Answer: 120.0

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 $x=\boxed{120}.$

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 $\boxed{120}.$

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: 2.0

Alina writes the numbers $1, 2, \dots , 9$ on separate cards, one number per card. She wishes to divide the cards into $3$ groups of $3$ cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

Solution: First, we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. $1 + 2 \cdots + 9 = \frac{9(10)}{2} = 45$ . Then, dividing by $3$ , we have $\frac{45}{3} = 15$ , so each group of $3$ must have a sum of 15. To make the counting easier, we will just see the possible groups 9 can be with. The possible groups 9 can be with 2 distinct numbers are $(9, 2, 4)$ and $(9, 1, 5)$ . Going down each of these avenues, we will repeat the same process for $8$ using the remaining elements in the list. Where there is only 1 set of elements getting the sum of $7$ $8$ needs in both cases. After $8$ is decided, the remaining 3 elements are forced in a group, yielding us an answer of $\boxed{2}$ as our sets are $(9, 1, 5) (8, 3, 4) (7, 2, 6)$ and $(9, 2, 4) (8, 1, 6) (7, 3 ,5)$

Solution: The group with $5$ must have the two other numbers adding up to $10$ , since the sum of all the numbers is $(1 + 2 \cdots + 9) = \frac{9(10)}{2} = 45$ . The sum of the numbers in each group must therefore be $\frac{45}{3}=15$ . We can have $(1, 5, 9)$ $(2, 5, 8)$ $(3, 5, 7)$ , or $(4, 5, 6)$ . With the first group, we have $(2, 3, 4, 6, 7, 8)$ left over. The only way to form a group of $3$ numbers that add up to $15$ is with $(3, 4, 8)$ or $(2, 6, 7)$ . One of the possible arrangements is therefore $(1, 5, 9) (3, 4, 8) (2, 6, 7)$ . Then, with the second group, we have $(1, 3, 4, 6, 7, 9)$ left over. With these numbers, there is no way to form a group of $3$ numbers adding to $15$ . Similarly, with the third group there is $(1, 2, 4, 6, 8, 9)$ left over and we can make a group of $3$ numbers adding to $15$ with $(1, 6, 8)$ or $(2, 4, 9)$ . Another arrangement is $(3, 5, 7) (1, 6, 8) (2, 4, 9)$ . Finally, the last group has $(1, 2, 3, 7, 8, 9)$ left over. There is no way to make a group of $3$ numbers adding to $15$ with this, so the arrangements are $(1, 5, 9) (3, 4, 8) (2, 6, 7)$ and $(3, 5, 7) (1, 6, 8) (2, 4, 9)$ . So，there are $\boxed{2}$ sets that can be formed.

Solution: The sum of the numbers across all equally valued sets is $(1 + 2 \cdots + 9) = \frac{9(10)}{2} = 45$ . The value of the numbers in each set would be $\frac{45}{3} = \textbf{15}$ . We know that the numbers $9$ $8$ , and $7$ must belong in different sets, as putting any $2$ numbers in $1$ set will either pass or match the limit of $15$ per set, and we would then still need to add $1$ more number after that. Note that these numbers must be distinct, as Alina only has $1$ of each number, and order does not matter in the sets. Starting with the set that includes the number $9$ , the next two numbers must add up to $6$ , and there are $\textbf{2}$ ways of doing this $(2,4) (1,5)$ . Note we cannot use any number past $6$ , as those numbers must be used in the other sets. The next set, which includes the number $8$ , must have two numbers that add up to $7$ , and there are $\textbf{3}$ ways to do this $(2,5) (1,6) (3,4)$ . The final set, which includes the number $7$ , must have $2$ numbers that sum up to $8$ , and there are $\textbf{2}$ ways to do this $(2,6) (3,5)$ . Now we have found the number of ways in which each set sums up to $15$ . To find the number of ways in which all three sets sum up to $15$ concurrently, we must take the minimum of $2$ $3$ , and $2$ , which gives us an answer of $\boxed{2}$ triplets of sets with 3 values, in which each set sum to the same amount.

Solution:
Note that each group of numbers should sum to $\frac{1+2+3+4+5+6+7+8+9}{3} = 15.$ Thus, this is equivalent to asking, “How many ways can you fill in a three by three magic square with the integers $1$ through $9$ ?” since we can take the three rows of the magic square as our three groups. If you have closely studied magic squares, you might know that in a three by three magic square that is to be filled in with the integers $1$ through $9$ , the center of the square would be $5$ (the average of the numbers), and the numbers in the corners should be even(*). The such pairs (disregarding order) are $(2,8)$ and $(4,6).$ Let’s fix the position of $2$ to be the top left corner. This would make $8$ in the bottom right corner. We can have either $4$ or $6$ to be in the top right corner, for a total of $\boxed{2}$ such groups of three. (The groups are $(8,3,4) (1,5,7) (6,9,2)$ and $(8,1,6) (3,5,7) (4,9,2).$ )
Note that if we had instead fixed the position of $4$ $6$ , or $8$ , they would correspond to one of the two cases, just in a different configuration.


Answer: 49.0

All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$

Solution: If you draw a clear diagram like the one below, it is easy to see that there are $\boxed{49}$ points.

Solution: Let the number of intersections be $x$ .  We know that $x\le \dbinom{8}{4} = 70$ , as every $4$ vertices on the octagon forms a quadrilateral with intersecting diagonals which is an intersection point.  However, four diagonals intersect in the center, so we need to subtract $\dbinom{4}{2} -1 = 5$ from this count, $70-5 = 65$ .  Note that diagonals like $\overline{AD}$ $\overline{CG}$ , and $\overline{BE}$ all intersect at the same point.  There are $8$ of this type with three diagonals intersecting at the same point, so we need to subtract $2$ of the $\dbinom{3}{2}$ (one is kept as the actual intersection). In the end, we obtain $65 - 16 = \boxed{49}$

Solution:
We know that the amount of intersection points is at most $\dbinom{8}{4} = 70$ , as in solution $2$ . There's probably going to be more than $5$ intersections counted multiple times (to get $\textbf{(B) }65$ ), leading us to the only reasonable answer, $\boxed{49}$ .
-Lcz

Solution: Like solution one, we may draw. Except note that the octagon has eight regions, and each region has an equal number of points, so drawing only one of the eight regions and the intersection points suffices. One of the eight regions contains $8$ points (not including the octagon center). However each adjacent region share one side in common and that side contains $2$ intersection points, so in actuality there are $8 - 2 = 6$ points per region. We multiply this by $8$ to get $6\cdot 8 = 48$ and add the one center point to get $48 + 1 = \boxed{49}$

Solution:
This problem is a counting problem of combinatoric geometry. There are 2 cases for the above diagram:
Case 1: Red Dots
The red dots are the intersection of 3 or more lines. It consists of 8 dots that make up an octagon and 1 dot in the center. Hence, there are $9$ red dots.
Case 2: Blue Dots
The blue dots are the intersection of 2 lines. Each vertex of the octagon has 2 purple lines, 2 green lines, and 1 orange line coming out of it. There are 5 dots of intersection on a purple line, 6 dots on a green line, and 5 dots on an orange line. There are $2 \cdot 5+2 \cdot 6+5=27$ dots that come out of 1 vertex, which includes 7 red dots already counted. So there are $27-7 = 20$ blue dots coming out of 1 vertex. There are 8 vertices, but each blue dot is the intersection of 2 lines, corresponding to $2 \cdot 2 = 4$ vertices. So there are $\frac{20 \cdot 8}{4} = 40$ blue dots.
The number of intersection dots are the sum of the number of red and blue dots. Hence, the answer is $40 + 9 = \boxed{49}$


Answer: 12.0

All lines with equation $ax+by=c$ such that $a,b,c$ form an arithmetic progression pass through a common point. What are the coordinates of that point?
$\textbf{(A) } (-1,2) \qquad\textbf{(B) } (0,1) \qquad\textbf{(C) } (1,-2) \qquad\textbf{(D) } (1,0) \qquad\textbf{(E) } (1,2)$

Solution: If all lines satisfy the condition, then we can just plug in values for $a$ $b$ , and $c$ that form an arithmetic progression. Let's use $a=1$ $b=2$ $c=3$ , and $a=1$ $b=3$ $c=5$ . Then the two lines we get are: \[x+2y=3\] \[x+3y=5\] Use elimination to deduce \[y = 2\] and plug this into one of the previous line equations. We get \[x+4 = 3 \Rightarrow x=-1\] Thus the common point is $\boxed{1,2}$

Solution: We know that $a$ $b$ , and $c$ form an arithmetic progression, so if the common difference is $d$ , we can say $a,b,c = a, a+d, a+2d.$ Now we have $ax+ (a+d)y = a+2d$ , and expanding gives $ax + ay + dy = a + 2d.$ Factoring gives $a(x+y-1)+d(y-2) = 0$ . Since this must always be true (regardless of the values of $a$ and $d$ ), we must have $x+y-1 = 0$ and $y-2 = 0$ , so $x,y = -1, 2,$ and the common point is $\boxed{1,2}$


Answer: 8.0

All of David's telephone numbers have the form $555-abc-defg$ , where $a$ $b$ $c$ $d$ $e$ $f$ , and $g$ are distinct digits and in increasing order, and none is either $0$ or $1$ . How many different telephone numbers can David have?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$

Solution: The only digits available to use in the phone number are $2$ $3$ $4$ $5$ $6$ $7$ $8$ , and $9$ . There are only $7$ spots left among the $8$ numbers, so we need to find the number of ways to choose $7$ numbers from $8$ . The answer is then $\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{8}$


Answer: 4.0

All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

Solution: The $6$ green marbles and yellow marbles form $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles. Now, suppose the total number of marbles is $x$ . We know the number of yellow marbles is $\frac{5}{12}x - 6$ and a positive integer. Therefore, $12$ must divide $x$ . Trying the smallest multiples of $12$ for $x$ , we see that when $x = 12$ , we get there are $-1$ yellow marbles, which is impossible. However when $x = 24$ , there are $\frac{5}{12} \cdot 24 - 6 = \boxed{4}$ yellow marbles, which must be the smallest possible.

