Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Each valve $A$ $B$ , and $C$ , when open, releases water into a tank at its own constant rate. With all three valves open, the tank fills in 1 hour, with only valves $A$ and $C$ open it takes 1.5 hours, and with only valves $B$ and $C$ open it takes 2 hours. The number of hours required with only valves $A$ and $B$ open is
$\textbf{(A)}\ 1.1\qquad\textbf{(B)}\ 1.15\qquad\textbf{(C)}\ 1.2\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 1.75$

Solution: Let the rate of water flowing through valve $A$ be $a$ , the rate of water flowing through valve $B$ be $b$ , and the rate of water flowing through valve $C$ be $c$ WLOG , let the volume of the tank be 1 liter, and let the units for the rates be liters per hour.  With this information, we can write three equations. \[\frac{1}{a+b+c} = 1\] \[\frac{1}{a+c} = \frac{3}{2}\] \[\frac{1}{b+c} = 2\] Manipulate each equation to get \[1 = a+b+c\] \[\frac{2}{3} = a+c\] \[\frac{1}{2} = b+c\] Solving for $a$ yields $a = \tfrac{1}{2}$ , and solving for $b$ yields $b = \tfrac{1}{3}$ .  The number of hours to fill the tub with only valves $A$ and $B$ on is $\frac{1}{\frac{1}{2} + \frac{1}{3}} = \frac{6}{5} = \boxed{1.2}$

Answer: C 1.2
Problem: If the sum of all the angles except one of a convex polygon is $2190^{\circ}$ , then the number of sides of the polygon must be
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 19 \qquad \textbf{(E)}\ 21$

Solution: Let $n$ be the number of sides in the polygon.  The number of interior angles in the polygon is $180(n-2)$ .  We know that the sum of all but one of them is $2190^{\circ}$ , so the sum of all the angles is more than that. \[180(n-2) > 2190\] \[n-2 > 12 \tfrac{1}{6}\] \[n > 14 \tfrac{1}{6}\]
The sum of the angles in a 15-sided polygon is $2340^{\circ}$ , making the remaining angle $150^{\circ}$ .  The angles of a convex polygon are all less than $180^{\circ}$ , and since adding one more side means adding $180^{\circ}$ to the measure of the remaining angle, we can confirm that there are $\boxed{15}$ sides in the polygon.

Answer: B 15
Problem: The number of sets of two or more consecutive positive integers whose sum is 100 is
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution: If the first number of a group of $n$ consecutive numbers is $a$ , the $n^\text{th}$ number is $a+n-1$ .  We know that the sum of the group of numbers is $100$ , so \[\frac{n(2a+n-1)}{2} = 100\] \[2a+n-1=\frac{200}{n}\] \[2a = 1-n + \frac{200}{n}\] We know that $n$ and $a$ are positive integers, so we check values of $n$ that are a factor of $200$ .  Of these values, the only ones that result in a positive integer $a$ is when $n = 5$ or when $n = 8$ , so there are $\boxed{2}$ sets of two or more consecutive positive integers whose sum is $100$

Answer: B 2
Problem: There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability (1/2) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac12 \qquad \textbf{(D)}\ \frac23 \qquad \textbf{(E)}\ \frac34$

Solution: There are three red faces, and two are on the card that is completely red, so our answer is $\frac{2}{3}$ , which is $\boxed{23}$

Answer: D 23
Problem: The check for a luncheon of 3 sandwiches, 7 cups of coffee and one piece of pie came to $$3.15$ . The check for a luncheon consisting of 4 sandwiches, 10 cups of coffee and one piece of pie came to $$4.20$ at the same place. The cost of a luncheon consisting of one sandwich, one cup of coffee, and one piece of pie at the same place will come to
$\textbf{(A)}\ $1.70 \qquad \textbf{(B)}\ $1.65 \qquad \textbf{(C)}\ $1.20 \qquad \textbf{(D)}\ $1.05 \qquad \textbf{(E)}\ $0.95$

