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When I am faced with the following:

$15 \equiv x$ (mod 10)

Ignore the simplicity of this. What I am interested is is the following in any way connected to the solution for $x$:

$\frac{15}{5} \equiv \frac{x}{5}$ (mod 2)

The congruence was divided by $5$. Does solving this for $x$ in any way connect to the first congruence?

Michael Munta
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1 Answers1

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$x \equiv ca \pmod{\!cn}\,\Rightarrow\, c\mid x,\ $ so $\,\ c(x/c)\equiv ca \pmod{\!cn} \iff x/c \equiv a \pmod{\! n}$

because $\,cn\mid cX-ca\iff n\mid X-a,\,$ since $\ \dfrac{cX-ca}{cn} = \dfrac{X-a}n\,$

Remark $ $ For the general case (and its fractional viewpoint) see this answer.

Bill Dubuque
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  • Here $,c\neq 0\ $ (usually by convention for moduli) – Bill Dubuque Jan 27 '19 at 17:23
  • So having this $\frac{15}{5} \equiv \frac{x}{5} \equiv$ -> $3 \equiv \frac{x}{5}$ (mod 2) essentially I would be searching for a number that when divided by $5$ gives $1$ in modulo $2$ and that number would be a solution in modulo $10$ as well. But if I were to utilize multiplicative inverse to solve for $x$ I would get $x \equiv 15$ (mod 2) and with it I would lose the connection to the congruence in modulo $10$. Is that right? – Michael Munta Jan 28 '19 at 10:39
  • Numbers divisible by $5$ congruent to $1$ modulo $2$ are: $5, 15, 25, 35...$ and they are all possible values of $x$. As soon as I get rid of the denominator it is no longer true. – Michael Munta Jan 28 '19 at 11:05
  • @Michael $x\equiv 15\equiv 5\pmod{!10}\iff x = 5+10n = 5(1+2n)\iff x,$ is $5$ times an odd integer. You can't use inverses because $5$ isn't invertible $!\bmod 10\ \ $ – Bill Dubuque Jan 28 '19 at 15:36
  • But I can multiply both sides by $5$ to get rid of denominators then I would just get $x$ to be $1$ modulo $2$. Getting rid of the denominator in this way would no longer have meaning to the first congruence? – Michael Munta Jan 28 '19 at 15:46
  • @Michael Scaling $,x/5 = 3 + 2n,$ by $5$ yields $, x = 15 + 10n,.,$ i.e. you need to scale the modulus too to get an equivalent congruence. – Bill Dubuque Jan 28 '19 at 15:53
  • Though when I said about using an inverse was for $3 \equiv \frac{x}{5}$ (mod $2$). I would get $3 \equiv x$ (mod $2$). Then the only fact we know about $x$ is that it is odd, but not that also it is a multiple of $5$. – Michael Munta Jan 29 '19 at 08:37
  • @Michael $,x/5,$ (& $,x/c,$ in my answer) denote a quotient in $,\Bbb Z,,$ not a modular fraction. When $,c,$ is coprime to the modulus it will be true that you get a true (but weaker) congruence by reducing it as a fraction, e.g. $, x/5\equiv 0\pmod{!2}\iff 2\mid x/5\iff 5\cdot 2\mid x,,$ but reducing it mod $2,$ yields a weaker condition $,x/(5\bmod 2)\equiv x\equiv 0\pmod{!2}\iff 2\mid x.,$ Further, generally $,c,$ need not be coprime to the modulus, e.g. the case $,c=n,$ of the answer is

    $$ x \equiv an!!!\pmod{!n^2},\Rightarrow, x/n \equiv a !!!\pmod{!n}$$

     – Bill Dubuque Jan 29 '19 at 17:50
  • We can view my answer operationally via the mod Distributive Law $, cb\bmod cn = c(b\bmod n) $ $$\begin{align} &c\mid x,\Rightarrow, x\bmod cn = c(x/c\bmod n)\ {\rm e.g.}\ \ \ &5\mid x,\Rightarrow, x\bmod 10 = 5(x/5\bmod 2) = 5(x\bmod 2) \end{align}$$ Consider what the prior means in terms of the decimal units digit of an integer $x.\ \ $ – Bill Dubuque Jan 29 '19 at 18:12