Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

Solution: Let $\underline{ABC}$ be one such odd positive $3$ -digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$
As $A\in\{1,2,4,5,6,7,8,9\}$ and $C\in\{1,5,7,9\},$ there are $8$ possibilities for $A$ and $4$ possibilities for $C.$ Note that each ordered pair $(A,C)$ determines the value of $B$ modulo $3,$ so $B$ can be any element in one of the sets $\{0,6,9\},\{1,4,7\},$ or $\{2,5,8\}.$ We conclude that there are always $3$ possibilities for $B.$
By the Multiplication Principle, the answer is $8\cdot4\cdot3=\boxed{96}.$

Answer: A 96
Problem: How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

Solution: Let $\underline{ABC}$ be one such odd positive $3$ -digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$
As $A\in\{1,2,4,5,6,7,8,9\},B\in\{0,1,2,4,5,6,7,8,9\},$ and $C\in\{1,5,7,9\},$ note that:
We apply casework to $A+B+C\equiv0\pmod3:$ \[\begin{array}{c|c|c||l} & & & \\ [-2.5ex] \boldsymbol{A\operatorname{mod}3} & \boldsymbol{B\operatorname{mod}3} & \boldsymbol{C\operatorname{mod}3} & \multicolumn{1}{c}{\textbf{Count}} \\ [0.5ex] \hline & & &  \\ [-2ex] 0 & 0 & 0 & 2\cdot3\cdot1=6 \\ 0 & 1 & 2 & 2\cdot3\cdot1=6 \\ 0 & 2 & 1 & 2\cdot3\cdot2=12 \\ 1 & 0 & 2 & 3\cdot3\cdot1=9 \\ 1 & 1 & 1 & 3\cdot3\cdot2=18 \\ 1 & 2 & 0 & 3\cdot3\cdot1=9 \\ 2 & 0 & 1 & 3\cdot3\cdot2=18 \\ 2 & 1 & 0 & 3\cdot3\cdot1=9 \\ 2 & 2 & 2 & 3\cdot3\cdot1=9 \end{array}\] Together, the answer is $6+6+12+9+18+9+18+9+9=\boxed{96}.$

Answer: A 96
Problem: How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

Solution: Analyze that the three-digit integers divisible by $3$ start from $102.$ In the $200$ 's, it starts from $201.$ In the $300$ 's, it starts from $300.$ We see that the units digits is $0, 1,$ and $2.$
Write out the $1$ - and $2$ -digit multiples of $3$ starting from $0, 1,$ and $2.$ Count up the ones that meet the conditions. Then, add up and multiply by $3,$ since there are three sets of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$ 's ll contain the digit $3.$
Together, the answer is $3(12+12+12)-12=\boxed{96}.$

Answer: A 96
Problem: How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

Solution: Consider the number of $2$ -digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$ -digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$ -digit number. However, we have $7 \equiv 1\pmod{3},$ and thus we need to count any $2$ -digit number $\equiv 2\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \boxed{96}.$

Answer: A 96
Problem: How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

Solution: We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$ -digit numbers. Exactly $\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\cdot\frac{1}{3}=150.$ Of these $150$ numbers, $\frac{4}{5}$ $\textbf{do not}$ have $3$ in their ones (units) digit, $\frac{9}{10}$ $\textbf{do not}$ have $3$ in their tens digit, and $\frac{8}{9}$ $\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$ -digit integers is \[900\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{5}\cdot\frac{9}{10}\cdot\frac{8}{9}=\boxed{96}.\]

Answer: A 96
Problem: How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

Solution: We will start with the numbers that could work. This numbers include _ _ $1$ , _ _ $5$ , _ _ $7$ , _ _ $9$ . Let's work case by case.
Case $1$ : _ _ $1$ : The two blanks could be any number that is $2$ mod $3$ that does not include $3$ . We have $24$ cases for this case (we could count every case).
Case $2$ : _ _ $5$ : The $2$ blanks could be any number that is $1$ mod $3$ that does not include $3$ . But we could see that this case has exactly the same solutions to case $1$ because we have a $1-1$ correspondence. We can do the exact same for case $3$
Cases $4$ : _ _ $9$ : We need the blanks to be a multiple of $3$ , but does not contain 3. We have $(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87, 90, 96, 99)$ which also contains $24$ numbers. Therefore, we have $24 \cdot 4$ which is equal to $\boxed{96}.$

Answer: A 96
Problem: How many odd positive $3$ -digit integers are divisible by $3$ but do not contain the digit $3$
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$

