Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]
$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

Solution: Extend $\overline{AD}$ and $\overline{BE}$ to meet at $X$ . Drop an altitude from $F$ to $\overline{CE}$ and call it $h$ . Also, call $\overline{CE}$ $x$ . As stated before, we have $\triangle ABF \sim \triangle CEF$ , so the ratio of their heights is in a $1:2$ ratio, making the altitude from $F$ to $\overline{AB}$ $2h$ . Note that this means that the side of the square is $3h$ . In addition, $\triangle XDE \sim \triangle XAB$ by AA Similarity in a $1:2$ ratio. This means that the side length of the square is $2x$ , making $3h=2x$
Now, note that $[ADEF]=[XAB]-[XDE]-[ABF]$ . We have $[\triangle XAB]=(4x)(2x)/2=4x^2,$ $[\triangle XDE]=(x)(2x)/2=x^2,$ and $[\triangle ABF]=(2x)(2h)/2=(2x)(4x/3)/2=4x^2/3.$ Subtracting makes $[ADEF]=4x^2-x^2-4x^2/3=5x^2/3.$ We are given that $[ADEF]=45,$ so $5x^2/3=45 \Rightarrow x^2=27.$ Therefore, $x= 3 \sqrt{3},$ so our answer is $(2x)^2=4x^2=4(27)=\boxed{108}.$

Answer: B 108
Problem: Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$
[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]
$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

Solution: Solution with Cartesian and Barycentric Coordinates:
We start with the following:
Claim: Given a square $ABCD$ , let $E$ be the midpoint of $\overline{DC}$ and let $BE\cap AC = F$ . Then $\frac {AF}{FC}=2$
Proof: We use Cartesian coordinates. Let $D$ be the origin, $A=(0,1),C=(0,1),B=(1,1)$ . We have that $\overline{AC}$ and $\overline{EB}$ are governed by the equations $y=-x+1$ and $y=2x-1$ , respectively. Solving, $F=\left(\frac{2}{3},\frac{1}{3}\right)$ . The result follows. $\square$
Now, we apply Barycentric Coordinates w.r.t. $\triangle ACD$ . We let $A=(1,0,0),D=(0,1,0),C=(0,0,1)$ . Then $E=(0,\tfrac 12,\tfrac 12),F=(\tfrac 13,0,\tfrac 23)$
In the barycentric coordinate system, the area formula is $[XYZ]=\begin{vmatrix} x_{1} &y_{1}  &z_{1} \\ x_{2} &y_{2}  &z_{2} \\   x_{3}& y_{3} & z_{3} \end{vmatrix}\cdot [ABC]$ where $\triangle XYZ$ is a random triangle and $\triangle ABC$ is the reference triangle. Using this, we find that \[\frac{[FEC]}{[ACD]}=\begin{vmatrix} 0&0&1\\ 0&\tfrac 12&\tfrac 12\\ \tfrac 13&0&\tfrac 23 \end{vmatrix}=\frac16.\] Let $[FEC]=x$ so that $[ACD]=45+x$ . Then, we have $\frac{x}{x+45}=\frac 16 \Rightarrow x=9$ , so the answer is $2(45+9)=\boxed{108}$

Answer:  108
Problem: From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]
$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$

Solution: Choose side "lengths" $a,b,c$ for the triangle, where "length" is how many vertices of the octagon are skipped between vertices of the triangle, starting from the shortest side, and going clockwise, and choosing $a=b$ if the triangle is isosceles: $a+b+c=5$ , where either [ $a\leq b$ and $a < c$ ] or [ $a=b=c$ (but this is impossible in an octagon)].
Options are: $a=0$ with $b,c$ in { 0,5 ; 1,4 ; 2,3 ; 3,2 ; 4,1 }, and $a=1$ with { 1,3 ;  2,2} $5/7$ of these have a side with length 1, which corresponds to an edge of the octagon. So, our answer is $\boxed{57}$

Answer: D 57
Problem: From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]
$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$

Solution: We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent.
If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and $4$ places to put the remaining point, so we have $8\cdot4$ choices. The total amount of choices is ${8 \choose 3} = 8\cdot7$
Thus, our answer is $\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{57}$

Answer: D 57
Problem: From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]
$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$

