How to solve $(x \mod 7) - (x \mod 8) = 5$?

Question

I'm trying to solve $(x \mod 7) - (x \mod 8) = 5$ but no idea where to start. Help appreciated!

Answer 1

Suppose $x\neq 0$ then assuming

$$x=z+7k$$

with $z\neq 0$ we obtain

$$z-z-7k\equiv 5 \pmod 8$$

$$k\equiv 5 \pmod 8$$

and if

$$x=7k$$

we obtain

$$-k-7k\equiv 5 \pmod 8$$

$$0\equiv 5 \pmod 8$$

Therefore without any limit in the residues we have infinitely many solutions in the form

$$x=z+35 \quad z\not\equiv 0 \pmod 7$$

for example for $x=36$

$$(36 \mod 7) - (36 \mod 8) =1-(-4)= 5$$

for $x=43$

$$(43 \mod 7) - (43 \mod 8) =8-3= 5$$

Otherwise if we assume the standard convention for a complete residue system the possible solutions are

$$x=36,37,38,39,40,41$$

and by inspection we find that only $x=40,41$ work.

Answer 2

I'm going to assume you are incorrectly assuming $\mod n$ refers to the remainder function and want to solve $(x \% 7) -(x\% 8) = 5$ where $a \% b$ is the unique remainder when $a$ is divided by $b$.

$x \% 7 = a$ means $x = 7k + a$ for some integer $k$ and integer $0 \le a < 7$. So $a = x - 7k$.

And $x \% 8 = b$ means $x = 8j + b$ for some integer $j$ and integer $0 \le b < 8$. So $b = x - 8j$

So $a - b = 8j - 7k = 5$. By Bezout Lemma solutions exist. Let's find them.

$8*1 - 7*1 = 1$ so $8*5 - 7*5 =5$ is such a solution and $x = 5*7 + a = 35+a$ and $x = 5*8 + b = 40+b$ which can work with $a = 5$ or $6$ and $b = 0$ or $1$.

So $x = 40$ or $41$ will be solutions.

Answer 3

If CRT = Chinese Remainder Theorem is known then we can reformulate it as

$$\begin{align} x&\equiv a\!+\!5\!\!\pmod{7},\ \ \ \overbrace{ 0\le a\!+\!5 \le 6}^{\Large \iff a\ =\ 0,1}\\ x&\equiv a\quad\pmod{8},\ \ \ 0\le a\le 7\end{align}\qquad \qquad\qquad $$

Solve the $\,a=0\,$ case $\,x\equiv (5,0)\bmod (7,8),\,$ then $\,x\!+\!1\equiv (6,1)\,$ yields the $\,a=1\,$ case.

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W/o $ $ CRT: $\,\ \overbrace{ x-7j}^{\large x\bmod 7}-(\overbrace{x-8k)}^{\large x\bmod 8} = 5\iff 8k - 7j = 5\,\Rightarrow\, j \equiv \color{#c00}{5}\pmod{\!8},\ k\equiv \color{#c00}{5}\pmod{\!7}$

Working in the intial range $\, 0\le x\le 55\ $ and enforcing the remainder bounds

$\qquad\quad \begin{align} 0\le \overbrace{x-7\cdot \color{#c00}5}^{\large x\bmod 7} \le 6\iff 35\le x \le \color{#0a0}{41}\\[.5em] 0\le \underbrace{x-8\cdot \color{#c00}5}_{\large x\bmod 8} \le 7\iff \color{#0a0}{40}\le x \le 47\end{align}$

Therefore $\, x\equiv \color{#0a0}{40},\color{#0a0}{41}\pmod{56}$