Answer: C

Small Hint:

Negative exponent means reciprocal of a power

Big Hint:

Compute $2-\frac14$

Solution:

Since $(-2)^{-2}=\frac{1}{(-2)^2}=\frac14$, the expression is $$2-\frac14=\frac74.$$

Thus, the correct answer is C.

Answer: B

Small Hint:

Two tasks take $100$ minutes

Big Hint:

Add one more $50$-minute task after $2\!:\!40$ PM

Solution:

The time it takes to do $2$ tasks is $100$ minutes. Thus, it takes $50$ more minutes after $2:40,$ which is $3:30.$

Thus, the correct answer is B .

Answer: A

Small Hint:

Test whether $28$ is repeated two or three times

Big Hint:

The remaining total must divide by the remaining count

Solution:

Let the number written twice be $x$, and let the number written three times be $y$. Then $2x+3y=100$.

If $x=28$, then $3y=100-56=44$, impossible for an integer $y$. Therefore $y=28$, and $2x=100-84=16$, so $x=8$.

Thus, the correct answer is A.

Answer: C

Small Hint:

Dan receives $1-\frac15-\frac13-\frac14$

Big Hint:

Compare all shares with denominator $60$

Solution:

Since $\frac 13 > \frac 14 > \frac 15,$ we know Beth ate more than Cyril and Cyril ate more than Alex. Thus, those three are in order.

The amount Dan ate is $$1- \dfrac 13 - \dfrac 14 - \dfrac 15 = \dfrac{13}{60}.$$ This is greater than $\frac 15$ and less than $\frac 14 ,$ so Dan is in between Cyril and Alex. This makes the order Beth, Cyril, Dan, Alex.

Thus, the correct answer is C .

Answer: B

Small Hint:

Work backward from Marta’s $6$th place

Big Hint:

Hikmet is $6$ places behind Rand

Solution:

Marta finished $6$th, so Jack finished $5$th. Since Jack finished $2$ places behind Todd, Todd finished $3$rd. Since Todd finished $1$ place behind Rand, Rand finished $2$nd.

Hikmet finished $6$ places behind Rand, so Hikmet finished $8$th.

Thus, the correct answer is B.

Answer: E

Small Hint:

Monday and Wednesday are fixed

Big Hint:

Tuesday must be a running day

Solution:

Marley plays basketball on Monday and golf on Wednesday. She cannot fit all three running days among Thursday, Friday, Saturday, and Sunday without having two consecutive running days, so Tuesday must be a running day.

From Thursday through Sunday, she must run twice, swim once, and play tennis once. Tennis cannot be the day after running or swimming, so tennis must be Thursday. Then the two remaining running days must be Friday and Sunday, leaving Saturday for swimming.

Thus, the correct answer is E.

Answer: A

Small Hint:

Compute $1\diamond2$ and $2\diamond3$ first

Big Hint:

Remember $a\diamond b=a-\frac1b$

Solution:

$$\begin{aligned} &((1\diamond2)\diamond3)-(1\diamond(2\diamond3)) \\ &= \left(\dfrac 12- \dfrac 13\right) - \left(1 \diamond \dfrac 53\right)\\ &= \left(\dfrac 12 - \dfrac 13\right) - \left(1-\dfrac 35\right)\\ &= \dfrac 16 - \dfrac 25 \\&= -\dfrac{7}{30} \end{aligned}$$

Thus, the correct answer is A .

Answer: E

Small Hint:

Track what happens to the axes under the three transformations

Big Hint:

A half turn after a $y$-reflection is a reflection in the $x$-axis

Solution:

The first rotation puts the F below the $x$-axis.

A half turn is equivalent to reflecting in both coordinate axes. Therefore the final half turn cancels the preceding reflection in the $y$-axis and leaves a reflection in the $x$-axis.

Reflecting the rotated figure in the $x$-axis produces choice E.

Thus, the correct answer is E .

Answer: B

Small Hint:

Subtract the smaller semicircle from the larger quarter circle

Big Hint:

The smaller circle has radius $\frac32$

Solution:

The larger boundary is a quarter circle of radius $3$, so its area is $\frac14\pi\cdot3^2=\frac{9\pi}{4}$.

The inner boundary is the right half of a circle of radius $\frac32$, so its area is $\frac12\pi\left(\frac32\right)^2=\frac{9\pi}{8}$.

