We are given the equation:

$k(k-1) = n$

where $k$ is a whole number, and $n$ is the smallest number divisible by 2024.

Step 1: Find the prime factorization of 2024

We first factor 2024:

$2024 = 2^3 \times 11 \times 23$

So, we are looking for $n$ such that $n = k(k-1)$ and $n$ is divisible by $2^3 \times 11 \times 23$.

Step 2: Explore the equation $k(k-1) = n$

We are tasked with finding $k$ such that $k(k-1)$ contains all the factors of 2024, i.e., $k(k-1)$ must be divisible by $2^3 \times 11 \times 23$.

Step 3: Find the smallest $k$ satisfying the divisibility condition

The product $k(k-1)$ consists of two consecutive numbers. For divisibility by 2024, one of the numbers must be divisible by $23$, another by $11$, and at least one by $2^3 = 8$.

Let’s find the smallest $k$ that meets these criteria.

• $k$ or $k-1$ must be divisible by 23.
• $k$ or $k-1$ must be divisible by 11.
• $k$ or $k-1$ must be divisible by $2^3 = 8$.

We check multiples of 23, as one of the numbers must be divisible by 23:

Case 1: $k = 23$

For $k = 23$, we have: $k(k-1) = 23 \times 22 = 506$ 506 is not divisible by 2024.

Case 2: $k = 24$

For $k = 24$, we have: $k(k-1) = 24 \times 23 = 552$ 552 is not divisible by 2024.

Case 3: $k = 46$

For $k = 46$, we have: $k(k-1) = 46 \times 45 = 2070$ 2070 is still not divisible by 2024.

Case 4: $k = 88$

For $k = 88$, we have: $k(k-1) = 88 \times 87 = 7656$ 7656 is divisible by 2024! Thus, $n = 7656$.

Final Answer:

The smallest $n$ divisible by 2024 is $\boxed{7656}$.

Would you like further details or have any other questions?

Here are five related questions:

1. What is the general method for solving quadratic Diophantine equations?
2. How do you find the prime factorization of a number?
3. Can $k(k-1)$ always be divisible by any number, or are there exceptions?
4. How would the solution change if 2024 were replaced by a different number?
5. Why is it necessary for both factors $k$ and $k-1$ to satisfy divisibility conditions?

Tip: When dealing with product-based equations involving consecutive numbers, consider the divisibility of both terms carefully to simplify calculations.