[hide=C stands for Cantor, G stands for Gauss]they had two problem sets under those two names[/hide]


[u]Set 4[/u]


[b]G.16[/b] A number $k$ is the product of exactly three distinct primes (in other words, it is of the form $pqr$, where $p, q, r$ are distinct primes). If the average of its factors is $66$, find $k$.


[b]G.17[/b] Find the number of lattice points contained on or within the graph of $\frac{x^2}{3} +\frac{y^2}{2}= 12$. Lattice points are coordinate points $(x, y)$ where $x$ and $y$ are integers.


[b]G.18 / C.23[/b]  How many triangles can be made from the vertices and center of a regular hexagon? Two congruent triangles with different orientations are considered distinct.


[b]G.19[/b] Cindy has a cone with height $15$ inches and diameter $16$ inches. She paints one-inch thick bands of paint in circles around the cone, alternating between red and blue bands, until the whole cone is covered with paint. If she starts from the bottom of the cone with a blue strip, what is the ratio of the area of the cone covered by red paint to the area of the cone covered by blue paint?


[b]G.20 / C.25[/b] An even positive integer $n$ has an odd factorization if the largest odd divisor of $n$ is also the smallest odd divisor of n greater than 1. Compute the number of even integers $n$ less than $50$ with an odd factorization.


[u] Set 5[/u]


[b]G.21[/b] In the magical tree of numbers, $n$ is directly connected to $2n$ and $2n + 1$ for all nonnegative integers n. A frog on the magical tree of numbers can move from a number $n$ to a number connected to it in $1$ hop. What is the least number of hops that the frog can take to move from $1000$ to $2018$?


[b]G.22[/b] Stan makes a deal with Jeff. Stan is given 1 dollar, and every day for $10$ days he must either double his money or burn a perfect square amount of money. At first Stan thinks he has made an easy $1024$ dollars, but then he learns the catch - after $10$ days, the amount of money he has must be a multiple of $11$ or he loses all his money. What is the largest amount of money Stan can have after the $10$ days are up?


[b]G.23[/b] Let $\Gamma_1$ be a circle with diameter $2$ and center $O_1$ and let $\Gamma_2$ be a congruent circle centered at a point $O_2 \in \Gamma_1$. Suppose $\Gamma_1$ and $\Gamma_2$ intersect at $A$ and $B$. Let $\Omega$ be a circle centered at $A$ passing through $B$. Let $P$ be the intersection of $\Omega$ and $\Gamma_1$ other than $B$ and let $Q$ be the intersection of $\Omega$ and ray $\overrightarrow{AO_1}$. Define $R$ to be the intersection of $PQ$ with $\Gamma_1$. Compute the length of $O_2R$.


[b]G.24[/b] $8$ people are at a party. Each person gives one present to one other person such that everybody gets a present and no two people exchange presents with each other. How many ways is this possible?


[b]G.25[/b] Let $S$ be the set of points $(x, y)$ such that $y = x^3 - 5x$ and $x = y^3 - 5y$. There exist four points in $S$ that are the vertices of a rectangle. Find the area of this rectangle.



PS. You should use hide for answers. C1-15/ G1-10 have been posted [url=https://artofproblemsolving.com/community/c3h2790674p24540132]here [/url] and C16-30/G10-15, G25-30 [url=https://artofproblemsolving.com/community/c3h2790676p24540145]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url]
1. Let \( k = pqr \) where \( p, q, r \) are distinct primes. The factors of \( k \) include \( 1, p, q, r, pq, pr, qr, \) and \( pqr \).
2. The average of these factors is given by:
   \[
   \frac{1 + p + q + r + pq + pr + qr + pqr}{8} = 66
   \]
3. Simplifying, we get:
   \[
   1 + p + q + r + pq + pr + qr + pqr = 528
   \]
4. Since \( k = pqr \), we can rewrite the equation as:
   \[
   1 + p + q + r + pq + pr + qr + k = 528
   \]
5. Substituting \( k = pqr \), we get:
   \[
   1 + p + q + r + pq + pr + qr + pqr = 528
   \]
6. We know that one of the primes must be 2 because if all primes were odd, the sum of the factors would be odd, which contradicts the even sum of 528. Let \( p = 2 \).
7. Substituting \( p = 2 \), we get:
   \[
   1 + 2 + q + r + 2q + 2r + qr + 2qr = 528
   \]
8. Simplifying, we get:
   \[
   3 + 3q + 3r + qr = 528
   \]
9. Subtracting 3 from both sides, we get:
   \[
   3q + 3r + qr = 525
   \]
10. Dividing the entire equation by 3, we get:
    \[
    q + r + \frac{qr}{3} = 175
    \]
11. Let \( q = 3 \) and solve for \( r \):
    \[
    3 + r + \frac{3r}{3} = 175
    \]
    \[
    3 + r + r = 175
    \]
    \[
    2r + 3 = 175
    \]
    \[
    2r = 172
    \]
    \[
    r = 86
    \]
12. Since 86 is not a prime number, we need to try another pair. Let \( q = 11 \):
    \[
    11 + r + \frac{11r}{3} = 175
    \]
    \[
    11 + r + \frac{11r}{3} = 175
    \]
    \[
    33 + 3r + 11r = 525
    \]
    \[
    14r = 492
    \]
    \[
    r = 43
    \]
13. Now, we have \( p = 2 \), \( q = 3 \), and \( r = 43 \). Therefore, \( k = 2 \cdot 3 \cdot 43 = 258 \).

The final answer is \( \boxed{258} \).