Solution: This problem is solvable by inclusion exclusion principle. There are $\frac{999-105}{6} + 1 = 150$ odd $3$ -digit numbers divisible by $3$ . We consider the number of $3$ -digit numbers divisible by $3$ that contain either $1, 2$ or $3$ digits of $3$
For $\underline{AB3}$ $AB$ is any $2$ -digit number divisible by $3$ , which gives us $\frac{99-12}{3} + 1 = 30$ . For $\underline{A3B}$ , for each odd $B$ , we have $3$ values of $A$ that give a valid case, thus we have $5(3) = 15$ cases. For $\underline{3AB}$ , we also have $15$ cases, but when $B=3, 9$ $A$ can equal $0$ , so we have $17$ cases.
For $\underline{A33}$ , we have $3$ cases. For $\underline{3A3}$ , we have $4$ cases. For $\underline{33A}$ , we have $2$ cases. Finally, there is just one case for $\underline{333}$
By inclusion exclusion principle, we get $150 - 30 - 15 - 17 + 3 + 4 + 2 - 1 = \boxed{96}$ numbers.

Answer: A 96
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: More generally, let $a,b,c,d,p,$ and $q$ be positive integers such that $bc-ad=1$ and \[\frac ab < \frac pq < \frac cd.\] From $\frac ab < \frac pq,$ we have $bp-aq>0,$ or \[bp-aq\geq1. \hspace{15mm} (1)\] From $\frac pq < \frac cd,$ we have $cq-dp>0,$ or \[cq-dp\geq1. \hspace{15mm} (2)\] Since $bc-ad=1,$ note that:
To minimize $q,$ we set $q=b+d,$ from which $p=a+c.$ Together, we can prove that \[\frac{a+c-1}{b+d}\leq\frac ab<\frac{a+c}{b+d}<\frac cd\leq\frac{a+c+1}{b+d}. \hspace{15mm} (\bigstar)\] For this problem, we have $a=5,b=9,c=4,$ and $d=7,$ so $p=a+c=9$ and $q=b+d=16.$ The answer is $q-p=\boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Define variables $a,b,c,d,p,$ and $q$ as Solution 1 does. Moreover, this solution refers to inequalities $(1)$ and $(2)$ in Solution 1.
Note that \begin{align*} \frac{1}{bd}&=\frac{bc-ad}{bd} \\ &=\frac cd - \frac ab \\ &=\left(\frac cd - \frac pq\right)+\left(\frac pq - \frac ab\right) \\ &=\frac{cq-dp}{dq}+\frac{bp-aq}{bp} \\ &\geq\frac{1}{dq}+\frac{1}{bq}. \end{align*} Multiplying both sides of $\frac{1}{bd}\geq\frac{1}{dq}+\frac{1}{bq}$ by $bdq,$ we get $q\geq b+d.$
For this problem, we have $a=5,b=9,c=4,$ and $d=7,$ so $q\geq b+d=16.$ At $q=16,$ we have \[\frac{8}{16}<\frac59<\frac{9}{16}<\frac47<\frac{10}{16},\] from which $p=9.$ Therefore, the answer is $q-p=\boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Inverting the given inequality we get \[\frac{7}{4} < \frac{q}{p} < \frac{9}{5},\] which simplifies to \[35p < 20q < 36p.\] We can now substitute $q = p + k.$ Note we need to find $k:$ \[35p < 20p + 20k < 36p,\] which simplifies to \[15p < 20k < 16p.\] Clearly $p>k.$ We will now substitute $p = k + x$ to get \[15k + 15x < 20k < 16k + 16x.\] The inequality $15k + 15x < 20k$ simplifies to $3x < k.$
The inequality $20k < 16k + 16x$ simplifies to $k < 4x.$
Combining the two inequalities, we get \[3x < k < 4x.\] Since $x$ and $k$ are integers, the smallest values of $x$ and $k$ that satisfy the above equation are $2$ and $7$ respectively. Substituting these back in, we arrive with an answer of $\boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Because $q$ and $p$ are positive integers with $p<q,$ we can let $q=p+k$ where $k\in{\mathbb{Z}}.$ Now, the problem condition reduces to \[\frac{5}{9}<\frac{p}{p+k}<\frac{4}{7}.\]
Our first inequality is $\frac{5}{9}<\frac{p}{p+k},$ which gives us $5p+5k<9p,$ or $\frac{5}{4}k<p.$
