Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$

Solution: Consider the arrangement of the prime factors of 96 in a line $(2,2, 2, 2, 2, 3)$ .  An arrangement of factors can be created by placing "dividers" to group primes.  For example, $(2, 2, |, 2, 2, 2, |, 3)$ is equivalent to the arrangement $4 \cdot 8 \cdot 3$ .  Because there are $6$ ways to order the prime factors, and $2^5$ ways to place dividers, this gives us an initial $6 \cdot  2^5$ ways to arrange divisors.
However, through this method, we overcount cases where $3$ is combined with another factor.  For example, the arrangement $4 \cdot 6 \cdot 4$ can be written as $(2, 2, |, 2, 3, |, 2, 2)$ or $(2, 2, |, 3, 2, |, 2, 2)$ .  Precisely, we double count any case with $6$ as a factor, triple count any case with $12$ , quadruple count any case with $24$ , etc.
Now, consider all cases where $3$ must be grouped with at least one $2$ .  This can be expressed in the same "line" format as $(2, 2, 2, 2, 6)$ , where dividers can again be placed to group divisors.  In this case, there are $5$ ways to order divisors, and $2^4$ ways to place dividers, so we have an $5 \cdot 2^4$ possible sequences for this case.  Notice that in this format, we double count cases where $12$ is a factor, we triple count cases where $24$ is a factor, etc.  Precisely, for any case counted $n$ times in the first step, it is counted $n - 1$ times in this step.  Thus, if we subtract, we count each case exactly once.
So, we get:
$6 \cdot  2^5 - 5 \cdot 2^4 = \boxed{112}$ . ~ hdai1122

Answer: A 112
Problem: Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$

Solution: First we factor $32$ into $m$ numbers $g_1, \cdots, g_m$ where $g_i>1,i=1,\ldots,m$ . By applying stars and bars there are $\binom{5-1}{m-1}$ ways. Then we can either insert $3$ into each of the $m+1$ spaces between (or beyond) $g_i$ 's, or multiply it to one of the $g_i$ 's, a total of $2m+1$ ways. Hence the answer to the problem is
\[\sum_{m=1}^5 (2m+1)\binom{5-1}{m-1} = \sum_{n=0}^4 (2n+3) \binom{4}{n} = 8\sum_{n=0}^4 \frac{n}{4} \binom{4}{n} + 3 \sum_{n=0}^4 \binom{4}{n} = 8 \sum_{n=0}^4 \binom{3}{n-1} + 3\sum_{n=0}^4 \binom{4}{n} = 8 \cdot 2^3 + 3\cdot 2^4 = \boxed{112}.\]

Answer: A 112
Problem: Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$

Solution: Note that $96 = 2^5 \cdot 3$ $D(n)$ depends on dividing $2^5$ into different terms, which is the integer partition of $5$
Divide $96$ into $1$ term:
There is only one way. $\underline{\textbf{1}}$
Divide $96$ into $2$ terms:
$2 + 8 = \underline{\textbf{10}}$
Divide $96$ into $3$ terms:
$12 + 18 = \underline{\textbf{30}}$
Divide $96$ into $4$ terms:
$24 + 16 = \underline{\textbf{40}}$
Divide $96$ into $5$ terms:
When dividing $5$ into $5$ parts there are $2$ cases.
$20 + 5 = \underline{\textbf{25}}$
Divide $96$ into $6$ terms:
$5 = 1 + 1 + 1 + 1 + 1$ . The number of arrangements of $(2, 2, 2, 2, 2, 3)$ is $\frac{6!}{5!} = \underline{\textbf{6}}$
$1 + 10 + 30 + 40 + 25 + 6 = \boxed{112}$

Answer: A 112
Problem: Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$

Solution: Ignore the $3$ first and first count $2^{x_1+x_2+...+x^n}=32$ which $x_1+x_2+...+x_n=5$ . This implies that $n$ is less than or equal to $5$ . Now, we can see that $3$ can lie between the two $x_i, x_{i+1}$ , or contribute to one of them. This gives $2k+1$ if $x_1+...+x_k=5$ . Now, just sum up gives $\binom{5-1}{5-1}\cdot 11+\binom{5-1}{4-1}\cdot 9+\binom{5-1}{3-1}\cdot 7+\binom{5-1}{2-1}\cdot 5+\binom{5-1}{1-1}\cdot 3=\boxed{112}$

