Solution(s):

Let the number of pennies be $p.$ Then, the number of nickels is $13-p$ and $p-1,$ so $13-p=p-1.$ This means we have $7$ pennies and $6$ nickels.

Therefore, the number of cents is $7+6\cdot 5=37.$

Thus, the correct answer is C.

Solution(s):

$$\begin{align*} \dfrac{2^3 + 2^3}{2^{-3} + 2^{-3}} &= \dfrac{2\cdot 2^3}{2\cdot 2^{-3}} \\&=2^6 \\&= 64 \end{align*}$$

Thus, the correct answer is E.

Solution(s):

Let the path distance be $t.$ Then, we get: $$\begin{align*}t &= \dfrac t5 + 20 + \dfrac t3 \\ t &= 20 + \dfrac {8t}{15} \\ \dfrac 7{15} t &= 20\\ t &= \dfrac{300}7.\end{align*}$$

Thus, the correct answer is E.

Solution(s):

Let the price for a muffin be $m$ and let the price for a banana be $b.$ Then, $$\begin{align*} 2(4m+3b) &= 16b + 2m\\ 8m+6b &= 16b + 2m \\ 10b &= 6m\\ m &= \dfrac 53 b.\end{align*}$$

Thus, the correct answer is B.

Solution(s):

Let the smaller side of a pane be a distance of $x.$

Then, the side length is $5\cdot 2 + 4x.$ Also, the other direction has pane lengths of $2.5x.$

This means the side length is $3\cdot 2 + 2(2.5x) .$ We solve for $x$ as follows: $$\begin{align*}10+4x &= 6+5x \\ x&= 4\end{align*}$$

Therefore, the side length is $$10+4\cdot 4 = 26.$$

Thus, the correct answer is A.

Solution(s):

Suppose we buy $6$ balloons. Then, we can buy $3$ at full price and $3$ at a price of $\frac 23$ of a ballon.

Therefore, we can buy it at a price of $5$ balloons. Thus, with the money to buy $30$ balloons, we could buy $30 \cdot \frac 65 = 36$ balloons.

Thus, the correct answer is C.

Solution(s):

By definition, we know $$A = \dfrac {x+ 100}{100}B$$$$= B + \dfrac x{100}B.$$

This implies, $$\begin{align*} (A-B) &= \dfrac x{100}B\\ 100\left(\dfrac{A-B} B\right) &=x.\end{align*}$$

Thus, the correct answer is A.

Solution(s):

This means it travels $\dfrac{b}{18}$ yards in $t$ seconds since a yard is $3$ feet. Then, in one second, it travels $\dfrac{b}{18t}$ yards.

Therefore, in $3$ minutes which is $180$ seconds, it travels $$\dfrac{b}{18t}\cdot 180 = \dfrac {10b}t.$$

Thus, the correct answer is E.

Solution(s):

Observe that: $$\begin{align*} \dfrac{wz}{wz}\cdot \dfrac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} &= 2014\\ \dfrac{w+z}{z-w} &= 2014\\ \dfrac{w+z}{w-z} &= -1\cdot 2014\\ \dfrac{w+z}{w-z}&=-2014.\end{align*}$$

Thus, the correct answer is A.

Solution(s):

From the leftmost digit, we know $A+B = D\leq 9$ since it has no carryover.

This means there is no carryover when adding the units digit, making $C + D = D.$ Thus, $C = 0.$ Then, since $A,B \neq 0$ and $A \neq B,$ we know $$D=A+B \geq 1+2 = 3.$$

Since $3 \leq D \leq 9,$ we have $7$ values for $D.$

Thus, the correct answer is C.