Our second inequality is $\frac{p}{p+k}<\frac{4}{7},$ which gives us $7p<4p+4k,$ or $p<\frac{4}{3}k.$
Hence, we have $\frac{5}{4}k<p<\frac{4}{3}k,$ or $15k<12p<16k.$
It is clear that we are aiming to find the least positive integer value of $k$ such that there is at least one value of $p$ that satisfies the inequality.
Now, simple casework through the answer choices of the problem reveals that $q-p=p+k-p=k\ge{\boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: We subtract $\frac{1}{2}$ from both sides of the equation, so \[\frac{1}{18} < \frac{p}{q}-\frac{1}{2} < \frac{1}{14}.\] Then for $q$ to be as small as possible, $\frac{p}{q}-\frac{1}{2}$ has to be $\frac{1}{16},$ so $\frac{p}{q}$ is $\frac{9}{16}$ and $q-p$ is $\boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Cross-multiply the inequality to get \[35q < 63p < 36q.\] Then, we have $0 < 63p-35q < q,$ or \[0 < 7(9p-5q) < q.\] Since $p$ and $q$ are integers, $9p-5q$ is an integer. To minimize $q,$ start from $9p-5q=1,$ which gives $p=\frac{5q+1}{9}.$ This limits $q$ to be greater than $7,$ so test values of $q$ starting from $q=8.$ However, $q=8$ to $q=14$ do not give integer values of $p.$
Once $q>14,$ it is possible for $9p-5q$ to be equal to $2,$ so $p$ could also be equal to $\frac{5q+2}{9}.$ The next value, $q=15,$ is not a solution, but $q=16$ gives $p=\frac{5\cdot 16 + 1}{9} = 9.$ Thus, the smallest possible value of $q$ is $16,$ and the answer is $16-9= \boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Start with $\frac{5}{9}.$ Repeat the following process until you arrive at the answer: if the fraction is less than or equal to $\frac{5}{9},$ add $1$ to the numerator; otherwise, if it is greater than or equal to $\frac{4}{7},$ add one $1$ to the denominator. We have \[\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}.\]
Therefore, the answer is $16 - 9 = \boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill.
The interval can also be written as $0.5556<x<0.5714.$ This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range.
The denominators to be considered are $9,10,11,12,\ldots.$ We check $\frac{6}{10}, \frac{6}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}.$ At this point we know that we've got our fraction and our answer is $16-9=\boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Graph the regions $y > \frac{5}{9}x$ and $y < \frac{4}{7}x.$ Note that the lattice point $(16,9)$ is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is $\frac{9}{16}$ and the answer is $16-9= \boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: As the other solutions do, the mediant $\frac{9}{16}$ is between the two fractions, with a difference of $\boxed{7}\approx 0.6.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: In ascending order, we can use answer choices, values for $q-p,$ as a method of figuring out our answer through the means of substitution. Let the assumed difference be $7.$ Then, $p=q-7.$ We thus have two inequalities: $\frac{5}{9} < \frac{q-7}{q}$ and $\frac{q-7}{q} < \frac{4}{7}.$
Solving for $q$ in these equalities, we get $\frac{63}{4} < q < \frac{49}{3}.$ So, $q$ is between $15.75$ and $16.\overline{3},$ making it $16$ as $q$ is a positive integer (again, at this point, this is still an assumption). This would set $p=16-7=9.$
Since $\frac{5}{9} < \frac{9}{16} < \frac{4}{7},$ the minimum difference is $\boxed{7}.$

