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$11x + 13 \equiv 4$ (mod 37)

My "solution" to the problem

$11x + 13 \equiv 4$ (mod 37) $\rightarrow$

$11x + 13 = 4 + 37y $

$11x - 37y = - 9$

Euclid's algorithm.

$37 = 11*3 + 4$

$11 = 4*2 + 3$

$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$

$3 = 1*3 + 0$

Write as linear equation

$1 = 4 - 1*3$

$3 = 11 - 2*4$

$4 = 37-3*11$

$1 = 4 -1*3 = 4-1(11-2*4) = 3*4-1*11 = 3(37-3*11)-1*11 = 3*37 -10*11$

$1 = 3*37 -10*11$

$11(10) - 37(3) = -1$

$11(90) - 37(27) = -9$

x = 90

That is my answer. But the correct answer should be:

x = 16

Daniel Andersson
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5 Answers5

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Note: $90\mod37 =16$, so the answer you found was correct, just not fully reduced.

Tyler6
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We have $$11x\equiv -9\equiv 28\mod 37$$ so we get $$x\equiv \frac{28}{11}\equiv \frac{65}{11}\equiv\frac{102}{11}\equiv \frac{139}{11}\equiv \frac{176}{11}\equiv 16\mod 11$$

Dr. Sonnhard Graubner
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  • We don't need to employ a brute-force search above. Instead we can use Inverse Reciprocity. Here the idea is that we seek $,\color{#c00}k,$ that makes an exact quotient from $, 28/11 \equiv (28!+!37\color{#c00}k)/11,,$ which is true iff $$\bmod 11!:,\ 0\equiv 28!+!37k\equiv {-}5!+!4k\iff 4k\equiv 5\equiv 16\iff \color{#c00}{k\equiv 4}$$ Therefore $,x\equiv (28 +37\cdot\color{#c00} 4)/11 \equiv 176/11 \equiv 16$ $\ \ \ $ – Bill Dubuque Apr 23 '19 at 17:38
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$\bmod 37\!:\,\ x\equiv 90\equiv 16\ $ so both are correct. Gauss's algorithm is simpler:

$\bmod 37\!:\,\ x\equiv\dfrac{-9}{11}\equiv\dfrac{-27}{33}\equiv\dfrac{10}{-4}\equiv\dfrac{5}{{-}2\,}\equiv\dfrac{-32}{-2}\equiv 16$

Bill Dubuque
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We are looking for a solution to $11x\equiv-9\pmod{37}$.

First, solve $11x+37y=1$. As shown in this answer, we use the Extended Euclidean Algorithm: $$ \begin{array}{r} &&3&2&1&3\\\hline 1&0&1&-2&3&-11\\ 0&1&-3&7&-10&37\\ 37&11&4&3&1&0 \end{array} $$ Thus, $$ \begin{array}{c} 3\cdot37-10\cdot11=1\\ \Downarrow&\text{multiply by $-9$}\\ 90\cdot11\equiv-9\pmod{37}\\ \Downarrow&\text{add $13$ to both sides}\\ 16\cdot11+13\equiv4\pmod{37}&90\equiv16\pmod{37} \end{array} $$ Therefore, we have $$ x\equiv16\pmod{37} $$

robjohn
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Mod 11, it reduces to $2\equiv 4y+4$ and $24=2\cdot11+2=4(5)+4$ which then gets $11x+13=189\implies 11x=176\implies x=16$

Note however, this is just one of infinitely many answers. Incrementing x by 35, and y by 11 yields new answers that work. All the x values fall on the line $35z+16$, all the y values fall on the line $11a+5$