Is there a possible way to solve an equation like this? it has only 1 variable, but it is in both sides. Constraint: n must be an integer
20-3n = 0 mod (10n+7)
one solution would be n=1, Can there be multiple solutions other than that?
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Welcome to MSE. Please include in your question what you have tried so far, and please use mathJax. In its current state, the question will most likely be closed for being off topic. – Anadactothe Mar 01 '19 at 19:52
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For $n=-2$ we have $20-3n=26\equiv 0 \bmod -13=10n+7$. – Dietrich Burde Mar 01 '19 at 19:57
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1What are the constraints on $n$? does it need to be an integer? does it need to be positive? – user413766 Mar 01 '19 at 19:57
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@JonHales it must be an integer – Omar Mar 01 '19 at 19:58
3 Answers
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Let $m = 10 n + 7$. Note that $m$ and $10$ are coprime, so $10$ is invertible mod $m$, and $n \equiv - 7 \cdot 10^{-1} \bmod m$. Then your equation says
$$ 20 + 3 \cdot 7 \cdot 10^{-1} \equiv 0 \bmod m$$
and multiplying by $10$,
$$ 200 + 3 \cdot 7 = 221 \equiv 0 \bmod m$$
so $m$ must be a divisor of $221$. There are four positive possibilities: $m = 1, 13, 17$, or $221$, of which only $17$ is of the form $10n+7$ (corresponding to $n=1$), and four negative possibilities: $m=-1, -13, -17, -221$, of which only $-13$ is of the form $10n+7$ (corresponding to $n=-2$).
Robert Israel
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Hint $ $ by here $\,10n\!+\!7 = \gcd(\color{#c00}{10}n\!+\!\color{#0a0}{7,-3}n\!+\!\color{#c00}{20})\,\mid\, \color{#c00}{10(20)}\!-\!\color{#0a0}{7(-3)}\, =\, 13(17)$
Bill Dubuque
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In order for $20-3n \equiv 0 \mod (10n+7)$ we must have that $10n+7$ divides $20-3n$. So $|10n+7| \leq |20-3n|$.
But clearly for $n$ large enough, (or negative enough) we have that $|10n+7| > |20-3n|$. Thus we can reduce this to a finite check, and see that the only solutions are $n=1$ or $n=-2$.
user413766
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