Answer: C

Small Hint:

Factor a $2$ out of the numerator and a $3$ out of the denominator

Big Hint:

Both share the factor $1-2+3-4+5-6+7$

Solution:

Factoring gives $$\dfrac{2(1-2+3-4+5-6+7)}{3(1-2+3-4+5-6+7)}.$$ The identical alternating sums cancel, leaving $\dfrac{2}{3}.$

Thus, the correct answer is C.

Answer: D

Small Hint:

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

Big Hint:

If a pink pill costs $p,$ then the green pill costs $p+1$ and $p+(p+1)=39$

Solution:

Each day’s pills cost $546 \div 14 = 39$ dollars. If $x$ is the cost of a green pill, then the pink pill costs $x-1,$ so $x+(x-1)=39.$ Solving gives $x=20.$

Thus, the correct answer is D.

Answer: B

Small Hint:

The first $8$ odd counting numbers sum to $8^2$

Big Hint:

Write the even integers as $n,$ $n+2,$ $n+4,$ $n+6,$ $n+8$ and set their sum equal to $64-4$

Solution:

The first $8$ odd counting numbers sum to $1+3+\cdots+15=64.$

Letting $n$ be the smallest even integer, $$\begin{gathered} n+(n+2)+(n+4) \\ {}+(n+6)+(n+8) \\ = 5n+20 = 60, \end{gathered}$$ so $n=8.$

Thus, the correct answer is B.

Answer: A

Small Hint:

Find the area of each of the five regions first

Big Hint:

To minimize cost, plant the most expensive flower in the smallest region

Solution:

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

Cost is minimized by placing the most expensive flower in the smallest region, so the total is $$\begin{gathered} (3)(4)+(2.5)(6)+(2)(15) \\ {}+(1.5)(20)+(1)(21) \\ = 108. \end{gathered}$$

Thus, the correct answer is A.

Answer: C

Small Hint:

The effective swath width is $28-4=24$ inches, which is $2$ feet

Big Hint:

Divide the lawn’s area by the area Moe mows per hour

Solution:

The lawn has area $90 \cdot 150 = 13{,}500$ square feet.

Each foot Moe walks mows an effective strip $2$ feet wide, so he mows $2 \cdot 5000 = 10{,}000$ square feet per hour. The time needed is $$\dfrac{13{,}500}{10{,}000}=1.35$$ hours.

Thus, the correct answer is C.

Answer: D

Small Hint:

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

Big Hint:

The diagonal corresponds to $5$ parts and equals $27,$ so the length is $\tfrac{4}{5}$ of $27$

Solution:

Since the length and height are in ratio $4:3,$ the length, height, and diagonal form a $4:3:5$ right triangle. The diagonal is $27,$ so the horizontal length is $$\dfrac{4}{5}(27)=21.6,$$ which is closest to $21.5.$

Thus, the correct answer is D.

Answer: B

Small Hint:

$\lfloor \sqrt{n} \rfloor = k$ exactly when $k^2 \le n \lt (k+1)^2$

Big Hint:

Count how many of $1$ through $16$ give each floor value $1,$ $2,$ $3,$ and $4$

Solution:

The value is $1$ for $n=1,2,3;$ it is $2$ for $n=4,\ldots,8;$ it is $3$ for $n=9,\ldots,15;$ and it is $4$ for $n=16.$ The sum is $$3\cdot 1 + 5\cdot 2 + 7\cdot 3 + 1\cdot 4 = 38.$$

Thus, the correct answer is B.

Answer: B

Small Hint:

Dividing the fourth term by the second term gives $r^2$

Big Hint:

With $ar=2,$ the first term is $a=\dfrac{2}{r},$ and $r$ may be negative

Solution:

Let the terms be $a,$ $ar,$ $ar^2,$ $ar^3,$ $\dots$ with $ar=2$ and $ar^3=6.$ Dividing gives $r^2=3,$ so $r=\pm\sqrt3.$

Then $a=\dfrac{2}{r}=\pm\dfrac{2}{\sqrt3}=\pm\dfrac{2\sqrt3}{3}.$ The choice $-\dfrac{2\sqrt3}{3}$ matches the negative case.

Thus, the correct answer is B.

