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

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

**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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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 \(p_{13} - 1 = 40\). – Lee Gao · 3 years, 8 months ago

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– Cody Johnson · 3 years, 8 months ago

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

– Lee Gao · 3 years, 8 months ago

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 :)Log in to reply

– Cody Johnson · 3 years, 8 months ago

First complete solution, though.Log in to reply

Note that the expression is undefined when \(m = 1.\) – Michael Tang · 3 years, 8 months ago

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– Cody Johnson · 3 years, 8 months ago

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

– Zi Song Yeoh · 3 years, 8 months ago

Right!Log in to reply

– 敬全 钟 · 3 years, 8 months ago

That's obvious....Log in to reply

From where do you get such questions? – Nupur Prasad · 3 years, 8 months ago

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– Ahaan Rungta · 3 years, 8 months ago

Cody makes them.Log in to reply

The answer is 40. – Michael Lee · 3 years, 8 months ago

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– Cody Johnson · 3 years, 8 months ago

YES! Solution?Log in to reply

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

Suspicious....... – Zi Song Yeoh · 3 years, 8 months ago

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– Ahaan Rungta · 3 years, 8 months ago

I noticed that. -.-Log in to reply

999? – Ahaan Rungta · 3 years, 8 months ago

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– Ahaan Rungta · 3 years, 8 months ago

Proof: "The result will blow your mind. -- Cody Johnson"Log in to reply

Wow, really nice! :) – Ahaan Rungta · 3 years, 8 months ago

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darn at first I thought that "CMC" meant "Canada Mathematics Competition". – Ryan Soedjak · 3 years, 8 months ago

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