# linear congruence - theory number - 2 questions

1. 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???

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

I found there is one solution x is 111.

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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 $11$.

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@wow thank you very much I was so complicated with this question for some reason!!!! – Mary Dec 7 '12 at 18:47
@Nusha: I added a simpler way to look at it. Did not remove the first way. – André Nicolas Dec 7 '12 at 19:18
thanks but can you explain how did you got 2x+6=0? – Mary Dec 7 '12 at 19:44
Was working mod $11$, so $24\equiv 2$ and $17\equiv 6$. – André Nicolas Dec 7 '12 at 19:49

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, $x=111y=111(9b+1)\equiv 111\pmod{999}$

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