Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$
[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6;  path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$

Solution: A decagon can be formed from the trapezoids and the base. The sum of the decagon's angles is $180(10-2)=1440^\circ$ . Letting the larger angle in each trapezoid be $x$ , the two angles formed by the line each measures $(180-x)^\circ$ . There are $8$ congruent angles left. Each of those angles measures $(360-2x)^\circ$ .  Putting it all together: $8(360-2x)+2(180-x)=1440\implies x=\boxed{100}$

Answer: A 100
Problem: The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$
[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6;  path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$

Solution: If we reflect the arch across the line, we form an 18-gon. $\frac{180*(18-2)}{18} = 160^\circ$ so each interior angle of the 18-gon is $160^\circ$ . Let $x$ be the degree measure of the larger interior angle of a trapezoid. From the diagram, we see that $2x + 160 = 360$ , so $2x = 200$ and $x = 100$ , or $\boxed{100}$

Answer: A 100
Problem: A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of 3 kilometers per hour, bikes at a rate of 20 kilometers per hour, and runs at a rate of 10 kilometers per hour. Which of the following is closest to the triathlete's average speed, in kilometers per hour, for the entire race?
$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 5\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 7$

Solution: Since the three segments are all the same length, the triathlete's average speed is the harmonic mean of the three given rates. Therefore, the average speed is \[\frac{3}{\frac{1}{3}+\frac{1}{20}+\frac{1}{10}}=\frac{3}{\frac{29}{60}}=\frac{180}{29}\approx6\Rightarrow\boxed{6}\]

Answer: D 6
Problem: The fraction
\[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\] simplifies to which of the following?
$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9$

Solution: Using Difference of Squares, $\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}){^2}}$ becomes
$\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}$
$= \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}$
$= \boxed{9}$

Answer: E 9
Problem: Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3$ . What is the area of $S_3$
$\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 4$

Solution: Since the length ratio is $\frac{1}{\sqrt{2}}$ , then the area ratio is $\frac{1}{2}$ (since the area ratio between two similar 2-dimensional objects is equal to the side ratio of those objects squared). This means that $S_2 = 8$ and $S_3 = \boxed{4}$

Answer: E 4
Problem: In a collection of red, blue, and green marbles, there are $25\%$ more red marbles than blue marbles, and there are $60\%$ more green marbles than red marbles. Suppose that there are $r$ red marbles. What is the total number of marbles in the collection?
$\mathrm{(A)}\ 2.85r\qquad\mathrm{(B)}\ 3r\qquad\mathrm{(C)}\ 3.4r\qquad\mathrm{(D)}\ 3.85r\qquad\mathrm{(E)}\ 4.25r$

Solution: Let the number of red marbles be 100. You have $b = 80,$ and $g = 160.$
The answer is $\frac{100+80+160}{100} = \boxed{3.4}$

Answer:  3.4
Problem: Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$

Solution: Set the time Ian traveled as $I$ , and set Han's speed as $H$ . Therefore, Jan's speed is $H+5.$
We get the following equation for how much Han is ahead of Ian: $H+5I = 70.$
The expression for how much Jan is ahead of Ian is: $2(H+5)+10I.$
This simplifies to: $2H+10+10I.$
However, this is just $2(H+5I)+10.$
Substitute, from the first equation, $H+5I$ as $70.$
Therefore, the answer is $140 + 10$ , which is $150$ , or $\boxed{150}$

Answer: D 150
Problem: Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$

Solution: We let Ian's speed and time equal $I_s$ and $I_t$ , respectively. Similarly, let Han's and Jan's speed and time be $H_s$ $H_t$ $J_s$ $J_t$ . The problem gives us 5 equations
\begin{align} H_s&=I_s+5 \\ H_t&=I_t+1 \\ J_s&=I_s+10 \\ J_t&=I_t+2 \\ H_s \cdot H_t & =I_s \cdot I_t+70 \end{align}
Substituting equations $(1)$ and $(2)$ into $(5)$ gives:
\[(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)\]
We are asked the difference between Jan's and Ian's distances, or
\[J_s J_t-I_s I_t=x,\]
Where $x$ is the difference between Jan's and Ian's distances and the answer to the problem. Substituting $(3)$ and $(4)$ equations into this equation gives:
\[(I_s+10)(I_t+2)-I_s I_t=x \Longrightarrow I_s I_t+2I_s+10I_t+20-I_s I_t=x \Longrightarrow\]
\[2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x\]
Substituting $(*)$ into this equation gives:
\[2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x\]
Therefore, the answer is $150$ miles or $\boxed{150}$