Solution: Since $\frac{1}{3}$ of the marbles are blue and $\frac{1}{4}$ are red, it is clear that the total number of marbles must be divisible by $12$ . If there are $12$ marbles, then $4$ are blue, $3$ are red, and $6$ are green, meaning that there are $-1$ yellow marbles. This is impossible. Trying the next multiple of $12$ $24$ , we find that $8$ are green, $6$ are red, and $6$ are green, meaning that the minimum number of yellow marbles is $\boxed{4}$


Answer: 6.0

All of the even numbers from 2 to 98 inclusive, excluding those ending in 0, are multiplied together.  What is the rightmost digit (the units digit) of the product?
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution:
All the tens digits of the product will be irrelevant to finding the units digit.  Thus, we are searching for the units digit of $(2\cdot 4\cdot 6 \cdot 8) \cdot (2 \cdot 4 \cdot 6 \cdot 8) \cdot (2\cdot 4\cdot 6 \cdot 8) \cdot ...$
There will be $10$ groups of $4$ numbers.  The number now can be rewritten as $(2\cdot 4 \cdot 6 \cdot 8)^{10}$
Multiplying, we get $384^{10}$
Again, we can disregard the tens and hundreds digit of $384$ , since we only want the units digit of the number, leaving $4^{10}$
Now, we try to find a pattern to the units digit of $4^n$ .  To compute this quickly, we once again discard all tens digits and higher.
$4^1 = 4$
$4^2 = 4\cdot 4 = 1\underline{6}$ , discard the $1$
$4^3 = 1 \cdot 4 = \underline{4}$
$4^4 = 4 \cdot 4 = 1\underline{6}$ , discard the $1$
$4^5 = 1 \cdot 4 = \underline{4}$
Those equalities are, in reality, congruences $\mod {10}$
Thus, the pattern of the units digits is $\{4, 6, 4, 6, 4, 6, 4, 6, 4, 6\}$ .  The cycle repeats so that term $n$ is the same as term $n+2$ .  The tenth number in the cycle is $6$ , giving an answer of $\boxed{6}$

Solution:
Again, the value we seek is equal to $(2\cdot 4\cdot 6\cdot 8)^{10}\mod 10$ .
We can use equivalence to simplify.
$(2\cdot 4\cdot 6\cdot 8)^{10}\mod 10$
$\equiv (2\cdot 4\cdot (-4)\cdot (-2))^{10} \mod 10$
$\equiv 64^{10} \mod 10$
$\equiv 4^{10} \mod 10$
$\equiv 2^{20} \mod 10$
$\equiv 8^6\cdot 2^2 \mod 10$
$\equiv (-2)^6\cdot 2^2 \mod 10$
$\equiv 2^8 \mod 10$
$\equiv 256 \mod 10$
$\equiv 6 \mod 10$
Thus our answer is $\boxed{6}$


Answer: 24.0

Solution:
Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is $1$ , and thus its base is $2.$
Let $h$ be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and $\triangle{ADE}$ . The top triangle has a base of $3\cdot{2}=6$ , and $DE=4\cdot{2}=8.$ The height of $\triangle{ADE}$ is $h+1$ , therefore our ratio is $\frac{h}{6}=\frac{h+1}{8}$ , which yields $h=3$ as our answer.
To find the area of the trapezoid, we can take the area of $\triangle{ABC}$ and subtract the area of $\triangle{ADE},$ whose base is $8$ and height $3+1=4$ . It follows that the area of $\triangle{ADE}=16$ , and subtracting this from $40$ gives us $40-16=\boxed{24}$


Answer: 24.0

Solution: Let $x$ be the area of $ADE$ . Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$ ). Thus, we have \[x=7+\dfrac{9}{16}x.\] This gives $x=16$ , so the area of $DBCE=40-x=\boxed{24}$


Answer: 24.0

Solution: Let the base length of the small triangle be $x$ . Then, there is a triangle $ADE$ encompassing the 7 small triangles and sharing the top angle with a base length of $4x$ . Because the area is proportional to the square of the side, let the base $BC$ be $\sqrt{40}x$ . The ratio of the area of triangle $ADE$ to triangle $ABC$ is $\left(\frac{4x}{\sqrt{40}x}\right)^2 = \frac{16}{40}$ . The problem says the area of triangle $ABC$ is $40$ , so the area of triangle $ADE$ is $16$ . So the area of trapezoid $DBCE$ is $40 - 16 = \boxed{24}$

Solution:
Notice $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$ .
Let the base of the small triangles of area 1 be $x$ , then the base length of $\Delta ADE=4x$ . Notice, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$ , then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]$ Thus, $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.$

Solution: The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$ . Therefore the area of the trapezoid is $40-16=\boxed{24}$

Solution: You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$ , so to find the area of such trapezoid $BCED$ , we just take $40-16=\boxed{24}$ , like so.

Solution: The combined area of the small triangles is $7$ , and from the fact that each small triangle has an area of $1$ , we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\frac{1}{3}$ , so will their areas have a proportion that is the square of the proportion of their sides, or $\frac {1}{9}$ ). Thus, the combined area of the top triangle and the trapezoid immediately below is $7 + 9 = 16$ . The area of trapezoid $BCED$ is thus the area of triangle $ABC-16 =\boxed{24}$

Solution: You can assume for the base of one of the smaller triangles to be $\frac{1}{a}$ and the height to be $2a$ , giving an area of 1. The larger triangle above the 7 smaller ones then has base $\frac{3}{a}$ and height $6a$ , giving it an area of $9$ . Then the area of triangle $ADE$ is $16$ and $40-16=\boxed{24}$

Solution: You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence $7\cdot\left(1+\left(\frac{9}{16}\right)+\left(\frac{9}{16}\right)^2+\left(\frac{9}{16}\right)^3\dots\right)$ . Using the infinite geometric series sum formula gives us $7\cdot\left(\frac{1}{1-\frac{9}{16}}\right)=7\cdot\frac{16}{7}=16$ . The triangle's area would thus be $40-16=\boxed{24}$


Answer: 24.0

Solution:
Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is $1$ , and thus its base is $2.$
Let $h$ be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and $\triangle{ADE}$ . The top triangle has a base of $3\cdot{2}=6$ , and $DE=4\cdot{2}=8.$ The height of $\triangle{ADE}$ is $h+1$ , therefore our ratio is $\frac{h}{6}=\frac{h+1}{8}$ , which yields $h=3$ as our answer.
To find the area of the trapezoid, we can take the area of $\triangle{ABC}$ and subtract the area of $\triangle{ADE},$ whose base is $8$ and height $3+1=4$ . It follows that the area of $\triangle{ADE}=16$ , and subtracting this from $40$ gives us $40-16=\boxed{24}$


Answer: 24.0

Solution: Let $x$ be the area of $ADE$ . Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$ ). Thus, we have \[x=7+\dfrac{9}{16}x.\] This gives $x=16$ , so the area of $DBCE=40-x=\boxed{24}$


Answer: 24.0

Solution: Let the base length of the small triangle be $x$ . Then, there is a triangle $ADE$ encompassing the 7 small triangles and sharing the top angle with a base length of $4x$ . Because the area is proportional to the square of the side, let the base $BC$ be $\sqrt{40}x$ . The ratio of the area of triangle $ADE$ to triangle $ABC$ is $\left(\frac{4x}{\sqrt{40}x}\right)^2 = \frac{16}{40}$ . The problem says the area of triangle $ABC$ is $40$ , so the area of triangle $ADE$ is $16$ . So the area of trapezoid $DBCE$ is $40 - 16 = \boxed{24}$

Solution:
Notice $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$ .
Let the base of the small triangles of area 1 be $x$ , then the base length of $\Delta ADE=4x$ . Notice, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$ , then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]$ Thus, $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.$

Solution: The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$ . Therefore the area of the trapezoid is $40-16=\boxed{24}$

Solution: You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$ , so to find the area of such trapezoid $BCED$ , we just take $40-16=\boxed{24}$ , like so.

Solution: The combined area of the small triangles is $7$ , and from the fact that each small triangle has an area of $1$ , we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\frac{1}{3}$ , so will their areas have a proportion that is the square of the proportion of their sides, or $\frac {1}{9}$ ). Thus, the combined area of the top triangle and the trapezoid immediately below is $7 + 9 = 16$ . The area of trapezoid $BCED$ is thus the area of triangle $ABC-16 =\boxed{24}$

Solution: You can assume for the base of one of the smaller triangles to be $\frac{1}{a}$ and the height to be $2a$ , giving an area of 1. The larger triangle above the 7 smaller ones then has base $\frac{3}{a}$ and height $6a$ , giving it an area of $9$ . Then the area of triangle $ADE$ is $16$ and $40-16=\boxed{24}$

Solution: You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence $7\cdot\left(1+\left(\frac{9}{16}\right)+\left(\frac{9}{16}\right)^2+\left(\frac{9}{16}\right)^3\dots\right)$ . Using the infinite geometric series sum formula gives us $7\cdot\left(\frac{1}{1-\frac{9}{16}}\right)=7\cdot\frac{16}{7}=16$ . The triangle's area would thus be $40-16=\boxed{24}$


Answer: 150.0

All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A= \angle B = 90^\circ.$ What is the degree measure of $\angle E?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 108 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 150$

Solution: $AB = EC$ because they are opposite sides of a square. Also, $ED = DC = AB$ because all sides of the convex pentagon are of equal length. Since $ABCE$ is a square and $\triangle CED$ is an equilateral triangle, $\angle AEC = 90$ and $\angle CED = 60.$ Use angle addition: \[\angle E = \angle AEC + \angle CED = 90 + 60 = \boxed{150}\]


Answer: 28.0

All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
$\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textbf{(D) } 27\qquad\textbf{(E) } 28$

Solution:
Since she has half as many red marbles as green, we can call the number of red marbles $x$ , and the number of green marbles $2x$ .
Since she has half as many green marbles as blue, we can call the number of blue marbles $4x$ .
Adding them up, we have $7x$ marbles. The number of marbles therefore must be a multiple of $7$ . The only possible answer is $\boxed{28}.$

Solution: Suppose Maria has $g$ green marbles and let $t$ be the total number of marbles. She then has $\frac{g}{2}$ red marbles and $2g$ blue marbles. Altogether, Maria has \[g + \frac{g}{2} + 2g = \frac{7g}{2} = t\] marbles, implying that $g = \dfrac{2t}{7},$ so $t$ must be a multiple of $7$ . The only multiple of $7$ is $\boxed{28}.$


Answer: 7.0

All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution: Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity:  But since there are only four even numbers in the set, the five darker squares must contain the odd numbers, which sum to $1+3+5+7+9=25.$ Therefore if the sum of the numbers in the corners is $18$ , the number in the center must be $\boxed{7}$

Solution: Quick testing shows that \[3~2~1\] \[4~7~8\] \[5~6~9\] is a valid solution. $3+1+5+9 = 18$ , and the numbers follow the given condition. The center number is found to be $\boxed{7}$ .. — @adihaya talk ) 12:27, 21 February 2016 (EST) ~edited

Solution: First let the numbers be \[1   ~8~   7\] \[2  ~ 9   ~6\] \[3  ~ 4~   5\] with the numbers $1-8$ around the outsides and $9$ in the middle. We see that the sum of the four corner numbers is $16$ . If we switch $7$ and $9$ , then the corner numbers will add up to $18$ and the consecutive numbers will still be touching each other. The answer is $\boxed{7}$ . ~edited

Solution:
Testing out the box with the center square taking on the value of 5 and 6, we find that they either do not satisfy the first or the second condition. Testing 7, we find that a valid configuration is \[1 ~8~ 9\] \[2 ~ 7 ~6\] \[3 ~ 4~ 5\]
$\boxed{7}$


Answer: 729.0

All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
$\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680$

Solution:
Let us call the six faces of our cube $a,b,c,d,e,$ and $f$ (where $a$ is opposite $d$ $c$ is opposite $e$ , and $b$ is opposite $f$ .
Thus, for the eight vertices, we have the following products: $abc,abe,bcd,bde,acf,cdf,aef,$ and $def$ .
Let us find the sum of these products: \[abc+abe+bcd+bde+acf+cdf+aef+def\] We notice $b$ is a factor of the first four terms, and $f$ is a factor of the last four terms. \[b(ac+ae+cd+de)+f(ac+ae+cd+de)\] Now, we can factor even more:
\begin{align*} & (b+f)(ac+ae+cd+de) \\ = &(b+f)(a(c+e)+d(c+e)) \\ = &(b+f)(a+d)(c+e) \end{align*} We have the product. Notice how the factors are sums of opposite faces. The greatest sum possible is $(7+2)$ $(6+3)$ , and $(5+4)$ all factors. \begin{align*} & (7+2)(6+3)(5+4) \\ = & 9 \cdot 9 \cdot 9 \\ = & 729. \end{align*} Thus our answer is $\boxed{729}$

