**Problem 5. (6 points)** Find the value of

\[\sum_{m=1}^{8192}\left\lfloor\left\lfloor\frac{13}{1+\sum_{j=2}^m\left\lfloor\frac{(j-1)!+1}{j}-\left\lfloor\frac{(j-1)!}{j}\right\rfloor\right\rfloor}\right\rfloor^{1/13}\right\rfloor\]

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## Comments

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TopNewestSuppose \(k\) is prime, then it's quite easy to argue that \(k \mid k!\) but \(k \not \mid n!\) for any \(n < k\), so \((k-1)! \equiv -1 \mod k\). In fact, it should be obvious that the other direction holds as well: \[ (k-1)! \equiv -1 \mod k \iff k \text{ is prime}\] Now, this suggests that only when \(k\) is prime will \(\frac{(k-1)!+1}{k}\) be an integer (since by definition \(k\) divides \((k-1)! + 1\)). Furthermore, since \(0 < \frac{(k-1)!+1}{k} - \left\lfloor \frac{(j-1)!}{j} \right\rfloor \le 1\), then this will only be one iff \(k\) is prime and zero otherwise: an indicator for primeness. In otherwords the summation on the denominator counts the number of primes up to \(m\), aka the prime \(\pi\) function.

From here it's easy sailing. Since the maximum of the expression \(\frac{13}{1+\pi(m)}\) is 13, and because \(1 < \sqrt[13]{13} < 2\), each term will contribute exactly one to the sum until \(\frac{13}{1+\pi(m)} < 1\). This will happen exactly when we hit the \(13^{th}\) prime, which happens to be \(41\), so the sum must have a value of \(p_{13} - 1 = 40\).

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Correct! As your solution was posted before Michael Lee's, you get the 6 points. Great solution.

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Aww thank you so much, but I don't feel like it's fair to award the point to me since Michael solved the problem over an hour before I did, so I insist that the points should go to him! Thanks for the fun problems :)

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Note that the expression is undefined when \(m = 1.\)

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By standard math protocol, \(\sum_{i=n+1}^ni=0\).

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Right!

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That's obvious....

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From where do you get such questions?

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Cody makes them.

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The answer is 40.

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YES! Solution?

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Note : \(8192 = 2^{13}\)

Suspicious.......

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I noticed that. -.-

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999?

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Proof: "The result will blow your mind. -- Cody Johnson"

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Wow, really nice! :)

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darn at first I thought that "CMC" meant "Canada Mathematics Competition".

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