Solution: We can decide $2$ adjacent points with $8$ choices. The remaining point will have $6$ choices. However, we have counted the case with $3$ adjacent points twice, so we need to subtract this case once. The case with the $3$ adjacent points has $8$ arrangements, so our answer is
$\frac{8\cdot6-8}{{8 \choose 3 }}$ $=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{57}$

Answer: D 57
Problem: From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]
$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$

Solution: Let $1$ point of the triangle be fixed at the top. Then, there are ${7 \choose 2} = 21$ ways to choose the other $2$ points. There must be $3$ spaces in the points and $3$ points themselves. This leaves $2$ extra points to be placed anywhere. By stars and bars, there are $3$ triangle points ( $n$ ) and $2$ extra points ( $k-1$ ) distributed so by the stars and bars formula, ${n+k-1 \choose k-1}$ , there are ${4 \choose 2} = 6$ ways to arrange the bars and stars. Thus, the probability is $\frac{(21 - 6)}{21} = \boxed{57}$

Answer: D 57
Problem: From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?
[asy] size(3cm); pair A[]; for (int i=0; i<9; ++i) {   A[i] = rotate(22.5+45*i)*(1,0); } filldraw(A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,gray,black); for (int i=0; i<8; ++i) { dot(A[i]); } [/asy]
$\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}$

Solution: We select a vertex of the octagon; this will be the first vertex of our triangle. Define the $distance$ of a vertex from another to be the minimum number of edges that one must travel on to get from one vertex to the other. There are three distinct cases; the second vertex is a distance of 1 away from the selected vertex (i.e. they are adjacent), the second vertex is a distance of 2 away from the selected vertex, or the second vertex is a distance of 3 or more away from the selected vertex. We consider each of these cases separately.
Case 1: The first two chosen vertices are adjacent.
There is a $\frac{2}{7}$ chance of selecting a point that is adjacent to the one we have chosen. In this case, any choice of the third vertex will result in a triangle that shares at least one side with the given octagon. Thus, this case has a $\frac{2}{7}$ case of giving us a triangle that fulfills the conditions given in the problem.
Case 2: There is a distance of 2 between the first two chosen vertices.
There is a $\frac{2}{7}$ chance of selecting a vertex that is a distance of 2 away from the first vertex. In this case, there are three vertices that will create a triangle that satisfies the condition in the problem (the one that is adjacent to both of the first two selected vertices and the two that are adjacent to only one of the first two selected vertices). There is a $\frac{1}{2}$ chance of selecting one of these three vertices from the remaining six. Thus, this case has a $\frac{2}{7} \cdot \frac{1}{2} = \frac{1}{7}$ chance of giving us a triangle that fulfills the conditions given in the problem.
Case 3: There is a distance of 3 or more between the first two chosen vertices.
There is a $\frac{3}{7}$ chance of selecting a vertex that is a distance of 3 or more away from the first vertex. In this case, there are four vertices (the two adjacent to each of the first two vertices) that may be chosen to satisfy the problem's condition. There is a $\frac{2}{3}$ chance of selecting one of these four vertices from the remaining six. Thus, this case has a $\frac{3}{7} \cdot \frac{2}{3} = \frac{2}{7}$ chance of giving us a triangle that fulfills the conditions given in the problem.
Summing the probabilities from each of the individual cases, we find that there is a $\frac{1}{7} + \frac{2}{7} + \frac{2}{7} = \boxed{57}$ chance of acquiring a triangle which shares at least one side with the octagon.

Answer: D 57
Problem: How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$

Solution: We compute $2^8+1=257$ . We're all familiar with what $6^3$ is, namely $216$ , which is too small. The smallest cube greater than it is $7^3=343$ $2^{18}+1$ is too large to calculate, but we notice that $2^{18}=(2^6)^3=64^3$ , which therefore will clearly be the largest cube less than $2^{18}+1$ . So, the required number of cubes is $64-7+1= \boxed{58}$

Answer: E 58
Problem: How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$

Solution: First, $2^8+1=257$ . Then, $2^{18}+1=262145$ . Now, we can see how many perfect cubes are between these two parameters. By guessing and checking, we find that it starts from $7$ and ends with $64$ . Now, by counting how many numbers are between these, we find the answer to be $\boxed{58}$