Solution: Let $s$ be the cost of one sandwich, $c$ be the cost of one cup of coffee, and $p$ be the price of one piece of pie.  With the information,
\[3s+7c+p=3.15\] \[4s+10c+p=4.20\]
Subtract the first equation from the second to get
\[s+3c=1.05\]
That means $s=1.05-3c$ .  Substituting it back in the second equation results in.
\[4.20-12c+10c+p=4.20\]
Solving for $p$ yields $p=2c$ .  With the substitutions, the cost of one sandwich, one cup of coffee, and one slice of pie is $(1.05-3c)+c+(2c) = \boxed{1.05}$

Answer: D 1.05
Problem: The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$

Solution: Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference.  That means the last term is $a+r(n-1)$
We can write an equation on the difference between the last and first term based on the conditions. \[a+r(n-1)-a =10.5\] \[rn-r=10.5\] Also, half of the terms add up to $24$ while the other half of the terms add up to $30$ , so \[24 + r\frac{n}{2} = 30\] \[nr = 12\] Substituting the value back to a previous equation, \[12-r=10.5\] \[r=1.5\] Substituting to a previous equation again, \[1.5n-1.5=10.5\] \[n=8\] Thus, there are $\boxed{8}$ terms in the arithmetic sequence.

Answer: E 8
Problem: Two boys start moving from the same point A on a circular track but in opposite directions. Their speeds are 5 ft. per second and 9 ft. per second. If they start at the same time and finish when they first meet at the point A again, then the number of times they meet, excluding the start and finish, is
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ \text{infinity} \qquad \textbf{(E)}\ \text{none of these}$

Solution: Let $d$ be the length of the track in feet and $x$ be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is $\tfrac{dx}{5}$ .  Since the time elapsed for both boys is equal, one boy ran $5(\tfrac{dx}{5})$ feet while the other boy ran $9(\tfrac{dx}{5})$ feet.  Because both finished at the starting point, both ran an integral number of laps, so $5(\tfrac{dx}{5})$ and $9(\tfrac{dx}{5})$ are multiples of $d$ .  Because both stopped when both met at the start for the first time, $x = 5$
Note that between the time a runner finishes a lap and a runner (can be same) finishes a lap, both runners must meet each other.  When $0 < x \le 5$ and either $5(\tfrac{dx}{5})$ or $9(\tfrac{dx}{5})$ is a multiple of $d$ , one of the runners completed a lap.  This is achieved when $x = \tfrac59, 1, \tfrac{10}{9}, \tfrac{15}{9}, 2, \tfrac{20}{9}, \tfrac{25}{9}, 3, \tfrac{30}{9}, \tfrac{35}{9}, 4, \tfrac{40}{9}, 5$ , so the two meet each other (excluding start and finish) a total of $\boxed{13}$ times.

Answer: A 13
Problem: In the following equation, each of the letters represents uniquely a different digit in base ten:
\[(YE) \cdot (ME) = TTT\]
The sum $E+M+T+Y$ equals
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 24$

Solution: The right side of the equation can be rewritten as $111T = 37 \cdot 3T$ .  With trial and error and prime factorization as a guide, we can test different digits of $T$ to see if we can find two two-digit numbers that have the same units digit and multiply to $111T$
The only possibility that works is $37 \cdot 27 = 999$ .  That means $E+M+T+Y = \boxed{21}$

Answer: C 21
Problem: The volume of a pyramid whose base is an equilateral triangle of side length 6 and whose other edges are each of length $\sqrt{15}$ is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 9/2 \qquad \textbf{(C)}\ 27/2 \qquad \textbf{(D)}\ \frac{9\sqrt3}{2} \qquad \textbf{(E)}\ \text{none of these}$

Solution: [asy] import three; unitsize(1cm); size(200); draw((0,0,0)--(6,0,0)--(3,5.196,0)--(0,0,0)); draw((3,1.732,1.732)--(0,0,0)); draw((3,1.732,1.732)--(6,0,0)); draw((3,1.732,1.732)--(3,5.196,0)); draw((3,1.732,1.732)--(3,1.732,0)--(0,0,0),dotted); label("6",(4.5,2.598,0),SW); label("$\sqrt{15}$",(4.5,0.866,0.866),N);  currentprojection=orthographic(1/6,1/2,1/3);  [/asy]
Draw an altitude towards the equilateral triangle base.  By symmetry (this can also be proved by HL), the base of the altitude is equidistant from the three points of the equilateral triangle.  This means that the distance from the base of the altitude to one of the points of the equilateral triangle is $2\sqrt{3}$
[asy] draw((0,1.732)--(0,0)--(3.464,0),dotted); draw((0,1.732)--(3.464,0)); label("$2\sqrt{3}$",(1.732,0),S); label("$\sqrt{15}$",(1.732,0.866),NE); [/asy]
Using the Pythagorean Theorem , the length of the altitude is $\sqrt{3}$ , so the volume of the triangular pyramid is $\tfrac13 \cdot \tfrac{6^2 \cdot \sqrt{3}}{4} \cdot \sqrt{3} = \boxed{9}$