Solution:
We will use parity. If we attempt to maximize this cube in any given way, for example making sure that the sides with 5,6 and 7 all meet at one single corner, the first two answers clearly are out of bounds. Now notice the fact that any three given sides will always meet at one of the eight points. Also note the fact that there are 3 odd numbers. This means that there must be one side that has an odd area! Any odd number added with even numbers is always odd. Given that both c) and e) are both even, d) is our only choice.
Thus our answer is $\boxed{729}$

Solution: We first find the factorization $(b+f)(a+d)(c+e)$ using the method in Solution 1. By using AM-GM, we get, $(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3$ . To maximize the factorization, we get the answer is $\left( \frac{27}{3} \right)^3 = \boxed{729}$

Solution: First, we intuitivly notice that multiplying large numbers together and smaller numbers together tends to produce larger sums, while multiplying large numbers with smaller numbers tends to produce smaller sums. From this, we guess that it is optimal to have $7, 6,$ and $5$ to be around one vertex to produce at least one large product of $210$ . We can immediately eliminate a) and b) as answer choices since they are very close to this product, as well as e) since it is exactly $8 \cdot 210$ and we know that every following product will be smaller than this. We also guess that that is is most likely optimal to create a pairing where a number and its complement (the number that's the difference of 9 and this number) are on opposite sides. Using these two guesses, we can construct a net of sides and brute force the solution (or estimate) leaving us with $\boxed{729}$


Answer: 729.0

Solution: First assign each face the letters $a,b,c,d,e,f$ . The sum of the product of the faces is $abc+acd+ade+aeb+fbc+fcd+fde+feb$ . We can factor this into $(a+f)(b+c)(d+e)$ which is the product of the sum of each pair of opposite faces. In order to maximize $(a+f)(b+c)(d+e)$ we use the numbers $(7+2)(6+3)(5+4)$ or $\boxed{729}$

Solution:
We'll proceed from the factoring process above.
By the AM-GM inequality,
\[\frac{a_1+a_2+a_3}{3}\geq\sqrt[3]{a_1a_2a_3}\]
Cubing both sides,
\[\left(\frac{a_1+a_2+a_3}{3}\right)^3\geq{a_1a_2a_3}\]
Let $a_1=(a+f)$ $a_2=(b+c)$ , and $a_3=(d+e)$ . Let's substitute in these values.
\[\left(\frac{a+b+c+d+e+f}{3}\right)^3\geq{(a+f)(b+c)(d+e)}\]
$a+b+c+d+e+f$ is fixed at 27.
\[\left(\frac{27}{3}\right)^3\geq{(a+f)(b+c)(d+e)}\]
\[\boxed{729}\]

Solution: First, we see that we want to try and maximize each vertex. Since the multiplication of each vertex is the product of three values, we want to maximize those three values. Doing so, we see that we want them to be as close as possible, giving $4.5^3$ (the average of all the values). This gives us the maximum for each vertex, multiplied by the 8 vertices, yields our answer $\boxed{729}$ Also note that if you cannot evaluate $4.5^3$ quickly, a rough approximation of $5*4.5*4*8$ will yield 720, very close to our answer. -rayprati

Solution:
It is obvious to put $5$ $6$ , and $7$ on the faces that share the same vertex. As $4$ is the next biggest number, the face with $4$ has to be next to the faces with $6$ and $7$ . As $4$ is the next biggest number, the face with $3$ has to be next to the faces with $5$ and $7$ . making the face with $2$ next to the faces with $6$ and $5$
Therefore the answer is $7 \cdot 5 \cdot 6 + 7 \cdot 4 \cdot 6 + 7 \cdot 3 \cdot 5 + 7 \cdot  3 \cdot 4 + 2 \cdot 5 \cdot 6 + 2 \cdot 4 \cdot 6 + 2 \cdot 3 \cdot 5 + 2 \cdot  3 \cdot 4 = \boxed{729}$


Answer: 88.0

All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$
$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$

Solution: By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$ . By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$ , so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$ . Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \boxed{88}$

Solution: Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$ . Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\binom {6}{3} = 20$ products $r_a \cdot r_b \cdot r_c,$ we obtain \[B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{88}.\] ~ike.chen


Answer: 88.0

Solution: By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$ . By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$ , so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$ . Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \boxed{88}$

Solution: Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$ . Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\binom {6}{3} = 20$ products $r_a \cdot r_b \cdot r_c,$ we obtain \[B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{88}.\] ~ike.chen


Answer: 13.0

All the students in an algebra class took a $100$ -point test. Five students scored $100$ , each student scored at least $60$ , and the mean score was $76$ . What is the smallest possible number of students in the class?
$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ 14$

Solution:
Let the number of students be $n\geq 5$ . Then the sum of their scores is at least $5\cdot 100 + (n-5)\cdot 60$ . At the same time, we need to achieve the mean $76$ , which is equivalent to achieving the sum $76n$
Hence we get a necessary condition on $n$ : we must have $5\cdot 100 + (n-5)\cdot 60 \leq 76n$ .
This can be simplified to $200 \leq 16n$ . The smallest integer $n$ for which this is true is $n=13$
To finish our solution, we now need to find one way how $13$ students could have scored on the test. We have $13\cdot 76 = 988$ points to divide among them. The five $100$ s make $500$ , hence we must divide the remaining $488$ points among the other $8$ students. This can be done e.g. by giving $61$ points to each of them.
Hence the smallest possible number of students is $\boxed{13}$


Answer: 8.0

All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$ , with $A$ at the origin and $\overline{BC}$ parallel to the $x$ -axis. The area of the triangle is $64$ . What is the length of $BC$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$

Solution: Let the point where the height of the triangle intersects with the base be $D$ . Now we can guess what $x$ is and find $y$ . If $x$ is $3$ , then $y$ is $9$ . The cords of $B$ and $C$ would be $(-3,9)$ and $(3,9)$ , respectively. The distance between $B$ and $C$ is $6$ , meaning the area would be $\frac{6 \cdot 9}{2}=27$ , not $64$ . Now we let $x=4$ $y$ would be $16$ . The cords of $B$ and $C$ would be $(-4,16)$ and $(4,16)$ , respectively. $BC$ would be $8$ , and the height would be $16$ . The area would then be $\frac{8 \cdot 16}{2}$ which is $64$ , so $BC$ is $\boxed{8}$


Answer: 8.0

Solution:
By: Albert471
Plotting points $B$ and $C$ on the graph shows that they are at $\left( -x,x^2\right)$ and $\left( x,x^2\right)$ , which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}*2x*x^2  =   64=x^3$ Making x=4, and the length of $BC$ is $2x$ , so the answer is $\boxed{8}$


Answer: 14.0

All three vertices of an equilateral triangle are on the parabola $y = x^2$ , and one of its sides has a slope of $2$ . The $x$ -coordinates of the three vertices have a sum of $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$
$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}}$

Solution:
Let the three points be at $A = (x_1, x_1^2)$ $B = (x_2, x_2^2)$ , and $C = (x_3, x_3^2)$ , such that the slope between the first two is $2$ , and $A$ is the point with the least $y$ -coordinate.
Then, we have $\textrm{Slope of }AC = \frac{x_1^2 - x_3^2}{x_1 - x_3} = x_1 + x_3$ . Similarly, the slope of $BC$ is $x_2 + x_3$ , and the slope of $AB$ is $x_1 + x_2 = 2$ . The desired sum is $x_1 + x_2 + x_3$ , which is equal to the sum of the slopes divided by $2$
To find the slope of $AC$ , we note that it comes at a $60^{\circ}$ angle with $AB$ . Thus, we can find the slope of $AC$ by multiplying the two complex numbers $1 + 2i$ and $1 + \sqrt{3}i$ . What this does is generate the complex number that is at a $60^{\circ}$ angle with the complex number $1 + 2i$ . Then, we can find the slope of the line between this new complex number and the origin: \[(1+2i)(1+\sqrt{3}i)\] \[= 1 - 2\sqrt{3} + 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}\] \[= \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}\] \[= \frac{8 + 5\sqrt{3}}{-11}\] \[= \frac{-8 - 5\sqrt{3}}{11}.\] The slope $BC$ can also be solved similarly, noting that it makes a $120^{\circ}$ angle with $AB$ \[(1+2i)(-1+\sqrt{3}i)\] \[= -1 - 2\sqrt{3} - 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{\sqrt{3} - 2}{-2\sqrt{3} - 1}\] \[= \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \cdot \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}\]
At this point, we start to notice a pattern: This expression is equal to $\frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}$ , except the numerators of the first fractions are conjugates! Notice that this means that when we multiply out, the rational term will stay the same, but the coefficient of $\sqrt{3}$ will have its sign switched. This means that the two complex numbers are conjugates, so their irrational terms cancel out.
Our sum is simply $2 - 2\cdot\frac{8}{11} = \frac{6}{11}$ , and thus we can divide by $2$ to obtain $\frac{3}{11}$ , which gives the answer $\boxed{14}$

Solution:
Using the slope formula and differences of squares, we find:
$a+b$ = the slope of $AB$
$b+c$ = the slope of $BC$
$a+c$ = the slope of $AC$
So the value that we need to find is the sum of the slopes of the three sides of the triangle divided by $2$ . Without loss of generality, let $AB$ be the side that has the smallest angle with the positive $x$ -axis. Let $J$ be an arbitrary point with the coordinates $(1, 0)$ . Translate the triangle so $A$ is at the origin. Then $\tan(BOJ) = 2$ . Since the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the answer is $\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}$
Using $\tan(BOJ) = 2$ , and the tangent addition formula, this simplifies to $\dfrac{3}{11}$ , so the answer is $3 + 11 = \boxed{14}$


Answer: 48.0

Along the route of a bicycle race, $7$ water stations are evenly spaced between the start and finish lines,
as shown in the figure below. There are also $2$ repair stations evenly spaced between the start and
finish lines. The $3$ rd water station is located $2$ miles after the $1$ st repair station. How long is the race
in miles? [asy] // Credits given to Themathguyd and Kante314 usepackage("mathptmx"); size(10cm); filldraw((11,4.5)--(171,4.5)--(171,17.5)--(11,17.5)--cycle,mediumgray*0.4 + lightgray*0.6); draw((11,11)--(171,11),linetype("2 2")+white+linewidth(1.2)); draw((0,0)--(11,0)--(11,22)--(0,22)--cycle); draw((171,0)--(182,0)--(182,22)--(171,22)--cycle);  draw((31,4.5)--(31,0)); draw((51,4.5)--(51,0)); draw((151,4.5)--(151,0));  label(scale(.85)*rotate(45)*"Water 1", (23,-13.5)); label(scale(.85)*rotate(45)*"Water 2", (43,-13.5)); label(scale(.85)*rotate(45)*"Water 7", (143,-13.5));  filldraw(circle((103,-13.5),.2)); filldraw(circle((98,-13.5),.2)); filldraw(circle((93,-13.5),.2)); filldraw(circle((88,-13.5),.2)); filldraw(circle((83,-13.5),.2));  label(scale(.85)*rotate(90)*"Start", (5.5,11)); label(scale(.85)*rotate(270)*"Finish", (176.5,11)); [/asy] $\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 96$

