[i]20 problems for 25 minutes.[/i]


[b]p1.[/b] Matt has a twenty dollar bill and buys two items worth $\$7:99$ each. How much change does he receive, in dollars?


[b]p2.[/b] The sum of two distinct numbers is equal to the positive difference of the two numbers. What is the product of the two numbers?


[b]p3.[/b] Evaluate $$\frac{1 + 2 + 3 + 4 + 5 + 6 + 7}{8 + 9 + 10 + 11 + 12 + 13 + 14}.$$


[b]p4.[/b] A sphere with radius $r$ has volume $2\pi$. Find the volume of a sphere with diameter $r$.


[b]p5.[/b] Yannick ran $100$ meters in $14.22$ seconds. Compute his average speed in meters per second, rounded to the nearest integer.


[b]p6.[/b] The mean of the numbers $2, 0, 1, 5,$ and $x$ is an integer. Find the smallest possible positive integer value for $x$.


[b]p7.[/b] Let $f(x) =\sqrt{2^2 - x^2}$. Find the value of $f(f(f(f(f(-1)))))$.


[b]p8.[/b] Find the smallest positive integer $n$ such that $20$ divides $15n$ and $15$ divides $20n$.


[b]p9.[/b] A circle is inscribed in equilateral triangle $ABC$. Let $M$ be the point where the circle touches side $AB$ and let $N$ be the second intersection of segment $CM$ and the circle. Compute the ratio $\frac{MN}{CN}$ .


[b]p10.[/b] Four boys and four girls line up in a random order. What is the probability that both the first and last person in line is a girl?


[b]p11.[/b] Let $k$ be a positive integer. After making $k$ consecutive shots successfully, Andy's overall shooting accuracy increased from $65\%$ to $70\%$. Determine the minimum possible value of $k$.


[b]p12.[/b] In square $ABCD$, $M$ is the midpoint of side $CD$. Points $N$ and $P$ are on segments $BC$ and $AB$ respectively such that $ \angle AMN = \angle MNP = 90^o$. Compute the ratio $\frac{AP}{PB}$ .


[b]p13.[/b] Meena writes the numbers $1, 2, 3$, and $4$ in some order on a blackboard, such that she cannot swap two numbers and obtain the sequence $1$, $2$, $3$, $4$. How many sequences could she have written?


[b]p14.[/b] Find the smallest positive integer $N$ such that $2N$ is a perfect square and $3N$ is a perfect cube.


[b]p15.[/b] A polyhedron has $60$ vertices, $150$ edges, and $92$ faces. If all of the faces are either regular pentagons or equilateral triangles, how many of the $92$ faces are pentagons?


[b]p16.[/b] All positive integers relatively prime to $2015$ are written in increasing order. Let the twentieth number be $p$. The value of $\frac{2015}{p}-1$ can be expressed as $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Compute $a + b$.


[b]p17.[/b] Five red lines and three blue lines are drawn on a plane. Given that $x$ pairs of lines of the same color intersect and $y$ pairs of lines of different colors intersect, find the maximum possible value of $y - x$.


[b]p18.[/b] In triangle $ABC$, where $AC > AB$, $M$ is the midpoint of $BC$ and $D$ is on segment $AC$ such that $DM$ is perpendicular to $BC$. Given that the areas of $MAD$ and $MBD$ are $5$ and $6$, respectively, compute the area of triangle $ABC$.


[b]p19.[/b] For how many ordered pairs $(x, y)$ of integers satisfying $0 \le x, y \le 10$ is $(x + y)^2 + (xy - 1)^2$ a prime number?


[b]p20.[/b] A solitaire game is played with $8$ red, $9$ green, and $10$ blue cards. Totoro plays each of the cards exactly once in some order, one at a time. When he plays a card of color $c$, he gains a number of points equal to the number of cards that are not of color $c$ in his hand. Find the maximum number of points that he can obtain by the end of the game.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the problem, we need to determine the number of ordered pairs \((x, y)\) of integers satisfying \(0 \le x, y \le 10\) such that \((x + y)^2 + (xy - 1)^2\) is a prime number.

