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CMC - Problem 5

Problem 5. (6 points) Find the value of

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

Note by Cody Johnson
3 years, 10 months ago

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Suppose \(k\) 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, 10 months ago

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

Cody Johnson - 3 years, 10 months ago

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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 :)

Lee Gao - 3 years, 10 months ago

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@Lee Gao First complete solution, though.

Cody Johnson - 3 years, 10 months ago

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

Michael Tang - 3 years, 10 months ago

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

Cody Johnson - 3 years, 10 months ago

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Right!

Zi Song Yeoh - 3 years, 10 months ago

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That's obvious....

敬全 钟 - 3 years, 10 months ago

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From where do you get such questions?

Nupur Prasad - 3 years, 10 months ago

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Cody makes them.

Ahaan Rungta - 3 years, 10 months ago

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The answer is 40.

Michael Lee - 3 years, 10 months ago

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YES! Solution?

Cody Johnson - 3 years, 10 months ago

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

Suspicious.......

Zi Song Yeoh - 3 years, 10 months ago

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I noticed that. -.-

Ahaan Rungta - 3 years, 10 months ago

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999?

Ahaan Rungta - 3 years, 10 months ago

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

Ahaan Rungta - 3 years, 10 months ago

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Wow, really nice! :)

Ahaan Rungta - 3 years, 10 months ago

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

Ryan Soedjak - 3 years, 10 months ago

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