Solution:
Suppose that the race is $d$ miles long. The water stations are located at \[\frac{d}{8}, \frac{2d}{8}, \ldots, \frac{7d}{8}\] miles from the start, and the repair stations are located at \[\frac{d}{3}, \frac{2d}{3}\] miles from the start.
We are given that $\frac{3d}{8}=\frac{d}{3}+2,$ from which \begin{align*} \frac{9d}{24}&=\frac{8d}{24}+2 \\ \frac{d}{24}&=2 \\ d&=\boxed{48} ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM


Answer: 849.0

Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$

Solution:
Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number he gets right. Let $b$ be the number he gets right on day 2.
These inequalities follow: \[\frac{a}{q} < \frac{160}{300} = \frac{8}{15}\] \[\frac{b}{500-q} < \frac{140}{200} = \frac{7}{10}\] Solving for a and b and adding the two inequalities: \[a + b < \frac{8}{15}q + (350 - \frac{7}{10}q)\] \[a + b < 350 - \frac{1}{6}q\] From here, we see the largest possible value of $a+b$ is $349$
Checking our conditions, we know that $a$ must be positive so therefore $q$ must be positive. A quick check shows that for $2\le q \le 5$ $q$ follows all the conditions and results in $a+b=349$
This makes Beta's success ratio $\frac{349}{500}$ . Thus, the answer is $m+n = 349 + 500 = \boxed{849}$


Answer: 4.0

Amelia has a coin that lands heads with probability $\frac{1}{3}\,$ , and Blaine has a coin that lands on heads with probability $\frac{2}{5}$ . Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $q-p$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P$ , since if she gets to her turn again, she is back where she started with probability of winning $P$ . The chance she wins on her first turn is $\frac{1}{3}$ . The chance she makes it to her turn again is a combination of her failing to win the first turn - $\frac{2}{3}$ and Blaine failing to win - $\frac{3}{5}$ . Multiplying gives us $\frac{2}{5}$ . Thus, \[P = \frac{1}{3} + \frac{2}{5}P\] Therefore, $P = \frac{5}{9}$ , so the answer is $9-5=\boxed{4}$

Solution: Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} \cdots$ This can be represented as an infinite geometric series: \[P=\frac{\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{3}{5}} = \frac{\frac{1}{3}}{1-\frac{2}{5}} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3}\cdot \frac{5}{3} = \frac{5}{9}.\] Therefore, $P = \frac{5}{9}$ , so the answer is $9-5 = \boxed{4}.$


Answer: 59.0

Solution:
We can solve this by using 'casework,' the cases being:
Case 1: Amelia wins on her first turn.
Case 2 Amelia wins on her second turn.
and so on.
The probability of her winning on her first turn is $\dfrac13$ . The probability of all the other cases is determined by the probability that Amelia and Blaine all lose until Amelia's turn on which she is supposed to win. So, the total probability of Amelia winning is: \[\dfrac{1}{3}+\left(\dfrac{2}{3}\cdot\dfrac{3}{5}\right)\cdot\dfrac{1}{3}+\left(\dfrac{2}{3}\cdot\dfrac{3}{5}\right)^2\cdot\dfrac{1}{3}+\cdots.\] Factoring out $\dfrac13$ we get a geometric series: \[\dfrac{1}{3}\left(1+\dfrac{2}{5}+\left(\dfrac{2}{5}\right)^2+\cdots\right) = \dfrac{1}{3}\cdot\dfrac{1}{3/5} = \boxed{59}.\]


Answer: 132.0

Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105$

Solution:
Let the ordered triple $(a,b,c)$ denote that $a$ songs are liked by Amy and Beth, $b$ songs by Beth and Jo, and $c$ songs by Jo and Amy. The only possible triples are $(1,1,1), (2,1,1), (1,2,1)(1,1,2)$
To show this, observe these are all valid conditions. Second, note that none of $a,b,c$ can be bigger than 3. Suppose otherwise, that $a = 3$ . Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either $b$ or $c$ to be at least 1. In fact, we require either $b$ or $c$ to equal 1, otherwise there will be a song liked by all three. Suppose $b = 1$ . Then we must have $c=0$ since no song is liked by all three girls, a contradiction.
Case 1 : How many ways are there for $(a,b,c)$ to equal $(1,1,1)$ ? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So $(a,b,c)=(1,1,1)$ in $4\cdot3\cdot2\cdot4 = 96$ ways.
Case 2 : To find the number of ways for $(a,b,c) = (2,1,1)$ , observe there are $\binom{4}{2} = 6$ choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are $6\cdot2\cdot3=36$ ways for the girls to like the songs.
That gives a total of $96 + 36 = 132$ ways for the girls to like the songs, so the answer is $\boxed{132}$

Solution:
Let $AB, BJ$ , and $AJ$ denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let $A, B, J,$ and $N$ denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least $1\: AB, BJ$ , and $AJ$ , they must be $3$ songs out of the $4$ that Amy, Beth, and Jo listened to. The fourth song can be of any type $N, A, B, J, AB, BJ$ , and $AJ$ (there is no $ABJ$ because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange $AB, BJ, AJ$ , and a song from the set $\{N, A, B, J, AB, BJ, AJ\}$
Case 1: Fourth song = $N, A, B, J$
Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.
Number of ways to rearrange = $(4!)$ rearrangements for each choice $*\: 4$ choices = $96$
Case 2: Fourth song = $AB, BJ, AJ$
Note that in Case $2$ , all three of the choices for the fourth song repeat somewhere in the first three songs.
Number of ways to rearrange = $(4!/2!)$ rearrangements for each choice $*\: 3$ choices = $36$
$96 + 36 = \boxed{132}$

Solution:
There are $\binom{4}{3}$ ways to choose the three songs that are liked by the three pairs of girls.
There are $3!$ ways to determine how the three songs are liked, or which song is liked by which pair of girls.
In total, there are $\binom{4}{3}\cdot3!$ possibilities for the first $3$ songs.
There are $3$ cases for the 4th song, call it song D.
Case $1$ : D is disliked by all $3$ girls $\implies$ there is only $1$ possibility.
Case $2$ : D is liked by exactly $1$ girl $\implies$ there are $3$ possibility.
Case $3$ : D is liked by exactly $2$ girls $\implies$ there are $3$ pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first $3$ songs liked by the same girls.
Counting the overlaps, there are $3$ ways to choose the pair with overlaps and $4\cdot3=12$ ways to choose what the other $2$ pairs like independently. In total, there are $3\cdot12=36$ overlapped possibilities.
Finally, there are $\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132$ ways for the songs to be likely by the girls. $\boxed{132}$


Answer: 132.0

Solution:
Let the ordered triple $(a,b,c)$ denote that $a$ songs are liked by Amy and Beth, $b$ songs by Beth and Jo, and $c$ songs by Jo and Amy. The only possible triples are $(1,1,1), (2,1,1), (1,2,1)(1,1,2)$
To show this, observe these are all valid conditions. Second, note that none of $a,b,c$ can be bigger than 3. Suppose otherwise, that $a = 3$ . Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either $b$ or $c$ to be at least 1. In fact, we require either $b$ or $c$ to equal 1, otherwise there will be a song liked by all three. Suppose $b = 1$ . Then we must have $c=0$ since no song is liked by all three girls, a contradiction.
Case 1 : How many ways are there for $(a,b,c)$ to equal $(1,1,1)$ ? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So $(a,b,c)=(1,1,1)$ in $4\cdot3\cdot2\cdot4 = 96$ ways.
Case 2 : To find the number of ways for $(a,b,c) = (2,1,1)$ , observe there are $\binom{4}{2} = 6$ choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are $6\cdot2\cdot3=36$ ways for the girls to like the songs.
That gives a total of $96 + 36 = 132$ ways for the girls to like the songs, so the answer is $\boxed{132}$

Solution:
Let $AB, BJ$ , and $AJ$ denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let $A, B, J,$ and $N$ denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least $1\: AB, BJ$ , and $AJ$ , they must be $3$ songs out of the $4$ that Amy, Beth, and Jo listened to. The fourth song can be of any type $N, A, B, J, AB, BJ$ , and $AJ$ (there is no $ABJ$ because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange $AB, BJ, AJ$ , and a song from the set $\{N, A, B, J, AB, BJ, AJ\}$
Case 1: Fourth song = $N, A, B, J$
Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.
Number of ways to rearrange = $(4!)$ rearrangements for each choice $*\: 4$ choices = $96$
Case 2: Fourth song = $AB, BJ, AJ$
Note that in Case $2$ , all three of the choices for the fourth song repeat somewhere in the first three songs.
Number of ways to rearrange = $(4!/2!)$ rearrangements for each choice $*\: 3$ choices = $36$
$96 + 36 = \boxed{132}$

Solution:
There are $\binom{4}{3}$ ways to choose the three songs that are liked by the three pairs of girls.
There are $3!$ ways to determine how the three songs are liked, or which song is liked by which pair of girls.
In total, there are $\binom{4}{3}\cdot3!$ possibilities for the first $3$ songs.
There are $3$ cases for the 4th song, call it song D.
Case $1$ : D is disliked by all $3$ girls $\implies$ there is only $1$ possibility.
Case $2$ : D is liked by exactly $1$ girl $\implies$ there are $3$ possibility.
Case $3$ : D is liked by exactly $2$ girls $\implies$ there are $3$ pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first $3$ songs liked by the same girls.
Counting the overlaps, there are $3$ ways to choose the pair with overlaps and $4\cdot3=12$ ways to choose what the other $2$ pairs like independently. In total, there are $3\cdot12=36$ overlapped possibilities.
Finally, there are $\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132$ ways for the songs to be likely by the girls. $\boxed{132}$


Answer: 180.0

An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.

