Small Hint:

Factor a common constant out of the numerator and the denominator

Big Hint:

The two parenthesized sums $1 - 2 + 3 - \cdots + 7$ are identical

Solution:

Factor $2$ from the numerator and $3$ from the denominator: $$\frac{2(1 - 2 + 3 - 4 + 5 - 6 + 7)}{3(1 - 2 + 3 - 4 + 5 - 6 + 7)}.$$

The equal parenthesized sums cancel, leaving $\dfrac{2}{3}.$

Thus, the correct answer is C.

Small Hint:

Two weeks is $14$ days, so find the cost of one day’s pair of pills first

Big Hint:

If a green pill costs $g,$ then the pink pill costs $g - 1,$ and $g + (g-1)$ is one day’s cost

Solution:

Over $14$ days the daily cost of the two pills is $$\frac{546}{14} = 39.$$

Let $g$ be the cost of a green pill. The pink pill costs $g - 1,$ so $$g + (g - 1) = 39,$$ giving $g = 20.$

Thus, the correct answer is D.

Small Hint:

Find the area of each of the five regions from the figure

Big Hint:

To minimize the total, put the most expensive flower in the smallest region and so on

Solution:

The five regions have areas $4,$ $6,$ $15,$ $20,$ and $21$ square feet.

To minimize the cost, plant the most expensive flowers in the smallest regions. The least possible cost is $$\begin{aligned} &3(4) + 2.5(6) + 2(15) \\ &\quad {}+ 1.5(20) + 1(21) = 108. \end{aligned}$$

Thus, the correct answer is A.

Small Hint:

Each pass effectively cuts a strip only $28 - 4 = 24$ inches wide because of the overlap

Big Hint:

A $24$-inch strip is $2$ feet wide, so each foot walked mows $2$ square feet

Solution:

Because of the overlap, each pass adds a strip $28 - 4 = 24$ inches $= 2$ feet wide. So each foot Moe walks mows $2$ square feet, that is, $2 \cdot 5000 = 10000$ square feet per hour.

The lawn has area $90 \cdot 150 = 13500$ square feet, so the time is $$\frac{13500}{10000} = 1.35$$ hours.

Thus, the correct answer is C.

Small Hint:

With sides in ratio $4 : 3,$ the height, length, and diagonal are in ratio $3 : 4 : 5$

Big Hint:

The diagonal is $27,$ so the horizontal length is $\dfrac{4}{5}$ of $27$

Solution:

A rectangle with side ratio $4 : 3$ has height, length, and diagonal in ratio $3 : 4 : 5.$ With diagonal $27,$ the horizontal length is $$\frac{4}{5}(27) = 21.6,$$ which is closest to $21.5.$

Thus, the correct answer is D.

Small Hint:

If the first term is $a$ and the ratio is $r,$ then $ar = 2$ and $ar^3 = 6$

Big Hint:

Dividing the two equations gives $r^2 = 3,$ so $r = \pm\sqrt{3}$

Solution:

Let the first term be $a$ and the common ratio $r.$ Then $ar = 2$ and $ar^3 = 6,$ so $r^2 = 3$ and $r = \pm\sqrt{3}.$

The first term is $$a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}.$$ The choice $-\dfrac{2\sqrt{3}}{3}$ appears among the options.

Thus, the correct answer is B.

Small Hint:

With $n$ nickels, $d$ dimes, $q$ quarters: $n + d + q = 100$ and $5n + 10d + 25q = 835$

Big Hint:

Eliminate $n$ to get $d + 4q = 67,$ then bound $q$ so that $n,d,q \ge 0$ are integers

Solution:

Let $n,$ $d,$ $q$ be the numbers of nickels, dimes, quarters. Then $n + d + q = 100$ and $n + 2d + 5q = 167$ (dividing the value equation by $5$).

Subtracting gives $d + 4q = 67,$ so $d = 67 - 4q.$

The largest $d$ is at $q = 0,$ giving $d = 67$ (with $n = 33$). The smallest occurs at $q = 16,$ giving $d = 3$ (with $n = 81$). The difference is $67 - 3 = 64.$

Thus, the correct answer is D.

