[tex] \large \color{indigo}\bold{SOLUTION:}[/tex]
It's should be solved by trial and error. Let me help you.
Let x, y, z be the number of 161-rod, 23-rod and 7-rod respectively.
[tex] \bold{161x + 23y + 7z = 583} \\ \bold{0 + 2y + 0 = 2(mod \: 7)} \\ \bold{y = 1(mod \: 7)} \\ \bold{so \: y = 1, \: y = 8 ,\: y = 15, \: or \:, y = 22} \\ \\ \large \color{indigo}\bold{for \: y = 1} \\ \\ \dashrightarrow\bold{161x + 7z = 560} \\ \dashrightarrow\bold{x = 3, \: z = 11} \\ \dashrightarrow{ \bold{x = 2, \: z = 34}} \\ \dashrightarrow{ \bold{x = 1, \: z = 57}} \\ \\ \large \color{indigo}{ \bold{for \: y = 8}} \\ \\ \dashrightarrow{ \bold{161x + 7x = 399}} \\ \dashrightarrow{ \bold{x = 2, \: z = 11}} \\ \dashrightarrow{ \bold{x = 1, \: z = 34}} \\ \\ \large \color{indigo}{ \bold{for \: y = 15}} \\ \\ \dashrightarrow{ \bold{161x + 7z = 238}} \\ \dashrightarrow{ \bold{x = 1, \: z = 11}} \\ \\ \large\ \color{indigo}{ \bold{for \: y = 22}} \\ \\ \dashrightarrow{\bold{161x + 7z = 77}} \\ \dashrightarrow\bold{x = 0(not \: a \: solution)} \\ \\ \bold{So \: all \: the \: Solution \: Set \: are:} \\ \ \bold{»\color{darkviolet} \: (1,15,11),(1,8,34),(2,8,11),(1,1,57),(2,1,3,4),(3,1,11)}[/tex]
[tex] \large \color{indigo} \bold{Carry\: On\: Latex}[/tex]
[tex]\purple{\begin{gathered} \gamma \\ \huge \boxed{ \ddot \smile}\end{gathered}} \: \pink{\begin{gathered} \gamma \\ \huge \boxed{ \ddot \smile}\end{gathered}} \: \red{\begin{gathered} \gamma \\ \huge \boxed{ \ddot \smile}\end{gathered}} \: \orange{\begin{gathered} \gamma \\ \huge \boxed{ \ddot \smile}\end{gathered}}[/tex]