Solution: By counting the number of green cubes $2$ different ways, we have $12a=20b$ , or $a=\dfrac{5}{3} b$ . Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \times b \times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \times a \times c$ layers. Therefore, we have $a(bc-21)=25b$ and $b(ac-45)=9a$ . We check a few small values of $a,b$ and solve for $c$ , checking $(a,b)=(5,3)$ gives $c=12$ with a volume of $180$ $(a,b)=(10,6)$ gives $c=6$ with a volume of $360$ , and $(a,b)=(15,9)$ gives $c=4$ , with a volume of $540$ . Any higher $(a,b)$ will $ab>180$ , so therefore, the minimum volume is $\boxed{180}$

Solution:
The total number of green cubes is given by $12a=20b\Longrightarrow a=\frac{5}{3}b$
Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$ . Substitute $a=\frac{5}{3}b$ gives $r=15$
Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ layers gives $y=15$
Therefore $bc=9+12+15=36$ and $ac=15+20+25=60$ . Multiplying yields $abc^2=2160$
Since $abc^2$ is fixed, $abc$ is minimized when $c$ is maximized, which occurs when $a$ $b$ are minimized (since each of $ac$ $bc$ is fixed). Thus $(a,b,c)=(3,5,12)\Longrightarrow abc=\boxed{180}$


Answer: 130.0

An $m\times n\times p$ rectangular box has half the volume of an $(m + 2)\times(n + 2)\times(p + 2)$ rectangular box, where $m, n,$ and $p$ are integers, and $m\le n\le p.$ What is the largest possible value of $p$

Solution:
\[2mnp = (m+2)(n+2)(p+2)\]
Let’s solve for $p$
\[(2mn)p = p(m+2)(n+2) + 2(m+2)(n+2)\] \[[2mn - (m+2)(n+2)]p = 2(m+2)(n+2)\] \[p = \frac{2(m+2)(n+2)}{mn - 2n - 2m - 4} = \frac{2(m+2)(n+2)}{(m-2)(n-2) - 8}\]
Clearly, we want to minimize the denominator, so we test $(m-2)(n-2) - 8 = 1 \Longrightarrow (m-2)(n-2) = 9$ . The possible pairs of factors of $9$ are $(1,9)(3,3)$ . These give $m = 3, n = 11$ and $m = 5, n = 5$ respectively. Substituting into the numerator, we see that the first pair gives $130$ , while the second pair gives $98$ . We now check that $130$ is optimal, setting $a=m-2$ $b=n-2$ in order to simplify calculations. Since \[0 \le (a-1)(b-1) \implies a+b \le ab+1\] We have \[p = \frac{2(a+4)(b+4)}{ab-8} = \frac{2ab+8(a+b)+32}{ab-8} \le \frac{2ab+8(ab+1)+32}{ab-8} = 10 + \frac{120}{ab-8} \le 130\] Where we see $(m,n)=(3,11)$ gives us our maximum value of $\boxed{130}$

Solution:
Similarly as above, we solve for $p,$ but we express the denominator differently:
\[p=\dfrac{2(m+2)(n+2)}{(m+2)(n+2)-4(m+n+2)} \implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(m+n+2)}{(m+2)(n+2)}.\] Hence, it suffices to maximize $\dfrac{m+n+2}{(m+2)(n+2)},$ under the conditions that $p$ is a positive integer.
Then since $\dfrac{m+n+2}{(m+2)(n+2)}>\dfrac{1}{2}$ for $m=1,2,$ we fix $m=3.$ \[\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+5)}{5(n+2)}=\dfrac{n-10}{10(n+2)},\] where we simply let $n=11$ to achieve $p=\boxed{130}.$


Answer: 9990.0

An ATM password at Fred's Bank is composed of four digits from $0$ to $9$ , with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?
$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$

Solution: For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{9990}$

Solution: Counting the prohibited cases, we find that there are 10 of them. This is because, when we start with 9,1, and 1, we can have any of the 10 digits for the last digit. So, our answer is $10^4-10=\boxed{9990}.$


Answer: 1875.0

An American traveling in Italy wishes to exchange American money (dollars) for Italian money (lire).  If 3000 lire = 1.60, how much lire will the traveler receive in exchange for 1.00?
$\text{(A)}\ 180 \qquad \text{(B)}\ 480 \qquad \text{(C)}\ 1800 \qquad \text{(D)}\ 1875 \qquad \text{(E)}\ 4875$

Solution:
$\frac{3000 \text{ lire}}{1.6 \text{ dollars}} = \frac{x \text{ lire}}{1 \text{ dollar}}$
$x = 1875$ , and the answer is $\boxed{1875}$


Answer: 40.0

An Annville Junior High School, $30\%$ of the students in the Math Club are in the Science Club, and $80\%$ of the students in the Science Club are in the Math Club.  There are 15 students in the Science Club.  How many students are in the Math Club?
$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 40$

Solution:
If $80\%$ of the people in the science club of $15$ people are in the Math Club, $\frac{4}{5}\times15=12$ people are in the both the Math Club and the Science Club.
These $12$ people make up $30\%$ of the Math Club.
Setting up a proportion:
$12\cdot 1.00 =0.30\cdot x$
$\frac{12}{0.3} = 40=x$
There are $40=\boxed{40}$ people in the Math Club.


Answer: 14.0

An amusement park has a collection of scale models, with a ratio of $1: 20$ , of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number?
$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

Solution: You can see that since the ratio of real building's heights to the model building's height is $1:20$ . We also know that the U.S Capitol is $289$ feet in real life, so to find the height of the model, we divide by 20. That gives us $14.45$ which rounds to 14. Therefore, to the nearest whole number, the duplicate is $\boxed{14}$

Solution:
We can compute $\frac{289}{20}$ and round our answer to get $\boxed{14}$ .
It is basically Solution 1 without the ratio calculation. However, Solution 1 is referring further to the problem.

Solution: We know that $20 \cdot 14 = 280 ,$ and that $20 \cdot 15 = 300 .$ These are the multiples of $20$ around $289 ,$ and the closest one of those is $280.$ Therefore, the answer is $\dfrac {280} {20} = \boxed{14} .$


Answer: 92.0

An angle $x$ is chosen at random from the interval $0^\circ < x < 90^\circ.$ Let $p$ be the probability that the numbers $\sin^2 x, \cos^2 x,$ and $\sin x \cos x$ are not the lengths of the sides of a triangle. Given that $p = d/n,$ where $d$ is the number of degrees in $\text{arctan}$ $m$ and $m$ and $n$ are positive integers with $m + n < 1000,$ find $m + n.$

Solution:
Note that the three expressions are symmetric with respect to interchanging $\sin$ and $\cos$ , and so the probability is symmetric around $45^\circ$ .  Thus, take $0 < x < 45$ so that $\sin x < \cos x$ .  Then $\cos^2 x$ is the largest of the three given expressions and those three lengths not forming a triangle is equivalent to a violation of the triangle inequality
\[\cos^2 x > \sin^2 x + \sin x \cos x\]
This is equivalent to
\[\cos^2 x - \sin^2 x > \sin x \cos x\]
and, using some of our trigonometric identities , we can re-write this as $\cos 2x > \frac 12 \sin 2x$ . Since we've chosen $x \in (0, 45)$ $\cos 2x > 0$ so
\[2 > \tan 2x \Longrightarrow  x < \frac 12 \arctan 2.\]
The probability that $x$ lies in this range is $\frac 1{45} \cdot \left(\frac 12 \arctan 2\right) = \frac{\arctan 2}{90}$ so that $m = 2$ $n = 90$ and our answer is $\boxed{092}$


Answer: 408.0

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
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Solution:
Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof
Let the set of parallel lines be perpendicular to the x-axis , such that they cross it at $0, 1, 2 \ldots$ . The base of region $\mathcal{A}$ is on the line $x = 1$ . The bigger base of region $\mathcal{D}$ is on the line $x = 7$ .
Let the top side of the angle be $y = x - s$ and the bottom side be x-axis, as dividing the angle doesn't change the problem.
Since the area of the triangle is equal to $\frac{1}{2}bh$
\[\frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5} = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}\]
Solve this to find that $s = \frac{5}{6}$
Using the same reasoning as above, we get $\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}$ , which is $\boxed{408}$

Solution:
Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be $x$ and the area of it be $x^2$ . Also, let all sections of the line on the same side as the side with length $x$ on a trapezoid be equal to $1$
Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is ${\left(\frac{x+1}{x}\right)}^2$ . Multiplying, we get $(x+1)^2$ as the area of the triangle, so the area of the trapezoid is $2x+1$ . Repeating this process, we get that the area of B is $2x+3$ , the area of C is $2x+7$ , and the area of D is $2x+11$
We can now use the given condition that the ratio of C and B is $\frac{11}{5}$
$\frac{11}{5} = \frac{2x+7}{2x+3}$ gives us $x = \frac{1}{6}$
So now we compute the ratio of D and A, which is $\frac{(2)(\frac{1}{6}) + 11}{(\frac{1}{6})^2} = \boxed{408.}$

Solution: Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\frac{ay+yx+11ab}{xy}.$ We know that $ab = 3ay+3bx$ so the ratio we seed is $\frac{33(ay+yx)}{11xy}.$ Finally note that by similar triangles $\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.$ Therefore the ratio we seek is $\frac{66(ay)}{11xy} =\frac{66a}{11x}.$ Finally note that $ab=3ay+3bx \implies ab = 6bx \implies a = 6x$ so the final ratio is $6 \cdot 68 = \boxed{408}.$


Answer: 2.0

An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$ , with $b>c$ . Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$ , and let $OY$ be the radius of the larger circle that contains $Z$ . Let $a=XZ$ $d=YZ$ , and $e=XY$ . What is the area of the annulus?
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$\mathrm{(A) \ } \pi a^2 \qquad \mathrm{(B) \ } \pi b^2 \qquad \mathrm{(C) \ } \pi c^2 \qquad \mathrm{(D) \ } \pi d^2 \qquad \mathrm{(E) \ } \pi e^2$

Solution:
The area of the large circle is $\pi b^2$ , the area of the small one is $\pi c^2$ , hence the shaded area is $\pi(b^2-c^2)$
From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$ , hence $b^2-c^2=a^2$ and thus the shaded area is $\boxed{2}$

Solution: Set $c=0,$ then the shaded area is just the area of a circle with radius $a,$ which is $\boxed{2}$


Answer: 2.0

Solution:
The area of the large circle is $\pi b^2$ , the area of the small one is $\pi c^2$ , hence the shaded area is $\pi(b^2-c^2)$
From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$ , hence $b^2-c^2=a^2$ and thus the shaded area is $\boxed{2}$

Solution: Set $c=0,$ then the shaded area is just the area of a circle with radius $a,$ which is $\boxed{2}$


Answer: 49.0

An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly $8$ moves that ant is at a vertex of the top face on the cube is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Solution:
For all positive integers $k,$ let
The base case occurs at $k=1,$ from which $\left(N(1,\mathrm{BB}),N(1,\mathrm{BT}),N(1,\mathrm{TB}),N(1,\mathrm{TT})\right)=(2,1,0,0).$
Suppose the ant makes exactly $k$ moves for some $k\geq2.$ We perform casework on its last move:
Alternatively, this recursion argument is illustrated below, where each dashed arrow indicates $1$ way, and each solid arrow indicates $2$ ways:  Therefore, we have the following relationships: \begin{align*} N(1,\mathrm{BB})&=2, \\  N(1,\mathrm{BT})&=1, \\ N(1,\mathrm{TB})&=0, \\ N(1,\mathrm{TT})&=0, \\ N(k+1,\mathrm{BB})&=N(k,\mathrm{BB})+2\cdot N(k,\mathrm{TB}), \\ N(k+1,\mathrm{BT})&=N(k,\mathrm{BB}), \\ N(k+1,\mathrm{TB})&=N(k,\mathrm{TT}), \\ N(k+1,\mathrm{TT})&=N(k,\mathrm{TT})+2\cdot N(k,\mathrm{BT}). \end{align*} Using these equations, we recursively fill out the table below: \[\begin{array}{c||c|c|c|c|c|c|c|c}    \hspace{7mm}&\hspace{6.5mm}&\hspace{6.5mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&& \\ [-2.5ex] \boldsymbol{k} & \boldsymbol{1} & \boldsymbol{2} & \boldsymbol{3} & \boldsymbol{4} & \boldsymbol{5} & \boldsymbol{6} & \boldsymbol{7} & \boldsymbol{8} \\  \hline \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{BB})} &2&2&2&6&18&38&66&118 \\ \hline   &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{BT})} &1&2&2&2&6&18&38&66 \\ \hline  &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{TB})} &0&0&2&6&10&14&26&62 \\ \hline    &&&&&&&& \\ [-2.25ex] \boldsymbol{N(k,\mathrm{TT})} &0&2&6&10&14&26&62&138 \\ \hline \hline &&&&&&&& \\ [-2.25ex] \textbf{Total}&\boldsymbol{3}&\boldsymbol{6}&\boldsymbol{12}&\boldsymbol{24}&\boldsymbol{48}&\boldsymbol{96}&\boldsymbol{192}&\boldsymbol{384} \end{array}\] By the Multiplication Principle, there are $3\cdot2^{k-1}$ ways to make exactly $k$ moves. So, we must get \[N(k,\mathrm{BB})+N(k,\mathrm{BT})+N(k,\mathrm{TB})+N(k,\mathrm{TT})=3\cdot2^{k-1}\] for all values of $k.$
Finally, the requested probability is \[\frac{N(8,\mathrm{BT})+N(8,\mathrm{TT})}{N(8,\mathrm{BB})+N(8,\mathrm{BT})+N(8,\mathrm{TB})+N(8,\mathrm{TT})}=\frac{66+138}{118+66+62+138}=\frac{204}{384}=\frac{17}{32},\] from which the answer is $17+32=\boxed{049}.$

