Problem 5. (6 points) Find the value of
∑m=18192⌊⌊131+∑j=2m⌊(j−1)!+1j−⌊(j−1)!j⌋⌋⌋1/13⌋\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\rfloorm=1∑8192⎣⎢⎢⎢⎢⎣⎢⎢⎢1+∑j=2m⌊j(j−1)!+1−⌊j(j−1)!⌋⌋13⎦⎥⎥⎥1/13⎦⎥⎥⎥⎥
Note by Cody Johnson 7 years, 6 months ago
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Suppose kkk is prime, then it's quite easy to argue that k∣k!k \mid k!k∣k! but k∤n!k \not \mid n!k∣n! for any n<kn < kn<k, so (k−1)!≡−1mod k(k-1)! \equiv -1 \mod k(k−1)!≡−1modk. In fact, it should be obvious that the other direction holds as well: (k−1)!≡−1mod k ⟺ k is prime (k-1)! \equiv -1 \mod k \iff k \text{ is prime}(k−1)!≡−1modk⟺k is prime Now, this suggests that only when kkk is prime will (k−1)!+1k\frac{(k-1)!+1}{k}k(k−1)!+1 be an integer (since by definition kkk divides (k−1)!+1(k-1)! + 1(k−1)!+1). Furthermore, since 0<(k−1)!+1k−⌊(j−1)!j⌋≤10 < \frac{(k-1)!+1}{k} - \left\lfloor \frac{(j-1)!}{j} \right\rfloor \le 10<k(k−1)!+1−⌊j(j−1)!⌋≤1, then this will only be one iff kkk is prime and zero otherwise: an indicator for primeness. In otherwords the summation on the denominator counts the number of primes up to mmm, aka the prime π\piπ function.
From here it's easy sailing. Since the maximum of the expression 131+π(m)\frac{13}{1+\pi(m)}1+π(m)13 is 13, and because 1<1313<21 < \sqrt[13]{13} < 21<1313<2, each term will contribute exactly one to the sum until 131+π(m)<1\frac{13}{1+\pi(m)} < 11+π(m)13<1. This will happen exactly when we hit the 13th13^{th}13th prime, which happens to be 414141, so the sum must have a value of p13−1=40p_{13} - 1 = 40p13−1=40.
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Correct! As your solution was posted before Michael Lee's, you get the 6 points. Great solution.
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 – First complete solution, though.
Note that the expression is undefined when m=1.m = 1.m=1.
By standard math protocol, ∑i=n+1ni=0\sum_{i=n+1}^ni=0∑i=n+1ni=0.
Right!
That's obvious....
Note : 8192=2138192 = 2^{13}8192=213
Suspicious.......
I noticed that. -.-
The answer is 40.
YES! Solution?
From where do you get such questions?
Cody makes them.
Wow, really nice! :)
999?
Proof: "The result will blow your mind. -- Cody Johnson"
darn at first I thought that "CMC" meant "Canada Mathematics Competition".
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Suppose k is prime, then it's quite easy to argue that k∣k! but k∣n! for any n<k, so (k−1)!≡−1modk. In fact, it should be obvious that the other direction holds as well: (k−1)!≡−1modk⟺k is prime Now, this suggests that only when k is prime will k(k−1)!+1 be an integer (since by definition k divides (k−1)!+1). Furthermore, since 0<k(k−1)!+1−⌊j(j−1)!⌋≤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 π function.
From here it's easy sailing. Since the maximum of the expression 1+π(m)13 is 13, and because 1<1313<2, each term will contribute exactly one to the sum until 1+π(m)13<1. This will happen exactly when we hit the 13th prime, which happens to be 41, so the sum must have a value of p13−1=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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First complete solution, though.
Note that the expression is undefined when m=1.
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By standard math protocol, ∑i=n+1ni=0.
Right!
That's obvious....
Note : 8192=213
Suspicious.......
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I noticed that. -.-
The answer is 40.
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YES! Solution?
From where do you get such questions?
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Cody makes them.
Wow, really nice! :)
999?
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Proof: "The result will blow your mind. -- Cody Johnson"
darn at first I thought that "CMC" meant "Canada Mathematics Competition".