Answer: A 9
Problem: When one ounce of water is added to a mixture of acid and water, the new mixture is $20\%$ acid. When one ounce of acid is added to the new mixture, the result is $33\frac13\%$ acid. The percentage of acid in the original mixture is
$\textbf{(A)}\ 22\% \qquad \textbf{(B)}\ 24\% \qquad \textbf{(C)}\ 25\% \qquad \textbf{(D)}\ 30\% \qquad \textbf{(E)}\ 33\frac13 \%$

Solution: Let $a$ be the original number of ounces of acid and $w$ be the original number of ounces of water.  We can write two equations since we know the percentage of acid after some water and acid. \[\frac{a}{a+w+1} = \frac{1}{5}\] \[\frac{a+1}{a+w+2} = \frac{1}{3}\] Cross-multiply to get rid of the fractions. \[5a = a+w+1\] \[3a+3=a+w+1\] Solve the system to get $a=1$ and $w=3$ .  The percentage of acid in the original mixture is $\tfrac{1}{1+3} = \boxed{25}$

Answer: C 25
Problem: If $x=\dfrac{1-i\sqrt{3}}{2}$ where $i=\sqrt{-1}$ , then $\dfrac{1}{x^2-x}$ is equal to
$\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1+i\sqrt{3}\qquad \textbf{(D) }1\qquad  \textbf{(E) }2$

Solution: Using DeMoivre's theorem, we can calculate $x^2=\frac{1+i\sqrt{3}}{2}$ The denominator is therefore $-1$ which makes the answer \[\boxed{1}.\] ~lopkiloinm

Answer: C 1
Problem: The number of solutions to $\{1,~2\}\subseteq~X~\subseteq~\{1,~2,~3,~4,~5\}$ , where $X$ is a subset of $\{1,~2,~3,~4,~5\}$ is
$\textbf{(A) }2\qquad \textbf{(B) }4\qquad \textbf{(C) }6\qquad \textbf{(D) }8\qquad  \textbf{(E) }\text{None of these}$

Solution: $X$ has to contain $\{1,~2\}$ , so only $\{3,~4,~5\}$ matters. There are two choices for the elements; the element is either in $X$ or outside of $X$ . With this combinatorics in mind, the answer is simply $2^3=\boxed{8}.$ ~lopkiloinm

Answer: D 8
Problem: problem_id
4ca2bf599f1693b695997dd52d9e4774    A man walked a certain distance at a constant ...
4ca2bf599f1693b695997dd52d9e4774    A man walked a certain distance at a constant ...
Name: Text, dtype: object

Solution: We can make three equations out of the information, and since the distances are the same, we can equate these equations.
\[\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{5}{2})(x-\frac{1}{2})\] where $x$ is the man's rate and $t$ is the time it takes him.
Looking at the first two parts of the equations,
\[\frac{4t}{5}(x+\frac{1}{2})=xt\]
we note that we can solve for $x$ . Solving for $x$ , we get $x=2.$
Now we look at the last two parts of the equation:
\[xt=(t+\frac{5}{2})(x-\frac{1}{2})\]
we note that we can solve for $t$ and we get $t=\frac{15}{2}.$ We want the find the distance, which is $xt= \boxed{15}.$

Answer:  15
Problem: Inscribed in a circle is a quadrilateral having sides of lengths $25,~39,~52$ , and $60$ taken consecutively. The diameter of this circle has length
$\textbf{(A) }62\qquad \textbf{(B) }63\qquad \textbf{(C) }65\qquad \textbf{(D) }66\qquad  \textbf{(E) }69$