Solution:
Let the state from bottom to top be $B2T,$ from top to top be $T2T,$ from top to bottom be $T2B,$ and from bottom to bottom be $B2B.$ We can draw the following State Transition Diagram with Markov Chain . The numbers on the transition arc are the transition probabilities.
The probabilities of being in a state after $n$ steps and after $n-1$ steps has the following relationships: \begin{align*} B2T(n) &= B2B(n-1) \cdot \frac12\\ T2T(n) &= B2T(n-1) + T2T(n-1) \cdot \frac12\\ T2B(n) &= T2T(n-1) \cdot \frac12\\ B2B(n) &= T2B(n-1) + B2B(n-1) \cdot \frac12 \end{align*} Those probabilities are calculated by Dynamic Programming in the following table: \[\begin{array}{c|cccc}  &  &  &  & \\ [-2ex] n & B2T(n) & T2T(n) & T2B(n) & B2B(n) \\ [1ex] \hline  &  &  &  & \\ [-1ex] 1 & \frac13 & 0 & 0 & \frac23\\  &  &  &  & \\ 2 & \frac23 \cdot \frac12 = \frac13 & \frac13 & 0 & \frac23 \cdot \frac12 = \frac13 \\  &  &  &  & \\ 3 & \frac13 \cdot \frac12 = \frac16 & \frac13 + \frac13 \cdot \frac12 = \frac12 & \frac13 \cdot \frac12 = \frac16 & \frac13 \cdot \frac12 = \frac16 \\  &  &  &  & \\ 4 & \frac16 \cdot \frac12 = \frac{1}{12} & \frac16 + \frac12 \cdot \frac12 = \frac{5}{12} & \frac12 \cdot \frac12 = \frac14 & \frac16 + \frac16 \cdot \frac12 = \frac14 \\  &  &  &  & \\ 5 & \frac14 \cdot \frac12 = \frac18 & \frac{1}{12} + \frac{5}{12} \cdot \frac{1}{2} = \frac{7}{24} & \frac{5}{12} \cdot \frac12 = \frac{5}{24} & \frac14 + \frac14 \cdot \frac12 = \frac38 \\  &  &  &  & \\ 6 & \frac38 \cdot \frac12 = \frac{3}{16} & \frac18 + \frac{7}{24} \cdot \frac12 = \frac{13}{48} & \frac{7}{24} \cdot \frac12 = \frac{7}{48} & \frac{5}{24} + \frac38 \cdot \frac12 = \frac{19}{48} \\  &  &  &  & \\ 7 & \frac{19}{48} \cdot \frac12 = \frac{19}{96}&  \frac{3}{16} + \frac{13}{48} \cdot \frac{1}{2} = \frac{31}{96}& \frac{13}{48} \cdot \frac{1}{2} = \frac{13}{96} & \frac{7}{48} + \frac{19}{48} \cdot \frac12 = \frac{11}{32}\\  &  &  &  & \\ 8 & \frac{11}{32} \cdot \frac{1}{2} = \frac{11}{64} & \frac{19}{96} + \frac{31}{96} \cdot \frac12 = \frac{23}{64} & \frac{31}{96} \cdot \frac12 = \frac{31}{192} & \frac{13}{96} + \frac{11}{32} \cdot \frac12 = \frac{59}{192} \\ [1ex] \end{array}\] Finally, the requested probability is $\frac{11}{64} + \frac{23}{64} = \frac{17}{32},$ from which the answer is $17 + 32 = \boxed{049}.$

Solution:
Note that we don't care which exact vertex the ant is located at, just which level (either top face or bottom face).
Consider the ant to be on any of the two levels and having moved at least one move. Define $p_i$ to be the probability that after $i$ moves, the ant ends up on the level it started on.
On the first move, the ant can stay on the bottom level with $\frac{2}{3}$ chance and $7$ moves left. Or, it can move to the top level with $\frac{1}{3}$ chance and $6$ moves left (it has to spend another on the top as it can not return immediately). So the requested probability is $P = \frac{2}{3}(1 - p_7) + \frac{1}{3}p_6$
Consider when the ant has $i$ moves left (and it's not the ant's first move). It can either stay on its current level with $\frac{1}{2}$ chance and $i - 1$ moves left, or travel to the opposite level with $\frac{1}{2}$ chance, then move to another vertex on the opposite level, to have $i - 2$ moves left. Thus we obtain the recurrence \[p_i = \frac{1}{2}p_{i - 1} + \frac{1}{2}(1 - p_{i - 2})\]
Computing $p_i$ with the starting conditions $p_0 = 1$ and $p_1 = \frac{1}{2}$ , we obtain $p_6 = \frac{33}{64}$ and $p_7 = \frac{59}{128}$ . Hence $P = \frac{2}{3}(1 - p_7) + \frac{1}{3}p_6= \frac{17}{32}$ as desired; $\boxed{049}$

Solution:
On each move, we can either stay on the level we previously were (stay on the bottom/top) or switch levels (go from top to bottom and vise versa). Since we start on the bottom, ending on the top means that we will have to switch an odd number of times; since we cannot switch twice in a row, over an eight-move period we can either make one or three switches. Furthermore, once we switch to a level we can choose one of two directions of traveling on that level: clockwise or counterclockwise (since we can't go back to our previous move, our first move on the level after switching determines our direction).
Our probability is then $\frac{176 + 28}{3 \cdot 2^7} = \frac{17}{32}$ , so the answer is $17+32=\boxed{049}$


Answer: 0.25

An aquarium has a rectangular base that measures $100$ cm by $40$ cm and has a height of $50$ cm. The aquarium is filled with water to a  depth of $37$ cm. A rock with volume $1000\text{cm}^3$ is then placed in the aquarium and completely submerged. By how many centimeters does the water level rise?
$\textbf{(A)}\ 0.25\qquad\textbf{(B)}\ 0.5\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 2.5$

Solution: The water level will rise $1$ cm for every $100 \cdot 40 = 4000\text{cm}^2$ . Since $1000$ is $\frac{1}{4}$ of $4000$ , the water will rise $\frac{1}{4}\cdot1 = \boxed{0.25}$


Answer: 16.0

An arbitrary circle can intersect the graph of $y = \sin x$ in
$\mathrm{(A) \ } \text{at most }2\text{ points} \qquad \mathrm{(B) \ }\text{at most }4\text{ points} \qquad \mathrm{(C) \  } \text{at most }6\text{ points} \qquad \mathrm{(D) \ } \text{at most }8\text{ points}$ $\mathrm{(E) \  }\text{more than }16\text{ points}$

Solution: Consider a circle whose center lies on the positive $y$ -axis and which passes through the origin. As the radius of this circle becomes arbitrarily large, its curvature near the $x$ -axis becomes almost flat, and so it can intersect the curve $y = \sin x$ arbitrarily many times (since the $x$ -axis itself intersects the curve infinitely many times). Hence, in particular, we can choose a radius sufficiently large that the circle intersects the curve at $\boxed{16}$


Answer: 17.0

An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$ , which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$ $B$ , and $C$ are $12$ $9$ , and $10$ meters, respectively. What is the height, in meters, of the pillar at $E$
$\textbf{(A) }9 \qquad\textbf{(B) } 6\sqrt{3} \qquad\textbf{(C) } 8\sqrt{3} \qquad\textbf{(D) } 17 \qquad\textbf{(E) }12\sqrt{3}$

Solution: The pillar at $B$ has height $9$ and the pillar at $A$ has height $12.$ Since the solar panel is flat, the inclination from pillar $B$ to pillar $A$ is $3.$ Call the center of the hexagon $G.$ Since $\overrightarrow{CG}\parallel\overrightarrow{BA},$ it follows that the solar panel has height $13$ at $G.$ Since the solar panel is flat, the heights of the solar panel at $B,G,$ and $E$ are collinear. Therefore, the pillar at $E$ has height $9+4+4=\boxed{17}.$

Solution: Let the height of the pillar at $D$ be $x.$ Notice that the difference between the heights of pillar $C$ and pillar $D$ is equal to the difference between the heights of pillar $A$ and pillar $F.$ So, the height at $F$ is $x+2.$ Now, doing the same thing for pillar $E$ we get the height is $x+3.$ Therefore, we can see the difference between the heights at pillar $C$ and pillar $D$ is half the difference between the heights at $B$ and $E,$ so \begin{align*} x+3-9&=2 \cdot (x-10) \\  x-6&=2 \cdot (x-10) \\ x&=14. \end{align*} The answer is $x+3=\boxed{17}.$

Solution: We can extend $BA$ and $BC$ to $G$ and $H$ , respectively, such that $AG = CH$ and $E$ lies on $\overline{GH}$  Because of hexagon proportions, $\frac{BA}{AG} = \frac{1}{3}$ and $\frac{BC}{CH} = \frac{1}{3}$ . Let $g$ be the height of $G$ . Because $A$ $B$ and $G$ lie on the same line, $\frac{12-9}{g-12} = \frac{1}{3}$ , so $g-12 = 9$ and $g = 21$ . Similarly, the height of $H$ is $13$ $E$ is the midpoint of $GH$ , so we can take the average of these heights to get our answer, $\boxed{17}$

Solution:
Denote by $h_X$ the height of any point $X$
Denote by $M$ the midpoint of $A$ and $C$ .
Hence, \[h_M = \frac{h_A + h_C}{2} = 11.\] Denote by $O$ the center of $ABCDEF$ . Because $ABCDEF$ is a regular hexagon, $O$ is the midpoint of $B$ and $E$ .
Hence, \[h_O = \frac{h_E + h_B}{2} = \frac{h_E + 9}{2}.\] Because $ABCDEF$ is a regular hexagon, $M$ is the midpoint of $B$ and $O$ .
Hence, \[h_M = \frac{h_B + h_O}{2} = \frac{9 + h_O}{2}.\] Solving these equations, we get $h_E = \boxed{17}$

Solution:
In this solution, we define rise as the change of height (in meters) from the solar panel to the ground. It follows that the rise from $B$ to $A$ is $12-9=3,$ and the rise from $B$ to $C$ is $10-9=1.$ Note that $\vec{BE}=2\vec{BA}+2\vec{BC},$ so the rise from $B$ to $E$ is $2\cdot3+2\cdot1=8.$
Together, the height of the pillar at $E$ is $9+8=\boxed{17}$ meters.

