To solve this, we apply the Pigeonhole Principle, which states that if $n$ items are distributed among $k$ containers and $n > k$, then at least one container must hold more than one item.

Step 1: Define the problem

• There are 50 states ($k = 50$).
• We need to ensure at least 100 students come from the same state.

Step 2: Distribute students evenly

• If students are distributed as evenly as possible among the 50 states, each state will have at most 99 students before exceeding 99.

Thus:

• If each state has 99 students, the total number of students is: $50 \times 99 = 4950$ At this point, no state has reached 100 students.

Step 3: Add one more student

To guarantee that at least one state has 100 students, we add one more student to the total: $4950 + 1 = 4951$

Final Answer:

The minimum number of students required to guarantee that at least 100 come from the same state is: $\boxed{4951}$

Would you like a detailed explanation of the Pigeonhole Principle or similar examples?

Related Questions:

1. What is the Pigeonhole Principle and how does it apply in probability problems?
2. How can this principle be generalized to other real-world scenarios?
3. What if the threshold was 200 students per state instead of 100? How would the solution change?
4. What are examples of exceptions where the Pigeonhole Principle does not apply?
5. How can you mathematically model distributions where additional constraints exist?

Tip:

The Pigeonhole Principle is a powerful tool for proving existence results in mathematics; practice applying it to diverse scenarios for mastery!