Solution(s):

We need to find the smallest possible $n$ such that $$1- \dfrac n{100} < (0.85)^2,$$ $$1- \dfrac n{100} < (0.9)^3,$$ $$1- \dfrac n{100} < (0.75)(0.95).$$ Note that $$0.75\cdot 0.95 = 0.85^2-0.1^2 < 0.85^2,$$ so we don't need to worry about the first condintion since the last condition is true. Then, the second condition yields $$1- \dfrac n{100} < 0.729.$$ $$n > 27.1.$$

Also, we also can see that $$1- \dfrac n{100} < 0.95\cdot 0.75$$$$1- \dfrac n{100} < \dfrac 34 \cdot \dfrac{19}{20}$$$$1- \dfrac n{100} < \dfrac{57}{80}$$$$1- \dfrac n{100} < \dfrac{285}{400}.$$

Therefore, $$n > \dfrac{115}{4} = 28.75.$$ Combining our conditions yields a smallest $n$ of $n=29.$

Thus, the correct answer is C.

Solution(s):

The fifth-largest divisor is $2,014,000,000$ divided by the fifth smallest divisor.

The prime factoriztion of $2,014,000,000$ is: $$2^7\cdot 5^6 \cdot 19\cdot 53.$$ This makes the first $5$ smallest divisors $1,2,4,5,8$ Therefore, the fifth smallest divisor is $8,$ and the fifth largest divisor must be: $$\dfrac{2014000000}8 =251750000.$$

Thus, the correct answer is C.

Solution(s):

Since $AB = BC = AC$ by rotational symmetry, we know it is an equilateral triangle.

Then, one-fourth of $AB$ can be found as a the leg of a right triangle with hypotenuse $1$ and is opposite to the $60^{\circ}$ angle, making it $$\sin(60^\circ) = \dfrac{\sqrt 3}2.$$

As such, $$AB = 4\cdot \dfrac{\sqrt 3}2 = 2\sqrt 3.$$

Then, since it is an equilateral triangle, it has area $$\dfrac{s^2 \sqrt 3}4 = \dfrac{12\sqrt 3}4 = 3 \sqrt 3.$$

Thus, the correct answer is B.

Solution(s):

We know that the difference of the numbers $cba$ and $abc$ is equal to: $$100c + 10b+a - 100a - 10b-c$$$$= 99(c-a)$$ We know that this number also must be a multiple of $55.$ As $\gcd(55,99)$ is $11,$ we know that $c-a$ is a multiple of $5,$ and $c > a.$

This makes $a = 1, b = 0, c = 6$ the only possible value with $a+ b+c \leq 7$ as every other combination has $a+b+c > 7.$ As such, $a^2+b^2+c^2 = 37.$

Thus, the correct answer is D.

Solution(s):

The area of $EFB$ is equal to $$\dfrac{EF\cdot AD}2 .$$

Similarly, The area of $ABCD$ is equal to $$AB\cdot AD .$$

Thus, their ratio is $$\dfrac{EF}{2\cdot AD} = \dfrac{EF}{4\cdot AB}.$$

Then, $$\begin{align*} EF &= AF - AE \\&= AB \tan(\angle ADE) \\&\quad - AB \tan(\angle ADF) \\&= AB( \tan(60^\circ) - \tan(30^ \circ)) \\&= AB\left(\sqrt 3 - \dfrac {\sqrt{3}}3\right) \\&= AB \left(\dfrac{2\sqrt 3}3\right)\end{align*}$$ This makes our result $$\dfrac{EF}{4\cdot AB} = \dfrac{AB(\frac{2 \sqrt 3}3)}{4\cdot AB} = \dfrac {\sqrt 3}6.$$

Thus, the correct answer is A.

Solution(s):

There are two cases: All of the numbers are the same or one of the numbers is different from the other ones.

Case 1 - All numbers are the same: The probability of this is $$\dfrac{6}{6^4}$$ since there are $6$ combinations where they are all the same due to the $6$ possible values for the dice.

Case 2 - One of the numbers are: The probability of this is $$\dfrac{4\cdot 5\cdot 6}{6^4}$$ since there are $4$ ways to choose the dice that is different, $6$ values for the dice that are the same, and $5$ values for the other dice.

