Find all numbers of form $10^k+1$ divisible by $49$ Basically, I've tried to take mods, and it hasn't been very successful. Also, if it helps, I noticed that the sequence can be recursively written as $a_{n+1}=10a_n-9$, starting with $a_1=11$. 
 A: You have to solve
$$10^k\equiv-1\pmod{49}\ .$$
Calculating powers,
$$\eqalign{
  10^2&=100\equiv2\cr
  10^3&\equiv20\cr
  10^4&\equiv(10^2)^2\equiv4\cr}$$
and so on.  You will eventually find $10^{21}\equiv-1$ as the first solution (I actually did this by asking Maple).  So the order of $10$ modulo $49$ is $42$, and the complete solution is
$$k=21+42t\ ,\quad t\in\Bbb Z\ .$$
Short cut.  If $10^k\equiv-1\pmod{49}$, then $10^k\equiv-1\pmod{7}$ and a much smaller amount of trial and error shows that $k=3+6t$.  So if you go back to mod $49$, you now only need to calculate
$$\eqalign{
  10^2&\equiv2\cr
  10^3&\equiv20\cr
  10^6&\equiv(10^2)^3\equiv8\cr
  10^9&=10^310^6\equiv160\equiv13\cr
  10^{15}&=10^910^6\equiv104\equiv6\cr
  10^{21}&=10^{15}10^6\equiv48\equiv-1\ .\cr}$$
A: Since $3$ is a primitive element of $\mathbb{F}_7^*$, 
$$7\mid (10^k+1)\Longleftrightarrow 3^k\equiv -1\pmod{7}\Longleftrightarrow k\equiv 3\pmod{6}$$
so $k=3(2m+1)$ is equivalent to $7\mid (10^k+1)$. The sequence given by:
$$ a_n = 10^n+1\pmod{49} $$
is regulated by the recurrence relation:
$$ a_{n+1} = 10a_n-9 $$
and it is eventually periodic since the residue classes $\pmod{49}$ are just $49$. If a zero occurs, it occurs for some $n$ that is an odd multiple of $3$ by the previous argument, so let:
$$ b_n = a_{3n} = 20^n+1\pmod{49},$$
$$ c_n = b_{2n+1} = 20\cdot 8^n+1\pmod{49},$$
$$ d_n = \frac{20\cdot 8^n+1}{7}\pmod{7}.$$
Now we have just to understand for which $n$s we have $d_n=0$. 
Since it is not difficult to check, by induction, that:
$$ d_n\equiv 3-n\pmod{7}, $$
we have:
$$ 49\mid (10^k+1)\Longleftrightarrow k=3(2(7h+3)+1)\Longleftrightarrow k\equiv 21\pmod{42}.$$
