^8\equiv3\pmod {11}...[5]$$ $$2^9\equiv6\pmod {11}...[6]$$ $$2^{10}\equiv1\pmod {11}...[7]$$ $$2^{11}\equiv2\pmod {11}...[8]$$ $$2^{12}\equiv4\pmod {11}...[9]$$ $$2^{13}\equiv8\pmod {11}...[10]$$ $$2^{14}\equiv5\pmod {11}...[11]$$

So,we observe that from $2^{14}$ the series starts repeating the remainders.

So form here we have $n=7$ as a solution.

But,what about a more generalised solution?

The remainders are in an AP.

We know nth term of an $AP$ is $a_n=a+(n-1)d$

Since we have not taken $n=1,2,3$ we do this correction to our formula-

$(a_n-3)=a+(n-1)d$

$\implies(a_n-3)=4+(n-1)10$.

(We take $a=4$ since when the next power of $2$ comes to the fouth place after completing the whole cycle of remainders from $[1]$ to $[10]$ and again repeating the remainders of $[1],[2],[3],[4]...$)

Solving we get,

$\displaystyle\frac{a_n+3}{10}=n$

Now,$n$ has to be a integer since term of a series cannot be in fraction.

So,by intuition we see that $a_n$ must end in $7$ ($7+3=10$ and we have $10$ in denominator).

So,$\color{red}{7,17,27...n7}$ is the set of solutions for this.(n is any integer).

Hope this helps and please comment me if you have difficulty in understanding any part of the solution!!

Community
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Soham
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