I need to find the smallest integer k>0k>0 such that the following inequality holds:

(1−1365)k≤12(1-\frac{1}{365})^k\le\frac{1}{2}

The answer is supposedly greater than 200200. How can I find kk?

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1

Can you please explain which variant of the “common birthday” problem this inequality comes from?

– z100

2 days ago

It comes from the complement of the “common birthday” problem.

– revolution9540

2 days ago

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1 Answer

1

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Hint: klog(1−1365)≤log12k\log(1-\frac{1}{365}) \leq \log\frac12

1

So 252.6519888. Thanks!

– revolution9540

2 days ago

remember to find the smallest integer.

– Siong Thye Goh

2 days ago

It appears to be 252.

– revolution9540

2 days ago

You are very close but that is still not the right answer.

– Siong Thye Goh

2 days ago

Note that log(1−1365)\log (1-\frac{1}{365}) is negative.

– Siong Thye Goh

2 days ago