Solution: We note that $25^2+60^2=65^2$ and $39^2+52^2=65^2$ so our answer is $\boxed{65}$

Answer: C 65
Problem: A circular disc with diameter $D$ is placed on an $8\times 8$ checkerboard with width $D$ so that the centers coincide. The number of checkerboard squares which are completely covered by the disc is
$\textbf{(A) }48\qquad \textbf{(B) }44\qquad \textbf{(C) }40\qquad \textbf{(D) }36\qquad  \textbf{(E) }32$

Solution: Consider the upper right half of the grid, which consists of a $4\times4$ section of the checkerboard and a quarter-circle of radius $4$ . We can draw this as a coordinate grid and shade in the complete squares. There are $8$ squares in the upper right corner, so there are $8 \cdot 4 = \boxed{32}$ whole squares in total.

Answer:  32
Problem: When the number $2^{1000}$ is divided by $13$ , the remainder in the division is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }7\qquad  \textbf{(E) }11$

Solution: By Fermat's Little Theorem , we know that $2^{100} \equiv 2^{1000 \pmod{12}}\pmod{13}$ . However, we find that $1000 \equiv 4 \pmod{12}$ , so $2^{1000} \equiv 2^4 = 16 \equiv 3 \pmod{13}$ , so the answer is $\boxed{3}$

Answer: C 3
Problem: The minimum value of the quotient of a (base ten) number of three different non-zero digits divided by the sum of its digits is
$\textbf{(A) }9.7\qquad \textbf{(B) }10.1\qquad \textbf{(C) }10.5\qquad \textbf{(D) }10.9\qquad  \textbf{(E) }20.5$

Solution: The answer we are looking for can be expressed as $\dfrac{100a+10b+c}{a+b+c}$ . This is equivalent to $1 + \dfrac{99a+9b}{a+b+c}$ . Because we are trying to minimize our solution, we set $c$ $9$ , so we have $1 + \dfrac{99a+9b}{a+b+9}$ . This is equal to $1 + \dfrac{9a+9b+81}{a+b+9} + \dfrac{90a-81}{a+b+9}$ , which simplifies to $10+ \dfrac{90a-81}{a+b+9}$ . Since each digit is unique, we set $b$ to $8$ , leaving us with $10 + \dfrac{90a-81}{a+17}$ . Clearly, $a$ should be minimized, so $a = 1$ and our answer is \[\boxed{10.5}.\]

Answer: C 10.5
Problem: Three times Dick's age plus Tom's age equals twice Harry's age. 
Double the cube of Harry's age is equal to three times the cube of Dick's age added to the cube of Tom's age. 
Their respective ages are relatively prime to each other. The sum of the squares of their ages is
$\textbf{(A) }42\qquad \textbf{(B) }46\qquad \textbf{(C) }122\qquad \textbf{(D) }290\qquad  \textbf{(E) }326$

Solution: \[t=2h-3d\] \[3d^3+t^3=2h^3\]
First, substitute in t into the second equation and get $3d^3+8h^3-36h^2d+54hd^2-27d^3=2h^3$ . That turns into $h^3-6h^2d+9hd^2-4d^3=0$ which is factored into $(h-4d)(h-d)^2 =0.$ WLOG, $d=1$ and consequently $h=4$ . Then $t=8-3=5$ . Everything appears to be relatively prime already. The answer is thus $1+16+25=\boxed{42}.$ ~lopkiloinm

Answer: A 42
Problem: If $b$ men take $c$ days to lay $f$ bricks, then the number of days it will take $c$ men working at the same rate to lay $b$ bricks, is
$\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad  \textbf{(E) }f/b^2$

Solution: We can use a modified version of the equation $\text{Distance} = \text{Rate} \times {\text{Time}}$ , which is $\text{Work Done} = \text{Rate of Work} \times{ \text{Time Worked}}$ . In this case, the work done is the number of bricks laid, the rate of work is the number of men working, and the time worked is the number of days. With these definitions, we see from the first equation that $f = b \cdot c$ . If we let $X$ be the number of days it will take $c$ men working at the same rate to $b$ bricks, then we have the equation $b = c \cdot x$ . So, $x = \frac{b}{c}$ . The first equation says that $\frac{f}{c^2} = \frac{b}{c}$ , which leads to $x = \frac{f}{c^2}$ . This doesn't match any of our answer choices though, so we have to fiddle around a bit before we realize that $c = \frac{f}{b}$ , a substitution we can make to see that $x = \frac{b^2}{f}$ . Thus, the answer is $\boxed{2}$

