×

# 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

Sort by:

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 $$p_{13} - 1 = 40$$.

- 3 years, 10 months ago

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

- 3 years, 10 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 :)

- 3 years, 10 months ago

First complete solution, though.

- 3 years, 10 months ago

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

- 3 years, 10 months ago

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

- 3 years, 10 months ago

Right!

- 3 years, 10 months ago

That's obvious....

- 3 years, 10 months ago

From where do you get such questions?

- 3 years, 10 months ago

Cody makes them.

- 3 years, 10 months ago

- 3 years, 10 months ago

YES! Solution?

- 3 years, 10 months ago

Note : $$8192 = 2^{13}$$

Suspicious.......

- 3 years, 10 months ago

I noticed that. -.-

- 3 years, 10 months ago

999?

- 3 years, 10 months ago

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

- 3 years, 10 months ago

Wow, really nice! :)

- 3 years, 10 months ago

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

- 3 years, 10 months ago