MMO2025 Round 2

1. Let us compute $(A\star B)\star C$:

First, $A\star B=\frac{A}{AB+1}$.

Now, $(A\star B)\star C=\frac{A\star B}{(A\star B)C+1}$.

Substitute $A\star B$: $$(A\star B)\star C=\frac{\frac{A}{AB+1}}{\frac{A}{AB+1}\cdot C+1}$$

Simplify the denominator: $$\frac{A}{AB+1}\cdot C+1=\frac{AC}{AB+1}+1=\frac{AC+AB+1}{AB+1}$$

So, $$(A\star B)\star C=\frac{\frac{A}{AB+1}}{\frac{AC+AB+1}{AB+1}}=\frac{A}{AC+AB+1}$$

Now, compute $A\star (B+C)$: $$A\star (B+C)=\frac{A}{A(B+C)+1}=\frac{A}{AB+AC+1}$$

Therefore, $$(A\star B)\star C=A\star (B+C)$$

2. Let us compute $((((\cdots (((1\star 2)\star 3)\star 4)\cdots )\star 59)\star 60)\star 61)$.

Let us denote $S=(((\cdots (((1\star 2)\star 3)\star 4)\cdots )\star 59)\star 60)\star 61$.

From part (1), we have $(A\star B)\star C=A\star (B+C)$. This means the operation is associative in the following sense: $$(((A\star B)\star C)\star D)=A\star (B+C+D)$$

Therefore, $$S=1\star (2+3+4+\cdots +59+60+61)$$

The sum $2+3+4+\cdots +61$ is an arithmetic progression:

The sum from $2$ to $61$ is: $$\sum _{k=2}^{61}k=\sum _{k=1}^{61}k-1=\frac{61\cdot 62}{2}-1=1891-1=1890$$

Therefore, $$S=1\star 1890=\frac{1}{1\cdot 1890+1}=\frac{1}{1891}$$