Answer: D 2
Problem: A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips, and at most one third the number of red chips. The number which are white or blue is at least $55$ . The minimum number of red chips is
$\textbf{(A) }24\qquad \textbf{(B) }33\qquad \textbf{(C) }45\qquad \textbf{(D) }54\qquad  \textbf{(E) }57$

Solution: Let the number of white be $2x$ . The number of blue is then $x-y$ for some constant $y$ . So we want $2x+x-y=55\rightarrow 3x-y=55$ . We take mod 3 to find y. $55=1\pmod{3}$ , so blue is 1 more than half white. The number of whites is then 36, and the number of blues is 19. So $19*3=\boxed{57}$

Answer:  57
Problem: Given the progression $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$ . 
The least positive integer $n$ such that the product of the first $n$ terms of the progression exceeds $100,000$ is
$\textbf{(A) }7\qquad \textbf{(B) }8\qquad \textbf{(C) }9\qquad \textbf{(D) }10\qquad  \textbf{(E) }11$

Solution: The product of the sequence $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$ is equal to $10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}$ since we are looking for the smallest value $n$ that will create $100,000$ , or $10^5$ . From there, we can set up the equation $10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}=10^5$ , which simplified to $\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}=5$ , or $1+2+3\dots n=55$ This can be converted to $\frac{n(1+n)}{2}=55$ This simplified to the quadratic $n^2+n-110=0$ Or $(n+11)(n-10)=0$ So $n=-11$ or $10$ Since only positive values of $n$ work, $n=10$ makes the expression equal $100000$ . However, we have to exceed $100000$ , so our answer is $\boxed{11}.$

Answer: E 11
Problem: The sum of an infinite geometric series with common ratio $r$ such that $|r|<1$ is $15$ , and the sum of the squares of the terms of this series is $45$ . The first term of the series is
$\textbf{(A) } 12\quad \textbf{(B) } 10\quad \textbf{(C) } 5\quad \textbf{(D) } 3\quad \textbf{(E) 2}$

Solution: We know that the formula for the sum of an infinite geometric series is $S = \frac{a}{1-r}$
So we can apply this to the conditions given by the problem.
We have two equations:
\begin{align*} 15 &= \frac{a}{1-r} \\ 45 &= \frac{a^{2}}{1-r^{2}} \end{align*}
We get
\begin{align*} a &= 15 - 15r \\ a^{2} &= 45 - 45r^{2} \\ \\ (15 - 15r)^{2} &= 45 - 45r^{2} \\ 270r^{2} - 450r + 180 &= 0 \\ 3r^{2} - 5r + 2 &= 0 \\ (3r - 2)(r - 1) &= 0 \end{align*}
Since $|r| < 1$ $r = \frac{2}{3}$ , so plug this into the equation above and we get $a = 15 - 15r = 15 - 10 = \boxed{5}$

Answer: C 5
Problem: Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$
$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$

Solution: We can find the sum using the following method. We break it down into cases. The first case is the numbers $1$ to $9$ . The second case is the numbers $10$ to $99$ . The third case is the numbers $100$ to $999$ . The fourth case is the numbers $1,000$ to $9,999$ . And lastly, the sum of the digits in $10,000$ . The first case is just the sum of the numbers $1$ to $9$ which is, using $\frac{n(n+1)}{2}$ $45$ . In the second case, every number $1$ to $9$ is used $19$ times. $10$ times in the tens place, and $9$ times in the ones place. So the sum is just $19(45)$ . Similarly, in the third case, every number $1$ to $9$ is used $100$ times in the hundreds place, $90$ times in the tens place, and $90$ times in the ones place, for a total sum of $280(45)$ . By the same method, every number $1$ to $9$ is used $1,000$ times in the thousands place, $900$ times in the hundreds place, $900$ times in the tens place, and $900$ times in the ones place, for a total of $3700(45)$ . Thus, our final sum is $45+19(45)+280(45)+3700(45)+1=4000(45)+1=\boxed{180001}.$