Answer: E 58
Problem: How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$

Solution: First, we realize that question writers like to trick us. We know that most people will be calculating the lowest and highest number whose cubes are within the range. The answer will be the highest number $-$ the lowest number $+ 1$ . People will forget the $+1$ so the only possibilities are C and E. We can clearly see that C is too small so our answer is $\boxed{58}$

Answer: E 58
Problem: How many perfect cubes lie between $2^8+1$ and $2^{18}+1$ , inclusive?
$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$

Solution: There is not so much guessing and checking after we find that it starts from $7$ because $7^3=343$ , which is over $2^8+1=257$ . We can start guessing with the 10, answer C, as it is the middle value. Adding 10 to 7 gives us 17 and $17^3 = 4,913$ , which is a bit low. So, we move "up" to 57, answer D. Adding 57 to 7 gives us 64 and $64^3 = 262,144$ , which is perfect for the $2^{18}+1=262,145$ . But we are not done. Since it is inclusive, we must add 1 to this solution, which give us $\boxed{58}$

Answer: E 58
Problem: Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?
[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label("Colby", (-0.5, 0.5)); label("25\%", (-0.5, 0.3)); label("Alicia", (0.7, 0.2)); label("45\%", (0.7, 0)); label("Brenda", (-0.5, -0.4)); label("30\%", (-0.5, -0.6)); [/asy]
$\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120$

Solution: Let $x$ be the total amount of votes casted. From the chart, Brenda received $30\%$ of the votes and had $36$ votes. We can express this relationship as $\frac{30}{100}x=36$ . Solving for $x$ , we get $x=\boxed{120}.$

Answer: E 120
Problem: Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?
[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label("Colby", (-0.5, 0.5)); label("25\%", (-0.5, 0.3)); label("Alicia", (0.7, 0.2)); label("45\%", (0.7, 0)); label("Brenda", (-0.5, -0.4)); label("30\%", (-0.5, -0.6)); [/asy]
$\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120$

Solution: We're being asked for the total number of votes cast -- that represents $100\%$ of the total number of votes. Brenda received $36$ votes, which is $\frac{30}{100} = \frac{3}{10}$ of the total number of votes. Multiplying $36$ by $\frac{10}{3},$ we get the total number of votes, which is $\boxed{120}.$

Answer: E 120
Problem: Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?
[asy] draw((-1,0)--(0,0)--(0,1)); draw((0,0)--(0.309, -0.951)); filldraw(arc((0,0), (0,1), (-1,0))--(0,0)--cycle, lightgray); filldraw(arc((0,0), (0.309, -0.951), (0,1))--(0,0)--cycle, gray); draw(arc((0,0), (-1,0), (0.309, -0.951))); label("Colby", (-0.5, 0.5)); label("25\%", (-0.5, 0.3)); label("Alicia", (0.7, 0.2)); label("45\%", (0.7, 0)); label("Brenda", (-0.5, -0.4)); label("30\%", (-0.5, -0.6)); [/asy]
$\textbf{(A) }70 \qquad \textbf{(B) }84 \qquad \textbf{(C) }100 \qquad \textbf{(D) }106 \qquad \textbf{(E) }120$

Solution: If $36$ votes is $\frac{3}{10}$ of all the votes, we can divide that by $3$ to get $12$ as 10%, and then we can multiply the $12$ by $10$ to get to $120$ . So, the answer is $\boxed{120}.$

Answer: E 120
Problem: What is the value of the expression $\sqrt{16\sqrt{8\sqrt{4}}}$
$\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16$

Solution: $\sqrt{16\sqrt{8\sqrt{4}}}$ $\sqrt{16\sqrt{8\cdot 2}}$ $\sqrt{16\sqrt{16}}$ $\sqrt{16\cdot 4}$ $\sqrt{64}$ $\boxed{8}$

Answer: C 8
Problem: When $0.000315$ is multiplied by $7,928,564$ the product is closest to which of the following?
$\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000$

Solution: We can approximate $7,928,564$ to $8,000,000$ and $0.000315$ to $0.0003.$ Multiplying the two yields $2400.$ Thus, it shows our answer is $\boxed{2400}.$

Answer: D 2400
Problem: What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$