Solution:
WLOG, let the side length of the hexagon be $6$
Establish a 3D coordinate system, in which $A=(0,0,0)$ . Let the coordinates of $B$ and $C$ be $(6,0,0)$ $\left(9,-3\sqrt{3},0\right)$ , respectively. Then, the solar panel passes through $P=(0,0,12), Q=(6,0,9), R=\left(9,-3\sqrt{3},10\right)$
The vector $\vec{PQ}=\langle 6,0,-3\rangle$ and $\vec{PR}=\left\langle 9,-3\sqrt{3}, -2\right\rangle$ . Computing $\vec{PQ} \times \vec{PR}$ by the matrix \[\begin{bmatrix} i & j & k \\ 6 & 0 & -3 \\ 9 & -3\sqrt{3} & -2 \end{bmatrix}\] gives the result $-9\sqrt{3}i -15j -18\sqrt{3} k$ . Therefore, a normal vector of the plane of the solar panel is $\left\langle -9\sqrt{3},-15,-18\sqrt{3}\right\rangle$ , and the equation of the plane is $-9\sqrt{3}x-15y-18\sqrt{3}z=k$ . Substituting $(x,y,z)=(0,0,12)$ , we find that $k=-216\sqrt{3}$
Since $E=\left(0,-6\sqrt{3}\right)$ , we substitute $(x,y)=\left(0,-6\sqrt{3}\right)$ into $-9\sqrt{3}x-15y-18\sqrt{3}z=-216\sqrt{3}$ , which gives $z=\boxed{17}$

Solution: Let the pillars be $AA', BB', \ldots, FF'$ . Since solar panel $A'B'C'D'E'F'$ is a hexagon, the line $B'E'$ hits the midpoint $M$ of $A'C'$ . So, the 3D slope (change in $x$ : change in $y$ : change in $z$ ) of $BE$ is same as $BD$ . If $a$ is side of the hexagonal solar panel, \[B'M' = \frac{1}{2}a, B'E' = a+2\cdot \frac{1}{2}a = 2a\] . So, $B'M:B'E'$ $1:4$ . Since the height of $M$ to ground $ABCDEF$ is $(10+12)/2 = 11$ , the rise (in z) from $B'$ to $M$ is 2 meaning the rise from $B'$ to $E'$ is $8$ . Thus, $EE' = 8+BB' = \boxed{17}$

Solution: Set the midpoint of $\overline{AC}$ as $M$  We know that the height of $M$ is $11$ as it is the midpoint of $\overline{AC}$ , so the height is the average of $A$ and $C$ , which is $\frac{10 + 12}{2}= 11$ . Since $ABCDEF$ is a regular hexagon, $BE = 4\cdot BM$ . Because the increase in height is proportional to the length of the line segments, and the increase in height from $B$ to $M$ is $2$ , the increase in height from $B$ to $E$ is $2\cdot4=8.$ Adding to the height of $B$ , we get $8+9=\boxed{17}$

Solution: Because the three points given are integers, it is likely that the answer is also an integer. This leaves us with $9$ or $17$ . Because both $A$ and $C$ are greater than $9$ and closer to $E$ than $B$ , we can assume that the height increases as the point gets closer to $E$ . Thus, we know the answer is greater than $9$ . The only choice that satisfies both these criteria is $\boxed{17}$


Answer: 31.0

An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$
$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5));  for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num)  { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); }  draw_num((0,0), 17); draw_num((4, 0), 81);  draw_num((0, 4), 1);  draw_num((4,4), 25);   void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); }  foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " ");   [/asy]

Solution:
We begin filling in the table.  The top row has a first term $1$ and a fifth term $25$ , so we have the common difference is $\frac{25-1}4=6$ .  This means we can fill in the first row of the table:
The fifth row has a first term of $17$ and a fifth term of $81$ , so the common difference is $\frac{81-17}4=16$ .  We can fill in the fifth row of the table as shown:
We must find the third term of the arithmetic sequence with a first term of $13$ and a fifth term of $49$ .  The common difference of this sequence is $\frac{49-13}4=9$ , so  the third term is $13+2\cdot 9=\boxed{31}$

Solution: The middle term of the first row is $\frac{25+1}{2}=13$ , since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is $\frac{17+81}{2}=49$ . Applying this again for the middle column, the answer is $\frac{49+13}{2}=\boxed{31}$

Solution: The value of $X$ is simply the average of the average values of both diagonals that contain $X$ . This is $\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{31}$


Answer: 20.0

An arithmetic sequence of positive integers has $\text{n} \ge 3$ terms, initial term $a$ , and common difference $d > 1$ . Carl wrote down all the terms in this sequence correctly except for one term, which was off by $1$ . The sum of the terms he wrote was $222$ . What is $a + d + n$
$\textbf{(A) } 24 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 28 \qquad \textbf{(E) } 26$

Solution:
Since one of the terms was either $1$ more or $1$ less than it should have been, the sum should have been $222-1=221$ or $222+1=223.$
The formula for an arithmetic series is $an+d\left(\dfrac{(n-1)n}2\right)=\dfrac n2\left(2a+d(n-1)\right).$ This can quickly be rederived by noticing that the sequence goes $a,a+d,a+2d,a+3d,\dots,a+(n-1)d$ , and grouping terms.
We know that $\dfrac n2(2a+d(n-1))=221$ or $223$ . Let us now show that $223$ is not possible.
If $\dfrac n2(2a+d(n-1))=223$ , we can simplify this to be $n(a+d(n-1))=223\cdot2.$ Since every expression in here should be an integer, we know that either $n=2$ and $a+d(n-1)=223$ or $n=223$ and $a+d(n-1)=2.$ The latter is not possible, since $n\ge3,d>1,$ and $a>0.$ The former is also impossible, as $n\ge3.$ Thus, $\dfrac n2(2a+d(n-1))\neq223\implies\dfrac n2(2a+d(n-1))=221$
We can factor $221$ as $13\cdot17$ . Using similar reasoning, we see that $221\cdot2$ can not be paired as $2$ and $221$ , but rather must be paired as $13$ and $17$ with a factor of $2$ somewhere.
Let us first try $n=13.$ Our equation simplifies to $2a+12d=34\implies a+6d=17.$ We know that $d>1,$ so we try the smallest possible value: $d=2.$ This would give us $a=17-2\cdot6=17-12=5.$ (Indeed, this is the only possible $d$ .)
There is nothing wrong with the values we have achieved, so it is reasonable to assume that this is the only valid solution (or all solutions sum to the same thing), so we answer $a+d+n=5+2+13=\boxed{20.}$

Solution:
There are $n$ terms, the $x$ th term is $a+(x-1)d$ , summation is $an+dn(n-1)/2=n(a+\frac{d(n-1)}{2})$
The summation of the set is $222 \pm 1 = 221,223$ . First, $221$ : its only possible factors are $1,13,17,221$ , and as said by the problem, $n\ge3$ , so $n$ must be $13,17,$ or $221$ . Let's start with $n=13$ . Then, $a+6d=17$ , and this means $a=5$ $d=2$ . Summing gives $13+5+2=\boxed{20}$ . We don't need to test any more cases, since the problem writes that all $a+d+n$ are the same.

Solution:
We must have the sum of terms of the arithmetic sequence is $222\pm 1$ , which is $221$ or $223$
Since we have $223$ is prime, it cannot be the sum of the arithmetic sequence.
We see that $221$ is just $13\times 17$
We can write any arithmetic sequence with an odd amount of terms like this: $b-md,\cdots ,b-2d,b-d,b,b+d,b+2d,\cdots b+md$ where b is the middle term and d is the common difference.
By the sum of an arithmetic sequence, we have $13b=221$ and $17b=221$ and therefore $(b,n)$ $=$ $(13,17)$ or $(17,13)$
Then $a+d+n=b-md+d+n=30-(m-1)d$
We must have that m is either $\frac{17-1}{2}$ or $\frac{13-1}{2}$ , so m is either $6$ or $8$
So $a+d+n=30-5d$ or $30-7d$
Taking $\mod 7$ , we have no answer choices that give $2$ , and then taking $\mod 5$ gives the only answer that works is $20$
Therefore we have $a+d+n=\boxed{20.}$

Solution:
The formula for the sum of an arithmetic sequence is $n(\frac{a_1+a_n}{2})$ , where $a$ is the first term, $a_n$ is the last term, and $n$ is the number of terms. Let $a$ be the first term, $d$ be the common difference, and $n$ be the number of terms of Carl's sequence. Since the sum the sequence is $1$ less or $1$ more than $222$ , we have \[n(\frac{a+d(n-1)}{2})\pm{1} = 222\] \[n(\frac{a+d(n-1)}{2}) = 222\pm{1}.\] The right-hand side is either $221$ or $223$ . We know that it has to be divisible by $n$ so we can find the factors of $221$ and $223$ . Checking all the primes less than $15$ , we find that $223$ is prime and $221=13\cdot17$
Because $n\ge3$ , the sum must be $221$ and the only possible values of $n$ are $13$ and $17$ . We can test both cases.
Case 1: $n=17$ Substituting for $n$ gives us $a+8d=17$ . Since the sequence consists of only positive integers, $d$ is an integer. We know that $d>1$ but if $d\ge2$ , then $8d>13$ . Hence, this case is not possible.
Case 2: $n=13$ Substituting for $n$ gives us $a+6d=17$ . Using the same logic from case 1, we get $3>d>1$ , so $d=2$ . Solving for $a$ , we get $a=5$ . Therefore, $a+d+n=5+2+13=20$ , so the answer is $\boxed{20.}$ .
~azc1027


Answer: 33.0

Solution:
We can consider the contributions of the sides of the three layers and the tops of the layers separately.
Layer $n$ (counting from the top starting at $1$ ) has $4$ side faces each with $n$ unit squares, so the sides of the pyramid contribute $4+8+12=24$ for the surface area.
The tops of the layers when combined form the same arrangement of unit cubes as the bottom of the pyramid, which is a $3\times 3$ square, hence this contributes $9$ for the surface area.
You do not have to count the side underneath, since it is not exposed.
Thus, the artist paints $24+9=33 \rightarrow \boxed{33}$ square meters.


Answer: 155.0

An athlete's target heart rate, in beats per minute, is $80\%$ of the theoretical maximum heart rate. The maximum heart rate is found by subtracting the athlete's age, in years, from $220$ . To the nearest whole number, what is the target heart rate of an athlete who is $26$ years old?
$\textbf{(A)}\ 134\qquad\textbf{(B)}\ 155\qquad\textbf{(C)}\ 176\qquad\textbf{(D)}\ 194\qquad\textbf{(E)}\ 243$

Solution: The maximum heart rate is $220-26=194$ beats per minute. The target heart rate is then $0.8*194 \approx \boxed{155}$ beats per minute.