Answer: A 180001
Problem: Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$
$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$

Solution: Consider the numbers from $0000-9999$ . We have $40000$ digits and each has equal an probability of being $0,1,2....9$ .
Thus, our desired sum is $4000\left( \frac{9 \cdot 10}{2} \right)+1=4000(45)+1=\boxed{180001}.$

Answer: A 180001
Problem: Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$
$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$

Solution: As in Solution 2, we consider the four digit numbers from $0000-9999.$ We see that we have $10000\times4=40000$ digits with each digit  appearing equally.
Thus, the digit sum will be the average of the digits multiplied by $40000.$ The digit average comes out to be $\frac{0+9}{2},$ since all digits are consecutive.
So, our answer will be $\frac{9}{2} \times 40000 = 180000.$ However, since we purposely did not include $10000,$ we add one to get our final answer as $\boxed{180001}.$

Answer: A 180001
Problem: Triangle $ABC$ is inscribed in a circle. The measure of the non-overlapping minor arcs $AB$ $BC$ and $CA$ are, respectively, $x+75^{\circ} , 2x+25^{\circ},3x-22^{\circ}$ . Then one interior angle of the triangle is:
$\text{(A) } 57\tfrac{1}{2}^{\circ}\quad \text{(B) } 59^{\circ}\quad \text{(C) } 60^{\circ}\quad \text{(D) } 61^{\circ}\quad \text{(E) } 122^{\circ}$

Solution: [asy] draw(circle((0,0),65)); draw((25,60)--(39,-52)--(-52,-39)--(25,60)); dot((25,60)); dot((39,-52)); dot((-52,-39)); dot((0,0)); draw((0,0)--(-52,-39)); draw((0,0)--(39,-52)); draw((0,0)--(25,60)); label("A",(-52,-39),SW); label("B",(25,60),NE); label("C",(39,-52),SE); [/asy] Because the triangle is inscribed, the sum of the minor arcs equals $360^\circ$ .  Thus, \[x+75+2x+25+3x-22=360\] \[6x+78=360\] Solving this yields $x = 47$ , so the inscribed angles are $122^\circ$ $99^\circ$ , and $119^\circ$ .  Noting that an angle of $\triangle ABC$ is half of its corresponding inscribed angle, so the angles of $\triangle ABC$ are $59.5^\circ$ $49.5^\circ$ , and $\boxed{61}$

Answer: D 61
Problem: The number of points equidistant from a circle and two parallel tangents to the circle is:
$\text{(A) } 0\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } \infty$

Solution: [asy] draw(circle((0,0),100)); draw((-300,100)--(300,100),Arrows); draw((-300,-100)--(300,-100),Arrows); draw((-300,0)--(300,0),dotted,Arrows); dot((-200,0)); dot((0,0)); dot((200,0)); draw((-200,100)--(-200,-100),dotted); draw((200,100)--(200,-100),dotted); [/asy] The distance between the two parallel tangents is the length of the circle's diameter, so the distance from a point that satisfies the conditions and the two tangents is the length of the circle's radius.  From the diagram, there are $\boxed{3}$ points that satisfies the conditions.

Answer: C 3
Problem: The number of points common to the graphs of $(x-y+2)(3x+y-4)=0 \text{    and     } (x+y-2)(2x-5y+7)=0$ is:
$\text{(A) } 2\quad \text{(B) } 4\quad \text{(C) } 6\quad \text{(D) } 16\quad \text{(E) } \infty$

Solution: By the Zero Product Property $x-y+2=0$ or $3x+y-4=0$ in the first equation and $x+y-2=0$ or $2x-5y+7=0$ in the second equation.  Thus, from the first equation, $y = x+2$ or $y =-3x+4$ , and from the second equation, $y=-x+2$ or $y = \frac{2}{5}x + \frac{7}{5}$
If a point is common to the two graphs, then the point must be in one of the lines in the first equation as well as one of the lines in the second equation.  Since the slopes of the lines are different, none of the lines are parallel.  Thus, there are $2 \cdot 2 = \boxed{4}$ points of intersection in the two graphs.

Answer: B 4