linear congruence - theory number - 2 questions 
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A Bus completes his path every $T$ hours. A man watched the bus for a few days .
  When he started watching the time was 00:00 and by the time he finished 17:00.
  The bus did 11 paths in the time the man watched it.
  Find $T$.

What I did:
$ 11T \equiv 1 \pmod {17}$
and then $T$ is 14.
Is that right???

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Need to solve
  $ 103x \equiv 444 \pmod{ 999}$

I found there is one solution  x is 111.
 A: The following is a probably unreasonable interpretation of the question. It is  motivated by the number-theoretic setting:  we are maybe implicitly asked to assume that $T$ is an integer.  
The man started watching, and watched continuously for several days, say $x$.  Then the time spent was $24x+17$. If $11$ complete rounds were made, then $24x+17\equiv 0\pmod{11}$. To solve this quickly, rewrite as $2x+6\equiv 0\pmod{11}$, giving $x\equiv -3\pmod{11}$. The smallest positive solution is $x=8$, giving $T=19$. Long drive!
Equivalently (and more simply!) we solve $11T\equiv 17\pmod{24}$. To solve quckly, note that $11^2\equiv 1\pmod{24}$, so multiply both sides of the congruence by $11$. 
A: 2.
So, $103x=444+999a$ for some integer $a$
or, $103x=111(4+9a)$ or $\frac{103x}{111}=4+9a$ an integer. 
$\implies 111\mid x$
Let $x=111y\implies 103y=4+9a$
$\implies 103y\equiv4\pmod 9$
$\implies 4y\equiv4\pmod 9$
$\implies y\equiv1\pmod 9$ as $(4,9)=1,y=9b+1$ for some integer $b$
So, $x=111y=111(9b+1)\equiv 111\pmod{999}$
