1. Define the problem and variables:
- Let n be the number of scientists.
- Let ti,j represent the time for which exactly one of the scientists i or j is in the cafeteria.
- Let T=∑ti,j be the total time for all pairs of scientists.
- We know that T≥2n(n−1)⋅x because for each pair of scientists, the time ti,j≥x.
2. Break down the 8-hour workday into intervals:
- Consider the workday divided into intervals where the set of scientists in the cafeteria remains constant.
- For each interval with k scientists and duration w, the interval contributes w⋅k⋅(n−k) to the sum T.
3. **Maximize the sum T:**
- To maximize T, we need to maximize k(n−k).
- The function k(n−k) is maximized when k=⌊2n⌋.
4. **Calculate the maximum T:**
- The total time is 8 hours, so the maximum T is 8⋅⌊2n⌋⋅(n−⌊2n⌋).
5. **Express min(ti,j):**
- We want to make all ti,j equal to maximize the minimum.
- Therefore, min(ti,j)=n(n−1)16⌊2n⌋(n−⌊2n⌋).
6. **Relate min(ti,j) to x:**
- Given min(ti,j)≥x, we have:
n(n−1)16⌊2n⌋(n−⌊2n⌋)≥x
7. **Solve for n in terms of x:**
- Simplify the inequality:
16⌊2n⌋(n−⌊2n⌋)≥x⋅n(n−1)
- Let k=⌊2n⌋, then:
16k(n−k)≥x⋅n(n−1)
- For n=2k:
16k2≥x⋅2k(2k−1)
16k2≥2xk(2k−1)
8k≥x(2k−1)
8k≥2kx−x
8k+x≥2kx
k≤2x−28+x
- For n=2k+1:
16k(k+1)≥x(2k+1)(2k)
16k2+16k≥2x(2k2+k)
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