[u]Set 1[/u]


[b]1.1[/b] How many positive integer divisors are there of $2^2 \cdot 3^3 \cdot 5^4$?


[b]1.2[/b] Let $T$ be the answer from the previous problem. For how many integers $n$ between $1$ and $T$ (inclusive) is $\frac{(n)(n - 1)(n - 2)}{12}$ an integer?


[b]1.3[/b] Let $T$ be the answer from the previous problem. Find $\frac{lcm(T, 36)}{gcd(T, 36)}$.


PS. You should use hide for answers. 
1. **Finding the number of positive integer divisors of \(2^2 \cdot 3^3 \cdot 5^4\)**

   To find the number of positive integer divisors of a number, we use the formula for the number of divisors of a number given its prime factorization. If a number \(N\) has the prime factorization \(N = p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_k^{e_k}\), then the number of positive divisors of \(N\) is given by:
   \[
   (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   For \(N = 2^2 \cdot 3^3 \cdot 5^4\):
   \[
   e_1 = 2, \quad e_2 = 3, \quad e_3 = 4
   \]
   Therefore, the number of positive divisors is:
   \[
   (2 + 1)(3 + 1)(4 + 1) = 3 \cdot 4 \cdot 5 = 60
   \]

2. **Finding the number of integers \(n\) between 1 and \(T\) (inclusive) such that \(\frac{(n)(n - 1)(n - 2)}{12}\) is an integer**

   We need to determine when \(\frac{(n)(n - 1)(n - 2)}{12}\) is an integer. This expression will be an integer if and only if the product \((n)(n - 1)(n - 2)\) is divisible by 12. 

   Since 12 factors as \(2^2 \cdot 3\), we need the product \((n)(n - 1)(n - 2)\) to be divisible by \(2^2 \cdot 3\). 

   - For divisibility by \(2^2\): Among any three consecutive integers, at least one is divisible by 2, and at least one is divisible by 4.
   - For divisibility by 3: Among any three consecutive integers, at least one is divisible by 3.

   Therefore, \((n)(n - 1)(n - 2)\) is always divisible by 12 for any integer \(n\). Since \(T = 60\), the number of integers \(n\) between 1 and 60 (inclusive) is 60.

3. **Finding \(\frac{\text{lcm}(T, 36)}{\text{gcd}(T, 36)}\)**

   Given \(T = 60\) and 36, we need to find the least common multiple (LCM) and greatest common divisor (GCD) of 60 and 36.

   - Prime factorization of 60: \(60 = 2^2 \cdot 3 \cdot 5\)
   - Prime factorization of 36: \(36 = 2^2 \cdot 3^2\)

   The GCD is found by taking the lowest power of all common prime factors:
   \[
   \text{gcd}(60, 36) = 2^2 \cdot 3 = 12
   \]

   The LCM is found by taking the highest power of all prime factors present in either number:
   \[
   \text{lcm}(60, 36) = 2^2 \cdot 3^2 \cdot 5 = 180
   \]

   Therefore, we have:
   \[
   \frac{\text{lcm}(60, 36)}{\text{gcd}(60, 36)} = \frac{180}{12} = 15
   \]

The final answer is \(\boxed{15}\)