Answer: D 150
Problem: Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$

Solution: Let Ian drive $D$ miles, at a speed of $R$ , for some time $T$ (in hours). Hence, we have $D=RT$ . We can find a similar equation for Han, who drove $D + 70$ miles, at a rate of $R+5$ , for $T+1$ hours, giving us $D + 70 = (R + 5)(T + 1)$ . We can do the same for Jan, giving us $D + x = (R + 10)(T + 2)$ , where $x$ is how much further Jan traveled than Ian. We now have three equations: \[D= RT\] \[D + 70 = (R+5)(T+1) = RT + R + 5T + 5\] \[D + x = (R + 10)(T + 2) = RT + 10 T + 2R + 20.\] Substituting $RT$ for $D$ in the second and third equations and cancelling gives us: \[70 = 5T + R + 5 \Longrightarrow 5T + R = 65\] \[x = 10T + 2R + 20  \Longrightarrow x = 2(5T + R ) + 20  \Longrightarrow x= 2(65) + 20 = 150.\] Since $x = 150$ , our answer is $\boxed{150}$

Answer: D 150
Problem: Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$

Solution: Let Ian drive $d$ miles, $t$ hours, and at speed $s$
Ian's Equation: $d=s \cdot t.$
Han drove 70 more miles, traveled 5 miles per hour faster and traveled 1 more hour than Ian.
Han's Equation: $d+70=(s+5) \cdot (t+1).$
Let Jan have driven $m$ miles. Jan also has driven 10 miles per hour faster and traveled 2 more hours than Ian.
Jan's Equation: $m=(s+10) \cdot (t+2).$
Let's group the equations together:
$(1) \phantom{a} d=s \cdot t$
$(2) \phantom{a} d+70=(s+5) \cdot (t+1)$
$(3) \phantom{a} m=(s+10) \cdot (t+2)$
Let's see what we want to find: We want to find $n$ . The equation is $m=d+n$ where $n$ is the number of more miles traveled by Jan than Ian.
Onto the calculating part.
Expanding the second equation, we get
$d+70=st+5t+s+5.$
Note that $st=d$ by the first equation, so substituting we get
$d+70=d+5t+s+5.$
Simplifying gets us
$(4) \phantom{a} s+5t=65.$
(Note for the above process: You could have substituted $d$ with $st$ but that would lead you to the same result since $d-d=st-st=0.$
Let's look at the third equation. Expanding, we get
$m=st+10t+2s+20.$
Since we want $n$ we want the equation $m=d+n$ . We write the expanded third equation into this form since $st=d.$
$(5) \phantom{a} m=d+(10t+2s+20)$
Let's take a closer look at the $n$ section of the equation:
$(6) \phantom{a} n=10t+2s+20$
This looks very similar to equation 4, if you multiply equation 4 by $2$ you get
$(7) \phantom{a} 10t+2s=130$
Plugging equation 7 into equation 6 we have
$n=(10t+2s)+20 \Rightarrow n=130+20$
Calculating gets us
$(8) \phantom{a} n=150.$
Substituting equation 8 into equation 5 gets us
$m=d+n \Rightarrow m=d+150.$
The $n$ term is $150$ which is what we want to find so the answer is $\boxed{150}.$

Answer: D 150
Problem: Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$

Solution: Since Han drove for $1$ hour and drove $70$ miles more than Ian during that hour, we know that Ian's speed is $65$ miles per hour since Han drove $5$ mph faster than him. Now Jan went $10$ mph faster than Ian for $2$ hours, so we can tell that she drove $75 \cdot 2$ miles more than Ian, therefore the answer is $\boxed{150}.$

Answer: D 150
Problem: Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$

Solution: First, choose the two letters to be repeated in each set. $\dbinom{5}{2}=10$ .  Now we have three remaining elements that we wish to place into two separate subsets. There are $2^3 = 8$ ways to do so because each of the three remaining letters can be placed either into the first or second subset. Both of those subsets contain the two chosen elements, so their intersection is the two chosen elements). Unfortunately, we have over-counted  (Take for example $S_{1} = \{a,b,c,d \}$ and $S_{2} = \{a,b,e \}$ ). Notice how $S_{1}$ and $S_{2}$ are interchangeable. Division by two will fix this problem. Thus we have:
$\dfrac{10 \times 8}{2} = 40 \implies \boxed{40}$