Answer: 200.0

An auditorium with $20$ rows of seats has $10$ seats in the first row.  Each successive row has one more seat than the previous row.  If students taking an exam are permitted to sit in any row, but not next to another student in that row, then the maximum number of students that can be seated for an exam is
$\text{(A)}\ 150 \qquad \text{(B)}\ 180 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 400 \qquad \text{(E)}\ 460$

Solution:
We first note that if a row has $n$ seats, then the maximum number of students that can be seated in that row is $\left\lceil \frac{n}{2} \right\rceil$ , where $\lceil x \rceil$ is the smallest integer greater than or equal to $x$ .  If a row has $2k$ seats, clearly we can only fit $k$ students in that row.  If a row has $2k+1$ seats, we can fit $k+1$ students by putting students at the ends and then alternating between skipping a seat and putting a student in.
For each row with $10+k$ seats, there is a corresponding row with $29-k$ seats.  The sum of the maximum number of students for these rows is \[\left\lceil \frac{10+k}{2}\right\rceil +\left\lceil \frac{29-k}{2} \right\rceil = 20.\] There are $20/2=10$ pairs of rows, so the maximum number of students for the exam is $20\times 10=200\rightarrow \boxed{200}$


Answer: 10.0

An automobile travels $a/6$ feet in $r$ seconds. If this rate is maintained for $3$ minutes, how many yards does it travel in $3$ minutes?
$\textbf{(A)}\ \frac{a}{1080r}\qquad \textbf{(B)}\ \frac{30r}{a}\qquad \textbf{(C)}\ \frac{30a}{r}\qquad \textbf{(D)}\ \frac{10r}{a}\qquad \textbf{(E)}\ \frac{10a}{r}$

Solution: Use dimensional analysis. \[\frac{a/6 \text{ feet}}{r \text{ seconds}} \cdot \frac{1 \text{ yard}}{3 \text{ feet}} \cdot \frac{60 \text{ seconds}}{1 \text{ minute}} \cdot 3 \text{ minutes}\] \[\frac{10a}{r} \text{ yards}\] The answer is $\boxed{10}$


Answer: 85.0

An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$ -plane and is tangent to the $x$ -axis. What is the length of its major axis

Solution:
An ellipse is defined to be the locus of points $P$ such that the sum of the distances between $P$ and the two foci is constant.  Let $F_1 = (9, 20)$ $F_2 = (49, 55)$ and $X = (x, 0)$ be the point of tangency of the ellipse with the $x$ -axis.  Then $X$ must be the point on the axis such that the sum $F_1X + F_2X$ is minimal.  (The last claim begs justification:  Let $F'_2$ be the reflection of $F_2$ across the $x$ -axis. Let $Y$ be where the line through $F_1$ and $F’_2$ intersects the ellipse.  We will show that $X=Y$ . Note that $X F_2 = X F’_2$ since $X$ is on the $x$ -axis.  Also, since the entire ellipse is on or above the $x$ -axis and the line through $F_2$ and $F’_2$ is perpendicular to the $x$ -axis, we must have $F_2 Y \leq F’_2 Y$ with equality if and only if $Y$ is on the $x$ -axis.  Now, we have \[F_1 X + F'_2 X = F_1 X + F_2 X = F_1 Y + F_2 Y \leq F_1 Y + F’_2 Y\] But the right most sum is the straight-line distance from $F_1$ to $F’_2$ and the left is the distance of some path from $F_1$ to $F_2$ ., so this is only possible if we have equality and thus $X = Y$ ). Finding the optimal location for $X$ is a classic problem: for any path from $F_1$ to $X$ and then back to $F_2$ , we can reflect (as above) the second leg of this path (from $X$ to $F_2$ ) across the $x$ -axis.  Then our path connects $F_1$ to the reflection $F_2'$ of $F_2$ via some point on the $x$ -axis, and this path will have shortest length exactly when our original path has shortest length.  This occurs exactly when we have a straight-line path, and by the above argument, this path passes through the point of tangency of the ellipse with the $x-$ axis.
The sum of the two distances $F_1 X$ and $F_2X$ is therefore equal to the length of the segment $F_1F_2'$ , which by the distance formula is just $d = \sqrt{(9 - 49)^2 + (20 + 55)^2} = \sqrt{40^2 + 75^2} = 5\sqrt{8^2 + 15^2} = 5\cdot 17 = 85$
Finally, let $A$ and $B$ be the two endpoints of the major axis of the ellipse.  Then by symmetry $AF_1 = F_2B$ so $AB = AF_1 + F_1B = F_2B + F_1B = d$ (because $B$ is on the ellipse), so the answer is $\boxed{085}$

Solution:
An ellipse is defined as the set of points where the sum of the distances from the foci to the point is fixed. The length of major axis is equal to the sum of these distances $(2a)$ . Thus if we find the sum of the distances, we get the answer. Let k be this fixed sum; then we get, by the distance formula:
$k = \sqrt{(x - 9)^2 + 20^2} + \sqrt{(x - 49)^2 + 55^2}$
This is the equation of the ellipse expressed in terms of $x$ . The line tangent to the ellipse at the given point $P(x, 0)$ will thus have slope $0$ . Taking the derivative gives us the slope of this line. To simplify, let $f(x) = (x - 9)^2 + 20^2$ and $g(x) = (x - 49)^2 + 55^2$ . Then we get:
$0 = \frac{f^\prime(x)}{2\sqrt{f(x)}} + \frac{g^\prime(x)}{2\sqrt{g(x)}}$
Next, we multiply by the conjugate to remove square roots. We next move the resulting $a^2 - b^2$ form expression into form $a^2 = b^2$
$\frac{(f^\prime(x))^2}{4\cdot f(x)} = \frac{(g^\prime(x))^2}{4\cdot g(x)}$
We know $f^\prime(x) = 2x - 18$ and $g^\prime(x) = 2x - 98$ . Simplifying yields:
$\frac{(x - 9)^2}{(x - 9)^2 + 20^2} = \frac{(x - 49)^2}{(x - 49)^2 + 55^2}$
To further simplify, let $a = (x - 9)^2$ and $b = (x - 49)^2$ . This means $\frac{a}{a + 400} = \frac{b}{b + 3025}$ . Solving yields that $16b = 121a$ .
Substituting back $a$ and $b$ yields:
$16 \cdot (x - 49)^2 = 121 \cdot (x - 9)^2$
Solving for $x$ yields $x = \frac{59}{3}$ . Substituting back into our original distance formula, solving for $k$ yields $\boxed{085}$


Answer: 3030.0

An empty $2020 \times 2020 \times 2020$ cube is given, and a $2020 \times 2020$ grid of square unit cells is drawn  on each of its six faces. A beam is a $1 \times 1 \times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:
What is the smallest positive number of beams that can be placed to satisfy these conditions?

Solution:
Place the cube in the xyz-coordinate, with the positive x-axis pointing forward, the positive y-axis pointing right, and the positive z-axis pointing up. Let the position of a unit cube be $(x, y, z)$ if it is centered at $(x, y, z)$ . Place the $2020 \times 2020 \times 2020$ cube so that the edges are parallel to the axes, and two of its corners are at $(1, 1, 1)$ and $(2020, 2020, 2020)$ . Now call a beam z-oriented if its endpoints differ only in z-coordinates, and similarly call it x-oriented or y-oriented if the endpoints differ in x- or y-coordinates, respectively.
We claim that the answer is $\boxed{3030}$ . First we will prove 3030 suffices. Place the first beam so its endpoints lie at $(1, 1, 1)$ , and $(2020, 1, 1)$ . Place the 2nd beam so its endpoints lie at $(1, 1, 2)$ and $(1, 2020, 2)$ . Place the 3rd beam so its endpoints lie at $(2, 2, 1)$ and $(2, 2, 2020)$ . Now it is clear that the only faces among these three beams that are not touching the faces of the $2020 \times 2020 \times 2020$ cube of the face of another beam are the top face of beam 2, and the front and right faces of beam 3.


Answer: 72.0

An equilangular octagon has four sides of length $1$ and four sides of length $\frac{\sqrt{2}}{2}$ , arranged so that no two consecutive sides have the same length.  What is the area of the octagon?
$\textbf{(A) } \frac72\qquad \textbf{(B) }  \frac{7\sqrt2}{2}\qquad \textbf{(C) }  \frac{5+4\sqrt2}{2}\qquad \textbf{(D) }  \frac{4+5\sqrt2}{2}\qquad \textbf{(E) }  7$

Solution:
The area of the octagon can be divided up into $5$ squares with side $\frac{\sqrt2}2$ and $4$ right triangles, which are half the area of each of the squares.
Therefore, the area of the octagon is equal to the area of $5+4\left(\frac12\right)=7$ squares.
The area of each square is $\left(\frac{\sqrt2}2\right)^2=\frac12$ , so the area of $7$ squares is $\boxed{72}$

Solution:
Refer to the following diagram:
AMC10 2005A P20.png
(Picture made on Geogebra)
Note that each square has area $\frac14$ , and each triangle has area $\frac18$ . The total area is $12\cdot\frac14+4\cdot\frac18=\frac72 \Longrightarrow \boxed{72}$


Answer: 6.0

An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}4\sqrt3\qquad\textbf{(E)}\hspace{.05in}6\sqrt3$

Solution:
Let the perimeter of the equilateral triangle be $3s$ . The side length of the equilateral triangle would then be $s$ and the sidelength of the hexagon would be $\frac{s}{2}$
A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio $1 : 4$ , since the sidelength of the small equilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is $1$ . The area of the hexagon is then $1 \times 6 = \boxed{6}$

Solution:
Let the side length of the equilateral triangle be $s$ and the side length of the hexagon be $y$ . Since the perimeters are equal, we must have $3s=6y$ which reduces to $s=2y$ . Substitute this value in to the area of an equilateral triangle to yield $\dfrac{(2y)^2\sqrt{3}}{4}=\dfrac{4y^2\sqrt{3}}{4}$
Setting this equal to $4$ gives us $\dfrac{4y^2\sqrt{3}}{4}=4\implies 4y^2\sqrt{3}=16\implies y^2\sqrt{3}=4$
Substitute $y^2\sqrt{3}$ into the area of a regular hexagon to yield $\dfrac{3(4)}{2}=6$
Therefore, our answer is $\boxed{6}$

Solution: Let the side length of the triangle be $s$ and the side length of the hexagon be $t$ . As explained in Solution 1, $s=2t$ , or $t=\frac{s}{2}$ . The area of the triangle is $\frac{s^2\sqrt3}{4}=4$ and the area of the hexagon is $\frac{t^2\sqrt3}{4} \cdot 6=\frac{3t^2\sqrt3}{2}$ . Substituting $\frac{s}{2}$ in for $t$ , we get \[\frac{\frac{3s^2\sqrt3}{4}}{2}=\frac{3s^2\sqrt3}{8}.\] $\frac{s^2\sqrt3}{4}=4 \implies \frac{s^2\sqrt3}{8}=2 \implies \frac{3s^2\sqrt3}{8}=\boxed{6}$


Answer: 6.0

An equilateral triangle is drawn with a side of length $a$ . A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is:
$\textbf{(A)}\ \text{Infinite} \qquad\textbf{(B)}\ 5\frac {1}{4}a \qquad\textbf{(C)}\ 2a \qquad\textbf{(D)}\ 6a \qquad\textbf{(E)}\ 4\frac {1}{2}a$

Solution:
The perimeter of the first triangle is $3a$ . The perimeter of the 2nd triangle is half of that, after drawing a picture. The 3rd triangle's perimeter is half the second's, and so on. Therefore, we are computing $3a+\frac{3a}{2}+\frac{3a}{4}+\cdots$
The starting term is $3a$ , and the common ratio is $1/2$ . Therefore, the sum is $\frac{3a}{1-\frac{1}{2}}=\boxed{6}$