Exercise 1. We have three operations on integers:

- Operation $A$ which consists of multiplying by 2 and adding 4 to the result.
- Operation $B$ which consists of multiplying by 4 and adding 16 to the result.
- Operation $C$ which consists of multiplying by 5 and adding 25 to the result.

a) Determine an integer $x$ such that starting from $x$ and successively applying the operations $A, B$, and $C$ in this order, the result is 1945.

b) Determine all integers $x$ such that starting from $x$ and successively applying two different operations, the result is 2015.
Solution to exercise 1 a) Starting from $x$, we successively obtain $2 x+4$, $4(2 x+4)+16=8 x+32$, and $5(8 x+32)+25=40 x+185$. Therefore, we have $40 x+185=1945$. By subtracting 185 from both sides, we deduce $40 x=1760$, so $x=\frac{1760}{40}=44$.

b) If the last operation is $A$ or $B$, then the resulting number is even and therefore cannot be 2015. Therefore, the last operation is $C$.

Let $y$ be the integer obtained after applying the first operation. We have $5 y+25=2015$ so $y=398$. Now, a number obtained after applying operation $B$ is divisible by 4, and 398 is not divisible by 4, so the first operation must necessarily be $A$. We deduce that $2 x+4=398$. Finally, $x=197$.