Problems need HardWork 3

We note that $4049-24 = 4025 = 5^\color{#D61F06}{2}\times 7^\color{#D61F06}{1} \times 23^\color{#D61F06}{1}$ . Therefore, the total number of factors of $4025$ is $(\color{#D61F06}{2}+1) (\color{#D61F06}{1}+1) (\color{#D61F06}{1}+1) = 12$ and there are $4$ factors $1$ , $5$ , $7$ and $23$ which are less than $24$ . Therefore, the number of positive integers that leave a remainder of 24 when divided into 4049 is $12-4 = \boxed{8}$ .

4049 - 24 = 4025

so prime factors of 4025 { 5 , 5, 7, 23 }, all less than 24. (other factors: 5 5, 5 7, 5 23, 7 23, 5 5 7, 5 5 23, 5 7 23, 5 5 7*23}

Factors of 4025 that are less than 23 ( 4! / 2! ) - 4 12 - 4 8

So, the answer is 8.