This makes the probability $$\dfrac{6\cdot 5\cdot 4+6} {6^4} = \dfrac{21}{216} = \dfrac 7{72}.$$

Thus, the correct answer is B.

Solution(s):

Observe that: $$\begin{align*} 10^{1002}-4^{501} &=10^{1002}-2^{1002} \\&= 2^{1002} (5^{1002}-1) \\&= 2^{1002} (25^{501}-1).\end{align*}$$ Then, since $$25^{501} \equiv 1^{501} \equiv 1 \mod 8,$$ We know that $25^{501}-1$ is a multiple of $2^3,$ making our number a multiple of $2^{1005}.$

However, $$\begin{align*} &25^{501} -1\\&\equiv 9^{501} -1\mod 16 \\&\equiv 9\left(81^{250}\right)-1 \mod 16\\&\equiv 8 \mod 16, \end{align*}$$ so $25^{501}-1$ is not a multiple of $2^4,$ and as such, our number isn't a multiple of $2^{1006} .$

Thus, the correct answer is D.

Solution(s):

Since $9$ is the median, there are $5$ numbers less than or equal to $9$ besides the middle.

Also, there are $5$ numbers greater than or equal to $9$ besides the middle.

Also, since $8$ is the unique mode, it must show up at least twice.

Furthermore, since we are trying to maximize the largest number and we have their mean, we need to minimize the sum of the first $10$ numbers. Finally, notice that the sum should be $11\cdot 10=110.$

Now, we could case on the number of times $8$ appears since every other number must show up less than it.

If it appears twice, every other number must show up at most once, so the maximum sum is $$1+2+3+8+8+9$$$$+10+11+12+13$$$$= 77.$$ This makes the largest number $33.$

If it appears three times, every other number must show up at most twice, so the maximum sum is $$1+1+8+8+8$$$$+9+9+10+10+11$$$$= 75.$$ This makes the largest number $35.$

If it appears four times, every other number must show up at most once, so the maximum sum is $$1+8+8+8+8$$$$+9+9+9+9+10$$$$= 79.$$ This makes the largest number $31.$

This makes the maximum $35.$

Thus, the correct answer is E.

Solution(s):

First, without loss of generality, we could choose some point on the outer circle. Then, the second point can be chosen in a region on the other circle.

This region is such that it has a line that intersects the circle, so the edge of the region is such that the chord is perpendicular with the inner circle.

If we look at the angle at the center, we can see that it has 2 right triangles where the adjacent side is $1$ and the hypotenuse is $2,$ making $$\cos \left(\dfrac \theta 2\right) = \dfrac 12.$$

Thus, $\dfrac \theta 2 = 60^\circ,$ making $\theta = 120 ^\circ .$

Therefore, the probability is $$\dfrac {120^\circ}{360^\circ} = \dfrac 13 .$$

Thus, the correct answer is D.

Solution(s):

First, note that $$x^4-51x^2+50$$$$= (x^2-50)(x^2-1).$$

If $(x^2-50)(x^2-1) < 0$ means that one of the terms is negative.

Since $$x^2-50 < x^2-1,$$ it must be that x^2-50 < 0, x^2-1 >0. This means $1 < x^2 < 50,$ making $$1< |x| \leq 7,$$ resulting in $12$ solutions.

Thus, the correct answer is C.

Solution(s):

Let the base of the altitude from $C$ to $AB$ be $E.$ and let the base of the altitude from $D$ to $AB$ be $F.$ Also, let $BC = 10$ since we can assign any value. This yields the following diagram:

Then, let $FB = x$ and the altitude be $h.$ This means $$\begin{align*}AE &= 33-x-EF\\&=33-x-21 \\&= 12-x.\end{align*}$$

This suggests that: $$14^2 = (12-x)^2 + h^2$$ $$10^2 = x^2 + h^2.$$ Subtracting the equations, we get: $$96 = 144 - 24x$$ $$x = 2.$$ Then, we want to find $$\sqrt{ (21+x)^2+h^2}$$$$= \sqrt{21^2 + 42x + (x^2 + h^2) }$$$$= \sqrt{441+42\cdot 2 + 100}$$$$= \sqrt{625}$$$$=25.$$

Thus, the correct answer is B.

