let k(k-1)= n ; where k is a whole number and n is smallest number divisible by 2024, find n

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.

Learn how to find the smallest value of n divisible by 2024 using the equation k(k-1) = n, where k is a whole number. This solution covers number theory concepts, including prime factorization and divisibility rules. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation