Realm of Numericon

This problem can be solved using Legendre’s Formula and Prime Factorization.

Legendre's Formula: Let $m$ be a prime number, and we call it $p,$ then the exponent of $p$ in the prime factorization of $n!$ is:

$f(n, p) = ⌊\frac{n}{p}⌋ + ⌊\frac{n}{p^2}⌋ + ⌊\frac{n}{p^3}⌋ +…$ until $p^x$ is less than or equal to $n.$

When $m$ is not a prime, then we need to express the composite number $m$ as a product of its prime factors (Prime factorization). Let's say, $m$ = $p_1^{c_1}\cdot p_2^{c_2}\cdot p_3^{c_3}\cdot …\cdot p_k^{c_k}$ where $p_1, p_2, p_3,…, p_k$ are distinct prime factors of $m$ and $c_1, c_2, c_3,…, c_k$ are the corresponding exponents.

For each prime factor $p_i$ (where $1≤i≤k$) we can use Legendre's formula to find the exponent of $p_i$ in the prime factorization of $n!.$ Then the largest exponent of $m$ that divide $n!$ would be the minimum of $ƒ(n, p_i)$ divided by $c_i.$

$min_{i = 1}^k⌊\frac{f (n,\space p_i)}{c_i}⌋$

Now, what is the largest exponent of $m$ that divides $^nC_r?$

For each prime factor $p_i$ (where $1≤i≤k$) we can apply Legendre's formula to find the exponent of $p_i$ in the prime factorization of $n!, r!$ and $(n - r)!$ separately. Then subtract the exponents we get for $r!$ and $(n - r)!$ from the exponent for $n!$ to find the exponent of $p_i$ in the prime factorization of $^nC_r.$

So, the largest exponent of $m$ that divides $^nC_r$ would be:

$min_{i = 1}^k⌊\frac{f(n,\space p_i) - f(r,\space p_i) - f(n - r,\space p_i)}{c_i}⌋$

Time and Space complexity: To avoid worst-case time complexity, use Sieve of Eratosthenes algorithm to store all primes up to $10^6,$ facilitating efficient prime factorization of $m.$

Let, $N = 10^6,$ then to store all prime up to $N$ using Sieve of Eratosthenes:

Time complexity, $O(N\cdot log(log(N)))$

Space complexity, $O(N)$

Let, $S$ be the number of primes which are less than or equal to $\sqrt m.$ Then in each test cases, prime factorization of $m$ using stored primes:

Time complexity, $O(T\cdot S)$

In the worst case, $S$ can be at most $78\, 498$ (Number of primes less than or equal to $10^6$). However, it’s still enough to pass within the given time limit.

So, the overall complexity of our solution:

Time complexity, $O(N\cdot log(log(N)) + T\cdot S)$

Space complexity, $O(N)$

All other complexities are small enough to be neglected.