Answer:  112
Problem: Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$

Solution: Consider how we partition the factors of $96 = 2^5\cdot 3$ . For each $k$ , there are two cases. Either we can put the $2$ s into $k$ nonzero parts, so that the $3$ shares a partition with some $2$ s, which can be done in $k\binom{5-(k-1)+(k-1)-1}{k-2} = k\binom{4}{k-2}$ ways, or we can put the $2$ s into $k-1$ nonzero parts and put the $3$ in its own partition, which can be done in $k\binom{5-k+k-1}{k-1} = k\binom{4}{k-1}$ ways. Summing over all $k$ , we have \[\sum_{k=1}^6 \binom{4}{k-2} + \sum_{k=1}^6 \binom{4}{k-1}.\]
But $\sum_{k=1}^6 \binom{4}{k-2} = 2\binom{4}{0}+3\binom{4}{1} + \dots + 6\binom{4}{4} = 4 \cdot \sum_{i=0}^4 \binom{4}{i} = 4 \cdot 2^4 = 64$ . Similarly, $\sum_{k=1}^6 \binom{4}{k-1} = 3\cdot 2^4 = 48$ . So our answer is $64+48 = \boxed{112}$

Answer: A 112
Problem: What is the value of \[2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?\]
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution: $2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9$
$=  1+1 = \boxed{2}$

Answer: C 2
Problem: Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$

Solution: Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,
\[A-1 = 5(B-1)\] and \[A = B^2.\]
Substituting the second equation into the first gives us
\[B^2-1 = 5(B-1).\]
By using difference of squares and dividing, $B=4.$ Moreover, $A=B^2=16.$
The answer is $16-4 = 12 \implies \boxed{12}$

Answer: D 12
Problem: Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$

Solution: Simple guess and check works. Start with all the square numbers - $1$ $4$ $9$ $16$ $25$ $36$ , etc. (probably stop at around $100$ since at that point it wouldn't make sense). If Ana is $9$ , then Bonita is $3$ , so in the previous year, Ana's age was $4$ times greater than Bonita's. If Ana is $16$ , then Bonita is $4$ , and Ana's age was $5$ times greater than Bonita's in the previous year, as required. The difference in the ages is $16 - 4 = 12. \boxed{12}$

Answer: D 12
Problem: Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$

Solution: The second sentence of the problem says that Ana's age was once $5$ times Bonita's age. Therefore, the difference of the ages $n$ must be divisible by $4.$ The only answer choice which is divisible by $4$ is $12 \rightarrow \boxed{12}.$

Answer: D 12
Problem: A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn?
$\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91$

Solution: We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee. 
Namely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls, without drawing $15$ balls of any one color. Drawing one more ball guarantees that we will get $15$ balls of one color — either red, green, or yellow. Thus, the answer is $75 + 1 = \boxed{76}$

Answer: B 76
Problem: What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$

Solution: We might at first think that the answer would be $9$ , because $1+2+3 \dots +n = 45$ when $n = 9$ . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \cdots, 44, 45$ cancels out except $45$ .  Thus, the answer is, intuitively, $\boxed{90}$ integers.

Answer: D 90
Problem: What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$

Solution: To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be $\frac12$ if the middle two numbers are $0$ and $1$ , so the answer is $\frac{45}{\frac12}=\boxed{90}$

Answer: D 90
Problem: For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

Solution: This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer is $\boxed{3}$

Answer: C 3
Problem: For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

Solution: We can use a process of elimination. Going down the list, we can see a square obviously works. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point that is equidistant, but isosceles trapezoids do have such a point, so the answer is $\boxed{3}$

Answer: C 3
Problem: For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

Solution: The perpendicular bisector of a line segment is the locus of all points that are equidistant from the endpoints. The question then boils down to finding the shapes where the perpendicular bisectors of the sides all intersect at a point. This is true for a square, rectangle, and isosceles trapezoid, so the answer is $\boxed{3}$

Answer: C 3
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\frac{x}{2} + b$ implies $2=\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ . 
Now we find the intersection points between each of the lines with $y=-x+10$ , which are $(6,4)$ and $(4,6)$ . Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base $2\sqrt{2}$ and height $3\sqrt{2}$ , whose area is $\boxed{6}$

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: $(2,2)$ $(6,4)$ $(4,6)$ . Now, using the Shoelace Theorem , we can directly find that the area is $\boxed{6}$

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at $(4, 6)$ and $(6, 4)$ . Then apply Heron's Formula: the semi-perimeter will be $s = \sqrt{2} + \sqrt{20}$ , so the area reduces nicely to a difference of squares, making it $\boxed{6}$

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . We can now draw the bounding square with vertices $(2, 2)$ $(2, 6)$ $(6, 6)$ and $(6, 2)$ , and deduce that the triangle's area is $16-4-2-4=\boxed{6}$

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . Using graph paper, we can see that this triangle has $6$ boundary lattice points and $4$ interior lattice points. By Pick's Theorem, the area is $\frac62 + 4 - 1 = \boxed{6}$

