, 2024

Solution:
Note that
$$(10 a + 9) \star (10 b + 9) = (100 a b + 90 a + 90 b + 81) - 2 \equiv 90(a + b) + 79 \pmod{100}$$
so throughout our process all numbers will end in $9$, so we will just track the tens digit. Then the "new operation" is
$$a \dagger b \equiv -(a + b) + 7 \pmod{10}$$
where $a$ and $b$ track the tens digits. Now
$$(a \dagger b) \dagger c \equiv (-(a + b) + 7) \dagger c \equiv a + b - c \pmod{10}.$$
Thus, our expression has tens digit congruent to
$$-0 + 1 - 2 + 3 - \cdots - 54 + 55 - 56 - 57 + 7 \equiv -28 - 57 + 7 \equiv 2 \pmod{10}$$
making the answer $29$.