This is a Number Theory problem about the extended Euclidean Algorithm I found:
Use the extended Euclidean Algorithm to find all numbers smaller than $2040$ so that $51 | 71n-24$.
As the eEA always involves two variables so that $ax+by=gcd(a,b)$, I am not entirely sure how it is applicable in any way to this problem. Can someone point me to a general solution to this kind of problem by using the extended Euclidean Algorithm? Also, is there maybe any other more efficient way to solve this than using the eEA?
(Warning: I'm afraid I'm fundamentally not getting something about the eEA, because that section of the worksheet features a number of similiar one variable problems, which I am not able to solve at all.)
I was thinking about using $71n-24=51x$, rearranging that into $$71n-51x=24.$$ It now looks more like the eEA with $an+bx=gcd(a,b)$, but $24$ isn‘t the $gcd$ of $71$ and $51$...