Small Hint:

Let $y = \clubsuit(x).$ For a two-digit $x,$ $y$ is between $1$ and $18$

Big Hint:

$\clubsuit(y) = 3$ forces $y = 3$ or $y = 12;$ count two-digit $x$ with each digit sum

Solution:

Let $y = \clubsuit(x).$ Since $x \le 99,$ we have $y \le 18.$ Then $\clubsuit(y) = 3$ requires $y = 3$ or $y = 12.$

The two-digit numbers with digit sum $3$ are $12, 21, 30$ ($3$ values), and those with digit sum $12$ are $39, 48, 57, 66, 75, 84, 93$ ($7$ values), for $10$ in all.

Thus, the correct answer is E.

Small Hint:

A linear function has a constant rate of change (slope)

Big Hint:

The slope is $\dfrac{f(6) - f(2)}{6 - 2},$ and the answer is that slope times $12 - 2$

Solution:

The slope of $f$ is $$\frac{f(6) - f(2)}{6 - 2} = \frac{12}{4} = 3.$$

Therefore $$f(12) - f(2) = 3(12 - 2) = 30.$$

Thus, the correct answer is D.

Small Hint:

By symmetry, assume one triangle is attached to a fixed side, then place the second

Big Hint:

Only the two sides adjacent versus non-adjacent to the fixed side give different shapes, up to reflection

Solution:

Assume one triangle is attached to side $AB.$ The second triangle can be attached to a side that is one step away or two steps away from $AB.$

Attaching it to $BC$ or $CD$ gives two figures; attaching it to $AE$ or $DE$ gives figures that are mirror images of these across the pentagon’s axis of symmetry.

So there are only $2$ non-congruent figures.

Thus, the correct answer is B.

Small Hint:

In $60$ real minutes the watch records only $57.6$ minutes

Big Hint:

If the watch shows $t$ minutes past noon, the real time is $\dfrac{60}{57.6}\,t = \dfrac{25}{24}\,t$

Solution:

In $60$ real minutes the watch advances only $57$ minutes $36$ seconds $= 57.6$ minutes. So when the watch shows $t$ minutes past noon, the real elapsed time is $\dfrac{60}{57.6}t = \dfrac{25}{24}t$ minutes.

The watch reads $10{:}00$ PM after $600$ recorded minutes, so the real elapsed time is $$\frac{25}{24}(600) = 625$$ minutes $= 10$ hours $25$ minutes past noon. The actual time is $10{:}25$ PM.

Thus, the correct answer is C.

Small Hint:

For even $n,$ the factors are five consecutive odd numbers

Big Hint:

Among any five consecutive odd numbers, one is a multiple of $3$ and one is a multiple of $5$

Solution:

For even $n,$ the five factors are consecutive odd numbers. Among any five consecutive odd numbers, at least one is divisible by $3$ and exactly one by $5,$ so the product is always divisible by $15.$

No larger divisor always works: the products for $n = 10$ and $n = 20$ are $11 \cdot 13 \cdot 15 \cdot 17 \cdot 19$ and $21 \cdot 23 \cdot 25 \cdot 27 \cdot 29,$ whose greatest common divisor is $15.$

Thus, the correct answer is D.

Small Hint:

The sphere and cone share the same radius $r;$ sphere volume is $\dfrac{4}{3}\pi r^3$

Big Hint:

Set $75\%$ of the sphere’s volume equal to the cone’s volume $\dfrac{1}{3}\pi r^2 h$

Solution:

Let $r$ be the common radius and $h$ the cone’s height. The melted ice cream fills the cone, so $$\frac{3}{4}\cdot\frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h.$$

This simplifies to $\pi r^3 = \dfrac{1}{3}\pi r^2 h,$ so $h = 3r,$ a ratio of $3 : 1.$

Thus, the correct answer is B.

