Find natural number $x=523...$ ($6$ digits) such that :$\begin{cases}x\equiv 0\pmod{7}\\x\equiv 0\pmod{8}\\x\equiv 0\pmod{9}\end{cases}$

Question

Problem :

Find natural number $x=523...$ ($6$ digits) such that :

$$\begin{cases}x\equiv 0\pmod{7}\\x\equiv 0\pmod{8}\\x\equiv 0\pmod{9}\end{cases}$$

This number is $x=523152,,523656$

But I don't know how I find it ??

From congruence $x\equiv 0\pmod{504}$

But how I complete ?

Answer 1

Your solution $x$ satisfies $x=523000+y$, with $0\le y<1000$.

As $523000\equiv 352\pmod{504}$, you should be able to finish.

Answer 2

$\begin{align}\text{Purely}\, {\textit mentally }\ \bmod 504\!:\ \ 0&\,\equiv\, x + \overbrace{\color{#c00}{523}(2\cdot \color{#0a0}{500})}^{\textstyle 523000}\\ &\,\equiv\, x\,+\, \color{#c00}{19}(2)(\color{#0a0}{-4})\\ &\,\equiv\, x\,-152\\[.1em] \iff &\ \ \ \ \ \bbox[5px,border:1px solid #c00]{x\equiv 152}\end{align}$

Answer 3

Let $N=523abc=504000+19000+abc$

$19000=352 mod 504$

$abc+352= 504 k$

$k=1$ ⇒ $abc=504-352=152$

$k=2$ ⇒ $abc=1008-352=656$

$k=3$ ⇒ $abc=1512-352=1106$

So abc=152 and 656 are solutions