Is there a way to express this congruence in terms of a and b?

[egal]

 

[pka]

View attachment 31181

Your post is extremely difficult to read. Is it, two sequence [imath]a_m~\&~b_n[/imath] both of which converge to [imath]x[/imath]???
What is [imath](m\cdot n)~?[/imath] or is it [imath](n-m)~?[/imath] OR what is it?
Please clarify the question.


[imath][/imath][imath][/imath]

 

[egal]

Your post is extremely difficult to read. Is it, two sequence [imath]a_m~\&~b_n[/imath] both of which converge to [imath]x[/imath]???
What is [imath](m\cdot n)~?[/imath] or is it [imath](n-m)~?[/imath] OR what is it?
Please clarify the question.


[imath][/imath][imath][/imath]

I apologize I forgot the mods in the picture, just I wonder if [math]x\equiv a \pmod m, \newline x\equiv b \pmod n.[/math] then can you express this congruence in terms of a and b? [math]x \equiv ? \mod m.n[/math] I thought this might be true [math]x \equiv ab \mod m.n[/math]

 

[pka]

[imath]68\equiv 2 \mod 3[/imath]
[imath]68\equiv 2\mod 6[/imath]
[imath]68\equiv 14 \mod 18[/imath]
[imath][/imath][imath][/imath]

 

[egal]

[imath]68\equiv 2 \mod 3[/imath]
[imath]68\equiv 2\mod 6[/imath]
[imath]68\equiv 14 \mod 18[/imath]
[imath][/imath][imath][/imath]

thanks a lot