The shaded area is the difference, $\frac{9\pi}{4}-\frac{9\pi}{8}=\frac{9\pi}{8}$.

Thus, the correct answer is B.

Answer: C

Small Hint:

Count how many odd negative factors there are

Big Hint:

An odd product with a factor of $5$ has units digit $5$

Solution:

There are $1007$ odd numbers greater than $-2015.$

Our product is of an odd number of negative numbers, so the result is negative.

Also, we multiply by $-5$ in there, so the product is a multiple of $5,$ making it end in $5$ or $0.$ None of our factors are even, so the product can’t be even.

Therefore, the product must end in $5.$

Thus, the correct answer is C .

Answer: B

Small Hint:

List one-digit primes and two-digit numbers using only $2,3,5,7$

Big Hint:

Two-digit primes ending in $3$ or $7$ must still pass divisibility tests

Solution:

The available digits are $2,3,5,7$. There are $4$ one-digit numbers and $4^2=16$ two-digit numbers, for $20$ total choices.

All $4$ one-digit choices are prime. A two-digit prime cannot end in $2$ or $5$, so checking endings $3$ and $7$ gives the two-digit primes $23,37,53,73$.

Thus $8$ of the $20$ choices are prime, and the probability is $\frac{8}{20}=\frac25$.

Thus, the correct answer is B.

Answer: A

Small Hint:

Substitute $(x,-x)$ into the distance formula from $(5,5)$

Big Hint:

Solve $2x^2+50\le100$

Solution:

The squared distance from $(x,-x)$ to $(5,5)$ is $$\begin{aligned} &(x-5)^2+(-x-5)^2 \\ &= 2x^2+50. \end{aligned}$$

Being inside or on the circle means $2x^2+50\le100$, so $x^2\le25$. Thus $-5\le x\le5$, giving $11$ integer values.

Thus, the correct answer is A.

Answer: E

Small Hint:

The intercepts are $5$ and $12$

Big Hint:

Use area to find the altitude to the hypotenuse

Solution:

The triangle is a right triangle with legs of $12$ and $5.$ This makes the hypotenuse $13.$

Two of the altitudes are then $12$ and $5.$ Also, for any side, $A = \frac{bh}2$ where $b$ is the base and $h$ is the altitude.

The area is $\frac {12\cdot 5}2 = 30,$ so the other altitude $h$ can be found with $30 = \frac{13h}2.$ Thus, this altitude is $\frac{60}{13}.$

Therefore, the sum is $$12+5+\dfrac{60}{13} = \dfrac{281}{13} .$$

Thus, the correct answer is E .

Answer: D

Small Hint:

Factor out $x-b$

Big Hint:

Maximize $b+\frac{a+c}{2}$ using distinct one-digit numbers

Solution:

Factoring the left-hand side gives $$(x-b)(2x-a-c)=0.$$ Thus the roots are $b$ and $\frac{a+c}{2},$ whose sum is $$b+\frac{a+c}{2}.$$

The coefficient of $b$ in this sum is twice the coefficient of either $a$ or $c,$ so assign the largest digit to $b.$ The next two largest distinct digits should be $a$ and $c.$

Taking $b=9$ and $\{a,c\}=\{7,8\}$ gives $$9+\frac{7+8}{2}=16.5.$$

Thus, the correct answer is D .

Answer: B

Small Hint:

Write the total as $13h+5c$

Big Hint:

Check each option modulo $5$ after subtracting multiples of $13$

Solution:

If there are $h$ horses and $c$ cows, then there are $3h$ people, $9h$ ducks, and $4c$ sheep. The total is therefore $13h+5c$.

The listed values except $47$ can be written in that form: $$\begin{gathered} 41=13\cdot2+5\cdot3, \\ 59=13\cdot3+5\cdot4, \\ 61=13\cdot2+5\cdot7, \\ 66=13\cdot2+5\cdot8. \end{gathered}$$ For $47$, subtracting $0,13,26,39$ leaves $47,34,21,8$, none of which is divisible by $5$.

Thus, the correct answer is B.

Answer: C

Small Hint:

Count ordered triples $(A,B,C)$ with $A$ a multiple of $B$ and $B$ a multiple of $C$

Big Hint:

Start with the smallest number $C$ in the divisibility chain

Solution:

Let $(A,B,C)$ be the numbers assigned to Al, Bill, and Cal. We need $A$ to be a multiple of $B$, and $B$ to be a multiple of $C$, with all three numbers distinct.

