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I have the ecuation:

$$42x \equiv 12 \text{ mod } 60$$

Using my logic the answer should be:

$x = \frac{43 \cdot ?}{12}=$ something $r$ $60$ but since that was an exam subject don't think that is so simple.

5xum
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user140102
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3 Answers3

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Since $(42,60)=6$, divide the equation by $6$: $7x\equiv2\pmod{10}$. Now these numbers are small enough to use a brute force approach, but we can also use the Euclid-Wallis implementation of the Euclidean Algorithm without first dividing: $$ \begin{array}{r} &&1&2&3\\\hline 1&0&1&-2&7\\ 0&1&-1&\color{#C00000}{3}&\color{#0000FF}{-10}\\ 60&42&18&\color{#00A000}{6}&0 \end{array} $$ This says that $(60,42)=6$ and $42\cdot\color{#C00000}{3}\equiv\color{#00A000}{6}\pmod{60}$. Multiply by $2$ to get $$ 42\cdot6\equiv12\pmod{60} $$ The last column says that we can add integer multiples of $\color{#0000FF}{-10}$ to this solution to get all the solutions.

Community
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robjohn
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Divide by 6 and get $7x \equiv 2 \pmod {10},\,$ with a solution $x\equiv 6 \pmod {10}$ - either by guessing or with $7^{-1}\equiv 3 \pmod {10}.\,$ Now verify that $6\times 42 \equiv 12 \pmod {60},\,$ i.e. $6$ is also a solution to your original problem (as indicated in the comments there are more solutions, actually $x\equiv 6+10k\pmod{60}, k=0\dots 5\,$).

gammatester
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  • $x\equiv6\pmod{10}$ is the solution (i.e. there are $6$ solutions $\bmod{60}$). No multiplying back necessary. Fortunately, $36\equiv 6\pmod{10}$, therefore that happened to work. – ccorn Apr 03 '14 at 12:38
  • @ccorn No, as written, the method does require verfiying the solution(s) because the proof's inferences are unidirectional, i.e. employ $,\Rightarrow,$ vs $,\iff\ \ $ – Bill Dubuque Apr 03 '14 at 13:34
  • @BillDubuque: Getting to $7x\equiv 2\pmod{10}$ is bidirectional. Remember that $gA\equiv gB\pmod{gN}$ means $gN\mid(gA-gB)$ which is equivalent to $N\mid(A-B)$ for $g\neq 0$, hence equivalent to $A\equiv B\pmod{N}$ – ccorn Apr 03 '14 at 14:00
  • @ccorn My point is: that is true only if you explicitly write the proof in such bidirectional form, which is not true above. Confusion between unidirectiona and bidirectional inferences is a very common source of errors for students, so one needs to be explicit about such. – Bill Dubuque Apr 03 '14 at 14:11
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First thing you can do here is simplify the problem by dividing through by 6.

i.e. if $42x \equiv 12$ mod $60$, then $7x \equiv 2$ mod $10$.

What does $2$ mod $10$ mean? It means that the number ends in a 2.

Hope this helps.

Maylor
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