Answer:
Let’s solve this problem step by step:
We know that when 2024 is divided by n, the remainder is 4.
Let p be the quotient when 2024 is divided by n. Therefore, we have: [np + 4 = 2024] Solving for np: [np = 2020]
Now let’s find the prime factorization of 2020: [2020 = 2^2 \cdot 5 \cdot 101]
Using these prime factors, we can determine all the possible pairs of positive integers that multiply to 2020:
(1 \cdot 2020)
(2 \cdot 1010)
(4 \cdot 505)
(5 \cdot 404)
(10 \cdot 202)
(20 \cdot 101)
Since n is a three-digit positive integer, we consider only the following values for n: 101, 202, 404, and 505.
The sum of these three-digit positive integers is: [101 + 202 + 404 + 505 = 1212]
Therefore, the sum of all such three-digit positive integers n is 1212
Step-by-step explanation:
