We are given a multiple-choice question about a geometric problem. The question is: "What is the minimum number of sides of a polygon such that the sum of the interior angles is 2024?" The options are A) 128, B) 128, C) 129, D) 130, E) 131. Note that options A and B are both 128, which might be a typo, but we will proceed with the given options.

We know the formula for the sum of the interior angles of a polygon with n sides is:

$S=(n-2)\times 180^{\circ }$

We are given that the sum is 2024 degrees. So we set up the equation:

$(n-2)\times 180=2024$

Solving for n:

$n-2=\frac{2024}{180}$

Calculate the division:

$\frac{2024}{180}=\frac{1012}{90}=\frac{506}{45}\approx 11.2444$

So:

$n-2\approx 11.2444$

$n\approx 13.2444$

But this is not an integer. However, the number of sides must be an integer. Moreover, we are looking for the minimum number of sides such that the sum is at least 2024.

We can rearrange the equation to:

$n=\frac{2024}{180}+2$

But since n must be an integer, we need to find the smallest integer n such that:

$(n-2)\times 180\ge 2024$

Let me compute the required n:

We have:

$(n-2)\times 180\ge 2024$

$n-2\ge \frac{2024}{180}$

$n\ge \frac{2024}{180}+2$

Calculate:

$\frac{2024}{180}=\frac{2024÷4}{180÷4}=\frac{506}{45}\approx 11.2444$

So:

$n\ge 11.2444+2=13.2444$

Therefore, the smallest integer n is 14.

Wait, but let me double-check with the formula. For a polygon with n sides, the sum of the interior angles is (n-2)*180. We want this to be at least 2024.

Let me compute for n=13:

$(13-2)\times 180=11\times 180=1980$

1980 < 2024, so not enough.

For n=14:

$(14-2)\times 180=12\times 180=2160$

2160 >= 2024, so it satisfies.

Therefore, the minimum number of sides is 14.

But wait, the options provided are 128, 128, 129, 130, 131. This suggests that I might have made a mistake. Let me reexamine the problem.

The problem says: "the sum of the interior angles is 2024". But note that the sum of interior angles for a polygon is always a multiple