Answer: A 7
Problem: Let $p$ and $q$ be positive integers such that \[\frac{5}{9} < \frac{p}{q} < \frac{4}{7}\] and $q$ is as small as possible. What is $q-p$
$\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19$

Solution: Assume that the difference $\frac{p}{q} - \frac{5}{9}$ results in a fraction of the form $\frac{1}{9q}.$ Then, \[9p - 5q = 1.\] Also assume that the difference $\frac{4}{7} - \frac{p}{q}$ results in a fraction of the form $\frac{1}{7q}.$ Then, \[4q - 7p = 1.\] Solving the system of equations yields $q=16$ and $p=9.$ Therefore, the answer is $\boxed{7}.$

Answer: A 7
Problem: In $\triangle{ABC}$ with side lengths $AB = 13$ $AC = 12$ , and $BC = 5$ , let $O$ and $I$ denote the circumcenter and incenter, respectively. A circle with center $M$ is tangent to the legs $AC$ and $BC$ and to the circumcircle of $\triangle{ABC}$ . What is the area of $\triangle{MOI}$
$\textbf{(A)}\ \frac52\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ \frac{13}{4}\qquad\textbf{(E)}\ \frac72$

Solution: In this solution, let the brackets denote areas.
We place the diagram in the coordinate plane: Let $A=(12,0),B=(0,5),$ and $C=(0,0).$
Since $\triangle ABC$ is a right triangle with $\angle ACB=90^\circ,$ its circumcenter is the midpoint of $\overline{AB},$ from which $O=\left(6,\frac52\right).$ Note that the circumradius of $\triangle ABC$ is $\frac{13}{2}.$
Let $s$ denote the semiperimeter of $\triangle ABC.$ The inradius of $\triangle ABC$ is $\frac{[ABC]}{s}=\frac{30}{15}=2,$ from which $I=(2,2).$
Since $\odot M$ is also tangent to both coordinate axes, its center is at $M=(a,a)$ and its radius is $a$ for some positive number $a.$ Let $P$ be the point of tangency of $\odot O$ and $\odot M.$ As $\overline{OP}$ and $\overline{MP}$ are both perpendicular to the common tangent line at $P,$ we conclude that $O,M,$ and $P$ are collinear. It follows that $OM=OP-MP,$ or \[\sqrt{(a-6)^2+\left(a-\frac52\right)^2}=\frac{13}{2}-a.\] Solving this equation, we have $a=4,$ from which $M=(4,4).$
Finally, we apply the Shoelace Theorem to $\triangle MOI:$ \[[MOI]=\frac12\left|\left(4\cdot\frac52+6\cdot2+2\cdot4\right)-\left(4\cdot6+\frac52\cdot2+2\cdot4\right)\right|=\boxed{72}.\] Remark

Answer: E 72
Problem: Consider polynomials $P(x)$ of degree at most $3$ , each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ . How many such polynomials satisfy $P(-1) = -9$
$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$

Solution: Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that \[P(-1)=-a+b-c+d=-9.\] Let $a'=9-a$ and $c'=9-c.$ Note that both of $a'$ and $c'$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a',b,c',d)$ have one-to-one correspondence.
We rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or \[a'+b+c'+d=9.\] By the stars and bars argument, there are $\binom{9+4-1}{4-1}=\boxed{220}$ ordered quadruples $(a',b,c',d).$

Answer: D 220
Problem: Consider polynomials $P(x)$ of degree at most $3$ , each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ . How many such polynomials satisfy $P(-1) = -9$
$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$

Solution: Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $P(-1)=-a+b-c+d=-9,$ which rearranges to \[b+d+9=a+c.\] Note that $b+d+9$ is an integer from $9$ through $27,$ and $a+c$ is an integer from $0$ through $18.$ Therefore, both of $b+d+9$ and $a+c$ are integers from $9$ through $18.$ We construct the following table: \[\begin{array}{c|c|c|c||c} & & & & \\ [-2.5ex] \boldsymbol{b+d} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(b,d)} & \boldsymbol{a+c} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(a,c)} & \boldsymbol{\#}\textbf{ of Ordered Quadruples }\boldsymbol{(a,b,c,d)} \\ [0.5ex] \hline & & & & \\ [-2ex] 0 & 1 & 9 & 10 & 1\cdot10=10 \\ 1 & 2 & 10 & 9 & \phantom{0}2\cdot9=18 \\ 2 & 3 & 11 & 8 & \phantom{0}3\cdot8=24 \\ 3 & 4 & 12 & 7 & \phantom{0}4\cdot7=28 \\ 4 & 5 & 13 & 6 & \phantom{0}5\cdot6=30 \\ 5 & 6 & 14 & 5 & \phantom{0}6\cdot5=30 \\ 6 & 7 & 15 & 4 & \phantom{0}7\cdot4=28 \\ 7 & 8 & 16 & 3 & \phantom{0}8\cdot3=24 \\ 8 & 9 & 17 & 2 & \phantom{0}9\cdot2=18 \\ 9 & 10 & 18 & 1 & 10\cdot1=10 \end{array}\] We sum up the counts in the last column to get the answer $2\cdot(10+18+24+28+30)=\boxed{220}.$

Answer: D 220
Problem: Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$
$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$

Solution: This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
Without the loss of generality, let $AC=BC=1.$ As shown below, we place Earth in the $xyz$ -plane with $C=(0,0,0)$ such that the positive $x$ -axis runs through $A,$ the positive $y$ -axis runs through $0^\circ$ latitude and $160^\circ \text{ W}$ longitude, and the positive $z$ -axis runs through the North Pole. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids;  currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^B--C--D--cycle); label("$A$",A,5*dir((2.5,-3,0))); label("$B$",B,3*dir(B)); label("$C$",C,1.5*(1,0,-1)); label("$D$",D,3*(-1/2,-1/2,0)); draw((-1.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-1.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-1.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); label("$x$",(1.5,0,0),2*dir((1.5,0,0))); label("$y$",(0,1.5,0),3*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); [/asy] It follows that $A=(1,0,0)$ and $D=(-t,t,0)$ for some positive number $t.$ Since $\triangle BCD$ is an isosceles right triangle, we have $B=\left(-t,t,\sqrt{2}t\right).$ By the Distance Formula, we get $(-t)^2+t^2+\left(\sqrt{2}t\right)^2=1,$ from which $t=\frac12.$
As $\vec{A} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $\vec{B} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix},$ we obtain \[\cos\angle ACB=\frac{\vec{A}\bullet\vec{B}}{\left\lVert\vec{A}\right\rVert\left\lVert\vec{B}\right\rVert}=-\frac12\] by the dot product, so $\angle ACB=\boxed{120}$ degrees.

Answer: C 120