A number theory problem by أحمد الحلاق

The problem can be solved using Chinese remainder theorem as follows. Let the required number be $n$ , then we have:

$\begin{aligned} n & \equiv 1 \text{ (mod 5)} & \small \color{#3D99F6}{\text{Let }n = 5p+1} \\ \implies 5p+1 & \equiv 2 \text{ (mod 7)} \\ 5p & \equiv 1 \text{ (mod 7)} \\ 5(7q + 3) & \equiv 1 \text{ (mod 7)} & \small \color{#3D99F6}{\implies p = 7q+3} \\ \implies 35q + 16 & \equiv 3 \text{ (mod 9)} \\ -q + 7 & \equiv 3 \text{ (mod 9)} \\ q & \equiv 4 \text{ (mod 9)} \\ 9r + 4 & \equiv 4 \text{ (mod 9)} & \small \color{#3D99F6}{\implies q = 9r+4} \\ \implies 9r + 3 & \equiv 4 \text{ (mod 11)} \\ 9r & \equiv 1 \text{ (mod 11)} \\ 9(5) & \equiv 1 \text{ (mod 11)} & \small \color{#3D99F6}{\implies r = 5} \end{aligned}$

$\begin{aligned} \implies r & = 5 \\ q & = 9r+4 = 49 \\ p & = 7q+3 = 346 \\ n & = 5p + 1 = \boxed{1731} \end{aligned}$

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