Solution: Directly calculating:
We evaluate both the top and bottom: $\frac{40320}{36}$ . This simplifies to $\boxed{1120}$

Answer: B 1120
Problem: What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$

Solution: It is well known that the sum of all numbers from $1$ to $n$ is $\frac{n(n+1)}{2}$ . Therefore, the denominator is equal to $\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 6$ . Now, we can cancel the factors of $2$ $3$ , and $6$ from both the numerator and denominator, only leaving $8 \cdot 7 \cdot 5 \cdot 4 \cdot 1$ . This evaluates to $\boxed{1120}$

Answer: B 1120
Problem: What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$
$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$

Solution: First, we evaluate $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8$ to get 36. We notice that 36 is 6 squared, so we can factor the denominator like $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}$ then cancel the 6s out,\ to get $\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}$ . Now that we have escaped fraction form, we multiply $4 \cdot 5 \cdot 7 \cdot 8$ . Multiplying these, we get $\boxed{1120}$

Answer: B 1120
Problem: If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$

Solution: The sum of the ratios is $10$ .  Since the sum of the angles of a triangle is $180^{\circ}$ , the ratio can be scaled up to $54:54:72$ $(3\cdot 18:3\cdot 18:4\cdot 18).$ The numbers in the ratio $54:54:72$ represent the angles of the triangle.  The question asks for the largest, so the answer is $\boxed{72}$

Answer: D 72
Problem: If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$

Solution: We can denote the angles of the triangle as $3x$ $3x$ $4x$ . Due to the sum of the angles in a triangle, $3x+3x+4x=180^{\circ}\implies x=18^{\circ}$ . The greatest angle is $4x$ and after substitution we get $\boxed{72}$

Answer: D 72
Problem: If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$

Solution: We know the longest side must be denoted by the 4 in the ratio. Since the ratio is 3:3:4, we know that the longest side must be $\frac{4}{3+3+4}$ of the degree total (which for all triangles is 180). Thus, \[\frac{4}{3+3+4} \cdot 180 = \frac{4}{10} \cdot 180 = \boxed{72}\]

Answer: D 72
Problem: If the degree measures of the angles of a triangle are in the ratio $3:3:4$ , what is the degree measure of the largest angle of the triangle?
$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$

Solution: Since we see the ratio is $3:3:4$ , we can rule out the answer of ${\textbf{(E) }90}$ because the numbers in the ratio are too big to have $90^\circ$ . Also, we are trying to find the largest angle and all the other angles except for 72 are too small to be the largest angle. Using all this, our answer is $\boxed{72}$

Answer: D 72
Problem: Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$

Solution: To check, if a number is divisible by 19, take its unit digit and multiply it by 2, then add the result to the rest of the number, and repeat this step until the number is reduced to two digits. If the result is divisible by 19, then the original number is also divisible by 19. Or we could just try to divide the example number the problem gave us by 19.
After we got 19 eliminated, we can see that the other options have a lot of 1's in them. The divisibility rule for 11 is add alternating digits up, then take the difference of them. The example number works like that. If we add variables, ABCDEF to make number ABCABC, we can see that (A+C+B) - (B+A+C) = 0. Which is divisible by 11, so our answer choice is $\boxed{11}$

Answer: A 11
Problem: Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$

Solution: We are given one of the numbers that can represent $Z$ , so we can just try out the options to see which one is a factor of $247247$ . We get $\boxed{11}$

Answer: A 11
Problem: Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$

Solution: To find out when a number is divisible by 11, place plus and minus signs alternatively in front of every digit, then calculate the result. If this result is divisible by 11 (including 0), the number is divisible by 11; otherwise, the number isn’t divisible by 11. In this case, $+2-4+7-2+4-7=0$ . Because the result is 0, the number 247247 is divisible by 11  and so we get $\boxed{11}$

Answer: A 11
Problem: Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$

Solution: Similar to solution 1, let $Z=ABCABC$ . To prove it is divisible by 11, we can compute its alternating sum, which is $A-B+C-A+B-C=0$ , which is divisible by 11. Therefore, the answer is $\boxed{11}$

Answer: A 11
Problem: Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$

Solution: We can find that all numbers like $Z$ are divisible by 1001. 1001 is divisible by 11 because when we divide it, we get a whole number. So, the answer is $\boxed{11}$

Answer: A 11