Answer: B 40
Problem: Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$

Solution: Another way of looking at this problem is to break it down into cases.
First, our two subsets can have 2 and 5 elements. The 5-element subset (aka the set) will contain the 2-element subset. There are $\dbinom{5}{2}=10$ ways to choose the 2-element subset. Thus, there are $10$ ways to create these sets.
Second, the subsets can have 3 and 4 elements. $3+4=7$ non-distinct elements. $7-5=2$ elements in the intersection. There are $\dbinom{5}{3}=10$ ways to choose the 3-element subset. For the 4-element subset, two of the elements must be the remaining elements (not in the 3-element subset). The other two have to be a subset of the 3-element subset. There are $\dbinom{3}{2}=3$ ways to choose these two elements, which means there are 3 ways to choose the 4-element subset. Therefore, there are $10\cdot3=30$ ways to choose these sets.
This leads us to the answer:
$10+30=40 \implies \boxed{40}$

Answer: B 40
Problem: Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?
$\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$

Solution: There are $\dbinom{5}{2}=10$ ways to choose the 2 shared elements. We now must place the 3 remaining elements into the subsets. Using stars and bars, we can notate this as: $I I I X \rightarrow \dbinom{4}{3}=4$ . Thus, $4*10=\boxed{40}$

Answer: B 40
Problem: Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$

Solution: $k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$
So, $k^2 \equiv 0 \pmod{10}$ . Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$
Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$ . So the units digit is $6 \Rightarrow \boxed{6}$

Answer: D 6
Problem: Let $k={2008}^{2}+{2}^{2008}$ . What is the units digit of $k^2+2^k$
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$

Solution: I am going to share another approach to this problem.
A units digit $k$ for an integer $n$ implies $n \equiv k \pmod{10}$
Let us take this step by step. First, we consider $k^2.$
Note that $k^2 = \left(2008^2 + 2^{2008}\right)^2 = 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016}.$ Now we calculate $k^2 \pmod{10}$
Before continuing, though, we must take note of the following:
\begin{align*} 2^1 &\equiv 2 \pmod {10} \\ 2^2 &\equiv 4 \pmod {10} \\ 2^3 &\equiv 8 \pmod {10} \\ 2^4 &\equiv 6 \pmod {10} \end{align*}
Now, we continue with the calculation.
\begin{align*} 2008^4 + 2 \cdot 2008^2 \cdot 2^{5016} &\equiv 8^4 + 2 \cdot 8^2 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^1 \cdot 2^6 \cdot 2^{2008} + 2^{5016} \pmod{10} \\ &\equiv 8^4 + 2^{1+6+2008} + 2^{5016} \pmod{10} \\ &\equiv 6 + 2^{2015} + 6 \pmod{10} \\ &\equiv 6 + 2^{3} + 6 \pmod{10} \\ &\equiv 6 + 8 + 6 \pmod{10} \\ &\equiv 20 \pmod{10} \\ &\equiv 0 \pmod{10} \\ \end{align*}
We do the same with $2^k.$ However, we just need to find $k \pmod 4$ in order to do this calculation since we have the table of $2^k \pmod 10.$
\begin{align*} 2008^2 + 2^{2008} &\equiv 8^2 \pmod{4} \\ &\equiv 64 \pmod 4\\ &\equiv 0 \pmod 4 \end{align*}
This implies that \begin{align*} 2^k &\equiv 2^{4} \pmod{10} \\ &\equiv 6 \pmod{10} \end{align*}
Thus, \begin{align*} k^2 + 2^k &\equiv 6+0 \pmod{10}\\ &\equiv \boxed{6}

Answer: D 6
Problem: A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?
$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$

Solution: The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is $\boxed{6}$

Answer: E 6
Problem: $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

Solution: After reversing the numbers on the second and fourth rows, the block will look like this:
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$
The positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{4}$

Answer: B 4
Problem: $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

Solution: Notice that at baseline the diagonals sum to the same number ( $52$ ).  Therefore we need only compute the effect of the swap.  The positive difference between $9$ and $10$ is $1$ and the positive difference between $22$ and $25$ is $3$ . Adding gives $1+3=\boxed{4}$

