MMO Round 2

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

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

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

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

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

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

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