CMC - Problem 5

Problem 5. (6 points) Find the value of

m=18192131+j=2m(j1)!+1j(j1)!j1/13\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
5 years, 11 months ago

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13 votes

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Suppose kk is prime, then it's quite easy to argue that kk!k \mid k! but k∤n!k \not \mid n! for any n<kn < k, so (k1)!1modk(k-1)! \equiv -1 \mod k. In fact, it should be obvious that the other direction holds as well: (k1)!1modk    k is prime (k-1)! \equiv -1 \mod k \iff k \text{ is prime} Now, this suggests that only when kk is prime will (k1)!+1k\frac{(k-1)!+1}{k} be an integer (since by definition kk divides (k1)!+1(k-1)! + 1). Furthermore, since 0<(k1)!+1k(j1)!j10 < \frac{(k-1)!+1}{k} - \left\lfloor \frac{(j-1)!}{j} \right\rfloor \le 1, then this will only be one iff kk is prime and zero otherwise: an indicator for primeness. In otherwords the summation on the denominator counts the number of primes up to mm, aka the prime π\pi function.

From here it's easy sailing. Since the maximum of the expression 131+π(m)\frac{13}{1+\pi(m)} is 13, and because 1<1313<21 < \sqrt[13]{13} < 2, each term will contribute exactly one to the sum until 131+π(m)<1\frac{13}{1+\pi(m)} < 1. This will happen exactly when we hit the 13th13^{th} prime, which happens to be 4141, so the sum must have a value of p131=40p_{13} - 1 = 40.

Lee Gao - 5 years, 11 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 - 5 years, 11 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 - 5 years, 11 months ago

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@Lee Gao First complete solution, though.

Cody Johnson - 5 years, 11 months ago

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Note that the expression is undefined when m=1.m = 1.

Michael Tang - 5 years, 11 months ago

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

Cody Johnson - 5 years, 11 months ago

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Right!

Zi Song Yeoh - 5 years, 11 months ago

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That's obvious....

敬全 钟 - 5 years, 11 months ago

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Note : 8192=2138192 = 2^{13}

Suspicious.......

Zi Song Yeoh - 5 years, 11 months ago

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I noticed that. -.-

Ahaan Rungta - 5 years, 11 months ago

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The answer is 40.

Michael Lee - 5 years, 11 months ago

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YES! Solution?

Cody Johnson - 5 years, 11 months ago

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From where do you get such questions?

Nupur Prasad - 5 years, 11 months ago

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Cody makes them.

Ahaan Rungta - 5 years, 11 months ago

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Wow, really nice! :)

Ahaan Rungta - 5 years, 11 months ago

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999?

Ahaan Rungta - 5 years, 11 months ago

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

Ahaan Rungta - 5 years, 11 months ago

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

Ryan Soedjak - 5 years, 11 months ago

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