Let's solve the problems one by one:

Problem a

Maria needs to walk from point P to point Q using the shortest path. Each time she can only move to adjacent squares (horizontally or vertically), but there are walls she cannot cross (the black lines). How many different paths are there?

We need to count the number of valid paths from P to Q avoiding the walls.

• Use dynamic programming or recursive backtracking to count paths.
• Start at P and explore all possible paths to Q, ensuring to avoid walls.

I'll use a dynamic programming approach to solve this problem:

1. Initialize a 5x5 grid to keep track of possible paths to each cell.
2. Set up base cases where you can only move right or down initially.
3. Propagate the path count considering the walls.

Let’s calculate this step-by-step.

Problem b

Suppose a positive integer has 48 positive factors. At most how many prime factors does this number have?

If $n$ has 48 factors, the total number of factors is given by $(e_1+1)(e_2+1)...(e_k+1) = 48$, where $e_i$ are the exponents of the prime factorization of $n$.

To maximize the number of prime factors, we should use the smallest exponents possible. Thus:

48 = (3+1)(3+1) = 4 \times 4 = 16 \\ 48 = (5+1)(2+1) = 6 \times 3 = 18$$ Using 1, 1, 1, and 1: $$(1+1)^4 = 16 = 2^4$$ Thus, at most, the integer can have 5 distinct prime factors. ### Problem c **Find the least integer $$n > 1$$ such that the product of all positive divisors of $$n$$ equals $$n^4$$.** If $$n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$$, the number of divisors of $$n$$ is $$(e_1+1)(e_2+1) \cdots (e_k+1)$$. The product of the divisors of $$n$$ is $$n^{T/2}$$, where $$T$$ is the total number of divisors of $$n$$. Given $$n^{T/2} = n^4$$, it follows that $$T/2 = 4$$, so $$T = 8$$. Hence, we need the smallest $$n$$ such that $$n$$ has 8 divisors. The smallest such number is $$2^7 = 128$$. ### Problem d **Suppose $$a, b, c,$$ and $$d$$ are four positive prime numbers (not necessarily distinct) such that the product of these four prime numbers is equal to the sum of 55 consecutive positive integers. Find the smallest possible value of $$a + b + c + d$$.** The sum of 55 consecutive integers is: $$\text{Sum} = 55 \cdot 28 = 1540$$ Now, we need to find the product of four primes equaling 1540. $$1540 = 2^2 \cdot 5 \cdot 7 \cdot 11$$ So, the smallest value of $$a + b + c + d = 2 + 2 + 5 + 7 + 11 = 27$$. ### Problem e **There is a 4-digit prime number $$p$$ such that $$p$$ is a factor of $$99^3 - 1$$. Find the value of $$p$$.** Factor $$99^3 - 1$$ using difference of cubes: $$99^3 - 1 = (99 - 1)(99^2 + 99 \cdot 1 + 1^2) = 98 \cdot 10001$$ 10001 factors: $$10001 = 73 \cdot 137$$ The 4-digit prime factor is $$137$$. If you have any questions or need further details on any solution, please let me know! Here are five additional questions you could ask: 1. How does the dynamic programming approach work for counting paths? 2. Can you explain more about prime factorization in problem b? 3. What are the properties of divisors used in problem c? 4. How do you determine the sum of consecutive integers for problem d? 5. Why does 10001 factor into primes 73 and 137? ### Tip When solving combinatorial problems, breaking down the problem into smaller subproblems and using dynamic programming can significantly simplify the process and ensure accuracy.