Small Hint:

$FG = 5 - 1 - 2 = 2,$ and $\triangle FEG$ is similar to $\triangle AEB$

Big Hint:

Let the distance from $E$ to line $CD$ be $k;$ the distance to $AB$ is $k + 3,$ with ratio $\dfrac{k}{k+3} = \dfrac{2}{5}$

Solution:

Since $FG = 5 - 1 - 2 = 2$ and $\overline{FG} \parallel \overline{AB},$ triangles $FEG$ and $AEB$ are similar with ratio $\dfrac{FG}{AB} = \dfrac{2}{5}.$

Let the distance from $E$ to line $CD$ be $k.$ Then the distance from $E$ to $AB$ is $k + 3,$ and $$\frac{k}{k + 3} = \frac{2}{5},$$ giving $k = 2.$

The height of $\triangle AEB$ is $k + 3 = 5,$ so its area is $$\frac{1}{2}(5)(5) = \frac{25}{2}.$$

Thus, the correct answer is D.

Small Hint:

Let $O$ be the center of the octagon, which is the midpoint of diagonal $AE$

Big Hint:

$\triangle AOB$ is one of the eight triangles from the center, so it has area $\dfrac{1}{8}$

Solution:

Let $O$ be the center of the octagon. Joining $O$ to the vertices splits the octagon into $8$ congruent triangles, so $\triangle AOB$ has area $\dfrac{1}{8}.$

Since $O$ is the midpoint of $\overline{AE},$ triangles $AOB$ and $BOE$ have equal areas, so $\triangle ABE$ has area $\dfrac{1}{4}.$

The rectangle $ABEF$ is split by diagonal $\overline{BE}$ into two congruent triangles, so $\triangle ABE$ is half of it. Hence $ABEF$ has area $\dfrac{1}{2}.$

Thus, the correct answer is D.

Small Hint:

The large semicircle has area $\dfrac{1}{2}\pi(2)^2 = 2\pi$

Big Hint:

Use inclusion-exclusion: subtract the two overlaps from the total area of the three small semicircles

Solution:

The large semicircle has area $\dfrac{1}{2}\pi(2)^2 = 2\pi.$

The three small semicircles have total area $\dfrac{3\pi}{2}$ before their overlaps are accounted for. The centers of each adjacent pair are $1$ unit apart, so their intersection is bounded by two $60^\circ$ sectors and an equilateral triangle. Each of the two overlaps therefore has area $\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}.$

The shaded area is $$\begin{aligned} &2\pi - \left[\frac{3\pi}{2} - 2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)\right] \\ &= \frac{7}{6}\pi - \frac{\sqrt{3}}{2}. \end{aligned}$$

Thus, the correct answer is E.

Small Hint:

Let $X = \log x$ and $Y = \log y;$ the equations become linear in $X$ and $Y$

Big Hint:

Solve $X + 3Y = 1$ and $2X + Y = 1,$ then compute $X + Y$

Solution:

Let $X = \log x$ and $Y = \log y.$ Then $$X + 3Y = 1$$ and $$2X + Y = 1.$$

Solving gives $X = \dfrac{2}{5}$ and $Y = \dfrac{1}{5},$ so $$\log(xy) = X + Y = \frac{3}{5}.$$

Thus, the correct answer is D.

Small Hint:

The minimum $x$ uses only the primes $7$ and $11,$ so write $x = 7^c 11^d$

Big Hint:

Match exponents of $7$ and $11:$ need $5c + 1 \equiv 0 \pmod{13}$ and $5d \equiv 1 \pmod{13}$

Solution:

For the minimum $x,$ neither $x$ nor $y$ has prime factors other than $7$ and $11.$ Write $x = 7^c 11^d,$ so $7x^5 = 7^{5c+1} 11^{5d}.$ Writing $y = 7^m 11^n,$ we need $7^{5c+1}11^{5d} = 7^{13m}11^{13n+1}.$

Matching exponents: $5c + 1 \equiv 0 \pmod{13}$ gives the least $c = 5,$ and $5d \equiv 1 \pmod{13}$ gives the least $d = 8.$ So $a = 7,$ $b = 11,$ and $$\begin{aligned} &a + b + c + d \\ &= 7 + 11 + 5 + 8 = 31. \end{aligned}$$

Thus, the correct answer is B.