Answer: B 4
Problem: A semipro baseball league has teams with 21 players each. League rules state that a player must be paid at least $15,000 and that the total of all players' salaries for each team cannot exceed $700,000. What is the maximum possible salary, in dollars, for a single player?
$\mathrm{(A)}\ 270,000\qquad\mathrm{(B)}\ 385,000\qquad\mathrm{(C)}\ 400,000\qquad\mathrm{(D)}\ 430,000\qquad\mathrm{(E)}\ 700,000$

Solution: The maximum salary for a single player occurs when the other 20 players receive the minimum salary. The total of all players' salaries is 700000. The answer is $700000-15000*20=400000\Rightarrow \boxed{400,000}$

Answer: C 400,000
Problem: An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$ . How many small triangles are required?
$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$

Solution: The area of the large triangle is $\frac{10^2\sqrt3}{4}$ , while the area of each small triangle is $\frac{1^2\sqrt3}{4}$ . Dividing these two quantities results in $100$ , therefore $\boxed{100}$ small triangles can fill the large one without overlap.

Answer: C 100
Problem: An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$ . How many small triangles are required?
$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$

Solution: [asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]
The number of triangles is $1+3+\dots+19 = \boxed{100}$

Answer:  100
Problem: A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$

Solution: The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $r\geq 0$ . Then we have $50-3\cdot 2r = 50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2 = 25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality $25-3r \geq 0$ which solves to $r\leq 8 \frac13$ $r$ must be an integer, so there are $\boxed{9}$ possible values of $r$ , and each gives us one solution.

Answer: C 9
Problem: A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$

Solution: Let $x$ and $y$ be the number of roses and carnations bought. The equation should be $3x+2y = 50$ . Since $50$ is an even number, the product of $3x$ must be even and smaller than $50$ . You can try nonnegative even integers for $x$ and you will end up with the numbers $0$ $2$ $4$ $6$ $8$ $10$ $12$ $14$ , and $16$ . There are $9$ numbers in total, so the answer is $\boxed{9}$

Answer: C 9
Problem: A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$

Solution: Let $r$ represent the number of roses, and let $c$ represent the number of carnations. Then, we get the linear Diophantine equation, $3r+2c=50$ .
Using the Euclidean algorithm, we get the initial solutions to be $r_0=50$ and $c_0=-50$ , meaning the complete solution will be, $r=50+\frac{2}{\gcd(2,3)}$ $k=50+2k$ $c=-50-\frac{3}{\gcd(2,3)}k=-50-3k$
The solution range for which both $r$ and $c$ are positive is $17$ $\leq k$ $\leq$ $25$ . There are $\boxed{9}$ possible values for $k$

Answer: C 9
Problem: Suppose that $(u_n)$ is a sequence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$
and that $u_3=9$ and $u_6=128$ . What is $u_5$
$\mathrm{(A)}\ 40\qquad\mathrm{(B)}\ 53\qquad\mathrm{(C)}\ 68\qquad\mathrm{(D)}\ 88\qquad\mathrm{(E)}\ 104$

Solution: If we plug in $n=4$ , we get
By plugging in $n=3$ , we get
This is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives $128=5u_4+18 \Longrightarrow u_4=22$ , therefore $u_5=\frac{128-22}{2}=53 \longrightarrow \textbf{\boxed{53}$ .
~ Mathkiddie

Answer: B 53
Problem: Postman Pete has a pedometer to count his steps. The pedometer records up to 99999 steps, then flips over to 00000 on the next step. Pete plans to determine his mileage for a year. On January 1 Pete sets the pedometer to 00000. During the year, the pedometer flips from 99999 to 00000 forty-four 
times. On December 31 the pedometer reads 50000. Pete takes 1800 steps per mile. Which of the following is closest to the number of miles Pete walked during the year?
$\mathrm{(A)}\ 2500\qquad\mathrm{(B)}\ 3000\qquad\mathrm{(C)}\ 3500\qquad\mathrm{(D)}\ 4000\qquad\mathrm{(E)}\ 4500$

Solution: Every time the pedometer flips from $99999$ to
$00000$ Pete has walked $100000$ steps.
So, if the pedometer flipped $44$ times
Pete walked $44*100000+50000=4450000$ steps.
Dividing by $1800$ steps per mile gives $2472.\overline{2}$
This is closest to answer $\boxed{2500}$

Answer: A 2500