1. Start by expanding the given expression:
   \[
   (x + y)^2 + (xy - 1)^2 = x^2 + 2xy + y^2 + x^2y^2 - 2xy + 1 = x^2 + y^2 + x^2y^2 + 1
   \]
   Simplify the expression:
   \[
   (x + y)^2 + (xy - 1)^2 = (x^2 + 1)(y^2 + 1)
   \]

2. For the product \((x^2 + 1)(y^2 + 1)\) to be a prime number, one of the factors must be 1 because a prime number has exactly two distinct positive divisors: 1 and itself.

3. Therefore, we need either \(x^2 + 1 = 1\) or \(y^2 + 1 = 1\):
   - If \(x^2 + 1 = 1\), then \(x^2 = 0\) which implies \(x = 0\).
   - If \(y^2 + 1 = 1\), then \(y^2 = 0\) which implies \(y = 0\).

4. Consider the case \(x = 0\):
   - Then \(y^2 + 1\) must be a prime number.
   - The possible values of \(y\) within the range \(0 \le y \le 10\) are those for which \(y^2 + 1\) is prime:
     \[
     y = 1 \implies y^2 + 1 = 2 \quad (\text{prime})
     \]
     \[
     y = 2 \implies y^2 + 1 = 5 \quad (\text{prime})
     \]
     \[
     y = 3 \implies y^2 + 1 = 10 \quad (\text{not prime})
     \]
     \[
     y = 4 \implies y^2 + 1 = 17 \quad (\text{prime})
     \]
     \[
     y = 5 \implies y^2 + 1 = 26 \quad (\text{not prime})
     \]
     \[
     y = 6 \implies y^2 + 1 = 37 \quad (\text{prime})
     \]
     \[
     y = 7 \implies y^2 + 1 = 50 \quad (\text{not prime})
     \]
     \[
     y = 8 \implies y^2 + 1 = 65 \quad (\text{not prime})
     \]
     \[
     y = 9 \implies y^2 + 1 = 82 \quad (\text{not prime})
     \]
     \[
     y = 10 \implies y^2 + 1 = 101 \quad (\text{prime})
     \]
   - Thus, the valid values of \(y\) are \(1, 2, 4, 6, 10\), giving 5 ordered pairs \((0, 1), (0, 2), (0, 4), (0, 6), (0, 10)\).

5. Consider the case \(y = 0\):
   - Then \(x^2 + 1\) must be a prime number.
   - The possible values of \(x\) within the range \(0 \le x \le 10\) are those for which \(x^2 + 1\) is prime:
     \[
     x = 1 \implies x^2 + 1 = 2 \quad (\text{prime})
     \]
     \[
     x = 2 \implies x^2 + 1 = 5 \quad (\text{prime})
     \]
     \[
     x = 3 \implies x^2 + 1 = 10 \quad (\text{not prime})
     \]
     \[
     x = 4 \implies x^2 + 1 = 17 \quad (\text{prime})
     \]
     \[
     x = 5 \implies x^2 + 1 = 26 \quad (\text{not prime})
     \]
     \[
     x = 6 \implies x^2 + 1 = 37 \quad (\text{prime})
     \]
     \[
     x = 7 \implies x^2 + 1 = 50 \quad (\text{not prime})
     \]
     \[
     x = 8 \implies x^2 + 1 = 65 \quad (\text{not prime})
     \]
     \[
     x = 9 \implies x^2 + 1 = 82 \quad (\text{not prime})
     \]
     \[
     x = 10 \implies x^2 + 1 = 101 \quad (\text{prime})
     \]
   - Thus, the valid values of \(x\) are \(1, 2, 4, 6, 10\), giving 5 ordered pairs \((1, 0), (2, 0), (4, 0), (6, 0), (10, 0)\).

6. Combining both cases, we have a total of \(5 + 5 = 10\) ordered pairs.

The final answer is \(\boxed{10}\).