Solution(s):

The distance from the center of the square to the center of the semicircles can be found as a hypotenuse of a right triangle.

One of the legs is from the center of the square to the center of one of the sides which is of distance $1.$

The other leg is from the center of the side to the center of one of the semicircles which is of distance $\dfrac 12.$ This also shows that the radius of the semicircles is $\dfrac 12.$

Therefore, the distance from the center of the square to the center of the semicircle is $$\sqrt{1^2 + \dfrac 12 ^2} = \dfrac {\sqrt 5}2.$$ Then , we subtract $\dfrac 12$ for the radius of the semicircle. This makes the radius of the circle $$\dfrac{\sqrt 5 -1}2 .$$

Thus, the correct answer is B.

Solution(s):

Let the radius of the sphere be $r.$ Then, let the top have a radius of $1$ and let the bottom have a radius of $x.$ Then, we could extend the cone to make a smaller cone on top with height $h.$ This makes a larger cone with radius $x$ and height $h + 2r.$

Note that $$\begin{align*}\dfrac{h+2r}h &= x\\ \dfrac {2r}h &= x-1\\h &= \dfrac {2r}{x-1}.\end{align*}$$

This makes the volume of the whole cone $$\dfrac {\pi(h+2r)(x^2)}{3},$$ and the volume of the top section $$\dfrac h3\pi.$$

As such, the volume of the truncated cone is $$\dfrac {hx^2+2rx^2-h}3\pi.$$

Then, by making a slice in the cone, we get the picture above. Then, by making two lines from the corners to the center, we get similar right triangles, which makes $$\dfrac r1 = \dfrac xr.$$ Thus, we get $r^2 = x.$

This makes the volume $$\dfrac {h(x^2-1)+2rx^2}3 \pi$$ and $$h = \dfrac {2r}{x-1} = \dfrac{2r(x+1)}{x^2-1}.$$ This further simplifies the volume to $$\dfrac {2r(x+1)+2rx^2}3 \pi$$$$= \dfrac{2r(x^2+x+1)}3 \pi .$$ The volume of the sphere is $\dfrac{4r^3\pi}3.$ This implies $$\dfrac{8r^3 \pi}3 = \dfrac {2r(x^2+x+1)}3 \pi .$$ Thus, $$\begin{align*}4x &= x^2+x+1\\ x^2-3x+1&=0\\x &= \dfrac {3 + \sqrt 5}2.\end{align*}$$ Since we made the top circle of radius $1,$ the ratio is $$\dfrac{3+\sqrt 5}2.$$

Thus, the correct answer is E.

Solution(s):

We wil always have the numbers $$1,2,3,4,5$$ by choosing the number, and we will always have $$14,13,12,11,10$$ by choosing the entire set except one. Then, $15$ is also chosen.

Therefore, we need to find $$6,7,8,9.$$ Furthermore, if we have $6$ or $7,$ we can get $8$ and $9$ by choosing the rest.

This means a configuration is only not bad if and only if we can find a set of numbers put together that adds to $6$ and $7.$

The only ways to make a sum $6$ is with a sequence of $$(1,2,3),(1,5),\text{ or }(2,4).$$ The only ways to make a sum $7$ is with a sequence of $$(1,2,4),(2,5),\text{ or } (3,4).$$ Note that the triples can be in any order.

This means that for any configuration without one of them, we need to prevent all of the combinations for one of the numbers. Thus, we can case on which number isn't accounted for.

