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I am preparing for an exam. I am dealing with this right now:

$$93x + 47 \equiv 61\pmod{101}$$

However, I can't figure it out. Can someone describe steps for this example, or provide a link to any free pdf, website describing this problem and the way it is solved on attached photo, please? I repeat it needs to be solved like on the photo.

Thanks in advance

Notes here

Community
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Andrew_Dublin
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    I'd encourage you to take a look at this for how to typeset your mathematics well. This will make your post much easier to read, and it will also enable users to more easily pinpoint where it is you are having an issue. – Daniel W. Farlow Apr 18 '15 at 15:55
  • Downvotes don't seem helpful here. To OP: I imagine the reason you were downvoted may be understood in light of my first comment. Some people just get annoyed by a huge picture and then just downvote for that reason alone even though it is obvious you put a good bit of effort into the question at hand (which the lack thereof is what usually induces one to downvote). Honestly, I don't even read questions where almost all of it is a picture and the content in the picture is not very easily read. Learning how to typeset everything well may be annoying now but it pays off bigtime in the future. – Daniel W. Farlow Apr 18 '15 at 16:10
  • I'm voting to close this question as off-topic because it is unclear as to why the problem needs to be solved exactly like in the photo. – Daniel W. Farlow Apr 18 '15 at 16:15

3 Answers3

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The equation is equivalent to $$ 93x\equiv 61-47\pmod{101} $$ so to $$ 93x\equiv14\pmod{101} $$ You just need to find the inverse of $93$ modulo $101$: \begin{align} 101&=93\cdot 1+8\\ 93&=8\cdot 11+5\\ 8&=5\cdot1+3\\ 5&=3\cdot1+2\\ 3&=2\cdot1+1 \end{align} so \begin{align} 1&=3+2\cdot(-1)\\ &=3+(5-3\cdot1)\cdot(-1)\\ &=3\cdot2+5\cdot(-1)\\ &=(8-5)\cdot2+5\cdot(-1)\\ &=8\cdot2+5\cdot(-3)\\ &=8\cdot2+(93-8\cdot11)\cdot(-3)\\ &=93\cdot(-3)+8\cdot35\\ &=93\cdot(-3)+(101-93)\cdot35\\ &=93\cdot(-38)+101\cdot35 \end{align} so the inverse is $-38\equiv63\pmod{101}$.

Thus $$ x\equiv14\cdot63\equiv74\pmod{101} $$

Yes, your computation seems right.

egreg
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  • Can you explain me what are you doing here please ? 1=3−2⋅1=3+3−5=3⋅2+5⋅(−1)=(8−5)⋅2+5⋅(−1)=8⋅2+5⋅(−3)=8⋅2+(93−8⋅11)⋅(−3)=93⋅(−3)+8⋅35=93⋅(−3)+(101−93)⋅35=93⋅(−38)+101⋅35 – Andrew_Dublin Apr 18 '15 at 16:28
  • @Andrew_Dublin Start from the last remainder ($1$ in this case) and write it as a linear combination of the two previous remainder. Go upwards and do the same for every remainder, substituting it in the relation you have. – egreg Apr 18 '15 at 16:30
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Rewrite our congruence as $93x\equiv 14\pmod{101}$.

Now we can use a general procedure, by finding the inverse $b$ of $93$ modulo $101$, and multiplying through by $b$.

Or else note that $93x\equiv -8x\pmod{101}$, and $14\equiv 14+202=216\pmod{101}$. That gives $x\equiv -27\pmod{101}$, or equivalently $x\equiv 74\pmod{101}$.

André Nicolas
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By mod $\rm\color{#c00}{\rm division}$ by $2$ $\ {\rm mod}\ 101\!:\ x\equiv \dfrac{67\!-\!41}{93}\equiv \dfrac{14}{-8}\equiv \dfrac{\color{#c00}7}{-4}\equiv \dfrac{\color{#c00}{108}}{-4}\equiv -27\equiv 74$

Bill Dubuque
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