Let's tackle each problem one by one.

Problem a: Shortest Path from P to Q

Maria needs to travel from P to Q following the shortest path while avoiding the black walls. Each step can only move to adjacent cells.

To find the number of paths:

1. Maria can move down and right to reach Q.
2. Maria cannot move through black walls.

Let's enumerate the possible paths:

• Start at P (0,0).
• Possible moves: Down (D) and Right (R).
• Maria needs to make 3 R and 3 D moves.
• However, the walls will restrict certain paths.

The total number of paths without any restriction is given by the binomial coefficient: $\binom{6}{3} = 20$

Now, let's consider paths that hit the walls:

• Paths that hit the first wall (at position (2, 1)) and get blocked.
• Paths that hit the second wall (at position (1, 3)) and get blocked.

Considering the walls, let's list the possible valid paths manually or through a combination approach.

Problem b: Positive Integer with 48 Positive Factors

Suppose a positive integer $n$ has 48 positive factors.

The number of factors of $n$ is given by: $(n_1+1)(n_2+1)\ldots(n_k+1) = 48$ We need the maximum number of prime factors.

By examining factorization forms: $48 = 2^4 \times 3$ This gives us multiple forms to maximize the number of prime factors:

• $n = p_1^{2} \times p_2^{2} \times p_3^{1} \Rightarrow (2+1)(2+1)(1+1) = 3 \times 3 \times 2 = 18$
• Therefore, the maximum number of prime factors is 3.

Problem c: Least Integer $n$ such that the product of all positive divisors of $n$ equals $n^4$

The product of the divisors of $n$ is: $n^{d(n)/2}$ Where $d(n)$ is the number of divisors of $n$.

Given $n^{d(n)/2} = n^4$, we get: $d(n)/2 = 4 \Rightarrow d(n) = 8$ The smallest $n$ with 8 divisors: $2^7 = 128$

Problem d: Sum and Product of Four Prime Numbers

Let $a, b, c, d$ be four prime numbers such that: $abc \cdot d = \sum_{k=1}^{55} k = \frac{55 \times 56}{2} = 1540$

Examining the smallest prime numbers: $2 \cdot 2 \cdot 5 \cdot 7 = 140$ Smallest sum: $2 + 2 + 5 + 7 = 16$

Problem e: Prime Number $p$ such that $p$ is a factor of $99^3 - 1$

The expression can be factored using the difference of cubes: $99^3 - 1 = (99-1)(99^2 + 99 \cdot 1 + 1^2) = 98 \cdot (99^2 + 99 + 1)$ Find the prime factor of 98 which is 4-digit: $(99-1) = 98 \implies 98 = 2 \times 7^2$

There are no 4-digit factors in the direct approach.

Would you like details or have any questions on specific parts? Here are five numbered questions for you to ask:

1. Can you explain the detailed solution to part (a) with path counting?
2. How do we manually count paths avoiding the black walls?
3. Can you show the factorization steps for part (b) in more detail?
4. Explain the divisor function in part (c) in more detail.
5. What is the detailed process for finding $p$ in part (e)?

Tip: Always double-check your calculations for accuracy, especially when dealing with combinatorics and prime factorization.