Case 1: There is no $6.$

This means the $1$ isn't next to the $5.$ Thus, it is next to the $2,3$ or $4.$ However, it can't have both $2$ and $3,$ so the $1$ must be next to $4$ and one of $2$ and $3.$

Also, this means the $2$ isn't next to the $4.$ Thus, it is next to the $1,3$ or $5.$ However, it can't have both $1$ and $3,$ so the $2$ must be next to $5$ and one of $1$ and $3.$

Then, $5$ must be next to the $2$ and $3$ since if it is connected to the $4,$ it would make a $$1-2-3$$ which has a sum of $6.$ Thus, the configuration is starts with $$3-5-2.$$ Since $2$ isn't next to $4,$ we have $$3-5-2-1-4$$ as the only configuration.

Case 2: There is no $7.$

This means the $2$ isn't next to the $5.$ Thus, it is next to the $1,3$ or $4.$ However, it can't have both $1$ and $4,$ so the $2$ must be next to $4$ and one of $1$ and $3.$

Also, this means the $4$ isn't next to the $3.$ Thus, it is next to the $1,2$ or $5.$ However, it can't have both $1$ and $2,$ so the $4$ must be next to $5$ and one of $1$ and $2.$

Then, $5$ must be next to the $1$ and $4$ since if it is connected to the $3,$ it would make a $$1-2-4$$ which has a sum of $6.$ Thus, the configuration is starts with $$1-5-4.$$ Since $4$ isn't next to $3,$ we have $$1-5-4-2-3$$ as the only configuration.

This means there are only two configurations.

Thus, the correct answer is B.

Solution(s):

Let the value $e_i$ be the probability of winning at each time. Then, for all $1 \leq i \leq 9,$ we have $$e_i = \dfrac{10-i}{10}e_{i+1} + \dfrac i{10} e_{i-1}.$$ Also, by symmetry, we have $e_5 = \dfrac 12.$ Our goal is to find $e_1.$

Then, $$\begin{align*}e_4 &= \dfrac 6{10} e_5 + \dfrac{4}{10}e_3 \\&=\dfrac 6{10} (\dfrac 12) + \dfrac{4}{10}e_3 \\&=\dfrac 3{10} + \dfrac{4}{10}e_3 .\end{align*}$$

Then, $$\begin{align*} e_3 &= \dfrac 7{10} e_4 + \dfrac{3}{10}e_2 \\&= \dfrac 7{10} (\dfrac 3{10} + \dfrac{4}{10}e_3) + \dfrac{3}{10}e_2 \\&=\dfrac {21}{100}+ \dfrac {28}{100}e_3 + \dfrac{30}{100}e_2\\&= \dfrac {21}{72} + \dfrac{30}{72}e_2 \\&= \dfrac 7{24} + \dfrac 5{12}e_2\end{align*}$$

Then, $$\begin{align*}e_2 &= \dfrac 8{10} e_3 + \dfrac{2}{10}e_1 \\&= \dfrac 8{10} ( \dfrac 7{24} + \dfrac 5{12}e_2) + \dfrac{2}{10}e_1 \\&= \dfrac {56}{240} + \dfrac{40}{120}e_2 + \dfrac 2{10}e_1 \\&=\dfrac 7{30}+ \dfrac 13 e_2 + \dfrac 15 e_1\\&=e_2 = \dfrac 7{20} + \dfrac 3{10}e_1.\end{align*}$$

Then, $e_1 = \dfrac 9{10} e_2 + \dfrac{1}{10}e_0 =$ $\dfrac 9{10} ( \dfrac 7{20} + \dfrac 3{10}e_1 ) =$$\dfrac {63}{200} + \dfrac {27}{100}e_1 .$

This suggests that $$\dfrac{73}{100}e_1 = \dfrac {63}{200}$$ $$e_1 = \dfrac{63}{146}.$$

Thus, the correct answer is C.