The valid triples are $$\begin{gathered} (4,2,1),(6,2,1),(8,2,1), \\ (10,2,1),(6,3,1),(9,3,1), \\ (8,4,1),(10,5,1),(8,4,2). \end{gathered}$$ There are $9$ favorable assignments.

The total number of assignments is $10\cdot9\cdot8=720$, so the probability is $\frac9{720}=\frac1{80}$.

Thus, the correct answer is C.

Answer: B

Small Hint:

View the octahedron as two pyramids with a central rhombus base

Big Hint:

The base rhombus has diagonals $4$ and $5$

Solution:

The octahedron can be viewed as two congruent pyramids whose shared base is the rhombus through the centers of the four side faces. This rhombus has diagonals $4$ and $5$, so its area is $\frac12\cdot4\cdot5=10$.

Each pyramid has height $\frac32$, half the prism’s height. Thus the total volume is $$2\cdot\frac13\cdot10\cdot\frac32=10.$$

Thus, the correct answer is B.

Answer: D

Small Hint:

A coin is tails at the end only after three tails

Big Hint:

Use linearity of expectation

Solution:

A coin ends as tails if and only if it has $3$ flips that are tails, which happens with probability $\frac 18.$ Thus, the probability of any coin being heads is $\frac 78 .$

Since each of the $64$ coins ends as heads with probability $\frac78,$ linearity of expectation gives the expected number of coins that are now heads: $$64 \cdot \dfrac 78 = 56.$$

Thus, the correct answer is D .

Answer: C

Small Hint:

The circle center is the circumcenter of the right triangle

Big Hint:

Put the circumcenter at the midpoint of the hypotenuse

Solution:

The center of the circle through $X,Y,Z,W$ lies on the perpendicular bisectors of $XY$ and $ZW$. These are also the perpendicular bisectors of $AB$ and $AC$, so the same point is the circumcenter of right triangle $ABC$.

Therefore the center is the midpoint $O$ of hypotenuse $AB$, so $OA=OB=OC=6$. Let $a=\frac12BC$ and $b=\frac12CA$. Then $a^2+b^2=6^2$.

From the square on $AB$, $OX^2=6^2+12^2=180$. From the square on $AC$, the corresponding radius also gives $OW^2=b^2+(a+2b)^2$. Hence $$b^2+(a+2b)^2=180.$$ Subtracting $a^2+b^2=36$ from this equation gives $b(a+b)=36$. But $a^2+b^2=36$ as well, so $ab=a^2$. Since $a>0$, we get $a=b=3\sqrt{2}$.

Thus $AC=BC=6\sqrt{2}$, and the perimeter is $12+12\sqrt{2}$.

Thus, the correct answer is C.

Answer: A

Small Hint:

After two moves, the next vertex on the starting face is forced

Big Hint:

Finish around the opposite face and keep only paths that cannot return

Solution:

The first two edges can be chosen in $3\cdot2=6$ ways. These two edges determine an initial face of the cube. After those moves, there is one unvisited vertex on that initial face.

That remaining vertex must be visited next; otherwise Erin would later reach it after all of its neighbors had already been visited, and the path could not continue. The last four vertices are then on the opposite face and can be visited in two cyclic orders.

Of those two orders, exactly one ends at a vertex not adjacent to the starting point. Hence there are $6$ valid paths.

Thus, the correct answer is A.

Answer: D

Small Hint:

Let Dash take $d+1$ jumps and Cozy take $d+20$

Big Hint:

Match $t$ between $5d+1,\ldots,5d+5$ and $2d+39,2d+40$

Solution:

Suppose Dash takes $d+1$ jumps. Then the number of steps $t$ is one of $$\begin{gathered} 5d+1,5d+2,5d+3, \\ 5d+4,5d+5. \end{gathered}$$ Cozy takes $19$ more jumps, so Cozy takes $d+20$ jumps, which means $t$ is either $2d+39$ or $2d+40$.

Matching these possibilities, the integer solutions are $$\begin{gathered} 5d+3=2d+39, \\ 5d+1=2d+40, \\ 5d+4=2d+40. \end{gathered}$$ They give $t=63,66,64$, respectively.

Thus $s=63+66+64=193$, and the sum of its digits is $13$.