An integer $n$ is such that $\lfloor n/9\rfloor$ is a three-digit number with equal digits, and $\lfloor (n-172)/4\rfloor$ is a four-digit number with the digits $2,0,2,4$ in some order. What is the remainder when $n$ is divided by $100$?

Step 1: Let $k=\lfloor (n-172)/4\rfloor$. Since $k$ is a four-digit number composed of the digits $2,0,2,4$ in some order, the possible values for $k$ are $2024,2042,2204,2240,2402,2420,4022,4202,$ and $4220$.

Step 2: From $k=\lfloor (n-172)/4\rfloor$, we have:

Step 3: We are also given that $m=\lfloor n/9\rfloor$ is a three-digit number with equal digits. Thus, $m\in \{111,222,333,444,555,666,777,888,999\}$. This implies:

Step 4: Combining the two inequalities for $n$:

which simplifies to:

Step 5: We test the possible values of $k$ to find one that yields one of the valid values for $m$. Let us try $m=999$:

Since $k$ must be one of the permitted permutations of $\{2,0,2,4\}$, we see that $k=2204$ lies perfectly in this range.

Step 6: With $k=2204$, the first inequality gives:

And with $m=999$, the second inequality gives:

Step 7: The only integer satisfying both conditions is $n=8991$.

Step 8: Finally, we find the remainder when $n$ is divided by $100$:

Thus, the remainder is $91$.