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45\] Therefore, $\sin A = \sqrt{1 - \cos^2 A} = \frac35$ , so the area is $\frac12 bc \sin A = \frac12 \cdot 2\sqrt5 \cdot 2\sqrt5 \cdot \frac35 = \boxed{6}$

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. \[\frac12\begin{Vmatrix} 2&2&1\\ 4&6&1\\ 6&4&1\\ \end{Vmatrix} = \frac12|-12| = \frac12 \cdot 12 = \boxed{6}\]

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . Then vectors $\overrightarrow{AB} = \langle 2, 4 \rangle$ and $\overrightarrow{AC} = \langle 4, 2 \rangle$ . The area of the triangle is half the magnitude of the cross product of these two vectors. \[\frac12\begin{Vmatrix} i&j&k\\ 2&4&0\\ 4&2&0\\ \end{Vmatrix} = \frac12|-12k| = \frac12 \cdot 12 = \boxed{6}\]

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . By the Pythagorean theorem, this is an isosceles triangle with base $2\sqrt2$ and equal length $2\sqrt5$ . The area of an isosceles triangle with base $b$ and equal length $l$ is $\frac{b\sqrt{4l^2-b^2}}{4}$ . Plugging  in $b = 2\sqrt2$ and $l = 2\sqrt5$ \[\frac{2\sqrt2 \cdot \sqrt{80-8}}{4} = \frac{\sqrt{576}}{4} = \frac{24}{4} = \boxed{6}\]

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45\] Therefore, $\sin A = \sqrt{1 - \cos^2 A} = \frac35$ . By the extended Law of Sines, \[2R = \frac{a}{\sin A} = \frac{2\sqrt2}{\frac35} = \frac{10\sqrt2}{3}\] \[R = \frac{5\sqrt2}{3}\] Then the area is $\frac{abc}{4R} = \frac{2\sqrt2 \cdot 2\sqrt5^2}{4 \cdot \frac{5\sqrt2}{3}} = \boxed{6}$

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: The area of a triangle formed by three lines, \[a_1x + a_2y + a_3 = 0\] \[b_1x + b_2y + b_3 = 0\] \[c_1x + c_2y + c_3 = 0\] is the absolute value of \[\frac12 \cdot \frac{1}{(b_1c_2-b_2c_1)(a_1c_2-a_2c_1)(a_1b_2-a_2b_1)} \cdot \begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\\ \end{vmatrix}^2\] Plugging in the three lines, \[-x + 2y - 2 = 0\] \[-2x + y + 2 = 0\] \[x + y - 10 = 0\] the area is the absolute value of \[\frac12 \cdot \frac{1}{(-2-1)(-1-2)(-1+4)} \cdot \begin{vmatrix} -1&2&-2\\ -2&1&2\\ 1&1&-10\\ \end{vmatrix}^2 = \frac12 \cdot \frac{1}{27} \cdot 18^2 = \boxed{6}\] Source: Orrick, Michael L. “THE AREA OF A TRIANGLE FORMED BY THREE LINES.” Pi Mu Epsilon Journal, vol. 7, no. 5, 1981, pp. 294–298. JSTOR, www.jstor.org/stable/24336991.

Answer: C 6
Problem: Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10  ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$

Solution: Like in other solutions, we find that our triangle is isosceles with legs of $2\sqrt5$ and base $2\sqrt2$ . Then, the semi - perimeter of our triangle is, \[\frac{4\sqrt5+2\sqrt2}{2} = 2\sqrt5 + \sqrt2.\] Applying Heron's formula, we find that the area of this triangle is equivalent to \[\sqrt{{(2\sqrt5+\sqrt2)}{(2\sqrt5-\sqrt2)}{(2)}} = \sqrt{{(20-2)}{(2)}} = \boxed{6}.\]

Answer: C 6
Problem: What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$

Solution: The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ , and we want this not to be a divisor of $n!$ (the product of the first $n$ positive integers). Notice that if and only if $n+1$ were composite, all of its factors would be less than or equal to $n$ , which means they would be able to cancel with the factors in $n!$ . Thus, the sum of $n$ positive integers would be a divisor of $n!$ when $n+1$ is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, $n+1$ must instead be prime. The greatest three-digit integer that is prime is $997$ , so we subtract $1$ to get $n=\boxed{996}$

Answer: B 996
Problem: What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$

Solution: As in Solution 1, we deduce that $n+1$ must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of $n$ . Choices $A$ $C$ , and $E$ don't work because $n+1$ is even, and all even numbers are divisible by two, which makes choices $A$ $C$ , and $E$ composite and not prime. Choice $D$ also does not work since $999$ is divisible by $9$ , which means it's a composite number and not prime. Thus, the correct answer must be $\boxed{996}$

Answer: B 996