Answer: B

Small Hint:

Rewrite every term as a power of $5$

Big Hint:

The right side becomes $5^{\frac{48-26-34}{x}};$ set the exponent equal to $-4$

Solution:

Writing everything base $5,$ the left side is $5^{-4}$ and the right side is $$5^{\frac{48-26-34}{x}}=5^{-\frac{12}{x}}.$$ Setting exponents equal, $-4=-\dfrac{12}{x},$ so $x=3.$

Thus, the correct answer is B.

Answer: C

Small Hint:

Count the plates in each scheme using the multiplication principle

Big Hint:

Divide the new count $26^3 \cdot 10^3$ by the old count $26 \cdot 10^4$

Solution:

The old scheme allows $26 \cdot 10^4$ plates and the new scheme allows $26^3 \cdot 10^3$ plates. The increase factor is $$\dfrac{26^3 \cdot 10^3}{26 \cdot 10^4}=\dfrac{26^2}{10}.$$

Thus, the correct answer is C.

Answer: A

Small Hint:

Write each line in point-slope form through $(10, 15)$

Big Hint:

Set $y=0$ in each equation to find where they cross the $x$-axis

Solution:

The lines are $y-15=3(x-10)$ and $y-15=5(x-10).$

Setting $y=0$ gives $x$-intercepts $x=5$ and $x=7.$ The distance between $(5,0)$ and $(7,0)$ is $2.$

Thus, the correct answer is A.

Answer: C

Small Hint:

Let Al, Betty, and Clare start with $a,$ $b,$ and $c,$ so $a+b+c=1000$

Big Hint:

The final total is $(a-100)+2(b+c)=1500;$ substitute $b+c=1000-a$

Solution:

Let $a,$ $b,$ and $c$ be the original portions. Then $a+b+c=1000$ and $(a-100)+2(b+c)=1500.$

Substituting $b+c=1000-a$ into the second equation, $$a-100+2(1000-a)=1500,$$ so $a=400.$

Thus, the correct answer is C.

Answer: E

Small Hint:

When $x$ has two digits, the digit sum $\clubsuit(x)$ is at most $18$

Big Hint:

$\clubsuit(y)=3$ with $y\le 18$ forces $y=3$ or $y=12;$ count $x$ for each

Solution:

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

The two-digit numbers with digit sum $3$ are $12,$ $21,$ and $30$ ($3$ of them). Those with digit sum $12$ are $39,$ $48,$ $57,$ $66,$ $75,$ $84,$ and $93$ ($7$ of them). In all there are $10.$

Thus, the correct answer is E.

Answer: D

Small Hint:

Since $5^2$ but not $5^3$ divides the number, $b$ can be at most $2$

Big Hint:

The larger $b$ is, the smaller $a$ becomes; try $b=2$ and write the number as a perfect square

Solution:

Because $a$ must be divisible by $5,$ and $3^8 \cdot 5^2$ is divisible by $5^2$ but not $5^3,$ we need $b\le 2.$

Taking $b=2$ gives $a=\sqrt{3^8 \cdot 5^2}=3^4 \cdot 5=405,$ so $a+b=407.$ This beats $b=1,$ which gives $a+b=164{,}026.$

Thus, the correct answer is D.

Answer: E

Small Hint:

Each match eliminates exactly one player

Big Hint:

All but the champion are eliminated, so count the number of eliminated players

Solution:

Each match eliminates exactly one player. Since $100$ players start and all but the champion are eliminated, there are $99$ matches.

Because $99=9 \cdot 11,$ it is divisible by $11$ but satisfies none of the other options.

Thus, the correct answer is E.

Answer: E

Small Hint:

With $m$ main courses there are $2m$ appetizers, giving $3 \cdot m \cdot 2m$ dinners

Big Hint:

The year $2003$ is not a leap year, so require $6m^2 \ge 365$

Solution:

With $m$ main courses, the number of dinners is $3 \cdot m \cdot 2m = 6m^2.$ This must be at least $365.$

So $m^2 \ge \dfrac{365}{6}\approx 60.8.$ Since $7^2=49$ is too small but $8^2=64$ works, $m=8.$

Thus, the correct answer is E.

Answer: B

Small Hint:

The cone and sphere share the same radius $r$

Big Hint:

Set $\tfrac34 \cdot \tfrac43 \pi r^3$ equal to the cone volume $\tfrac13 \pi r^2 h$

Solution:

The melted volume equals the cone’s volume, so $$\dfrac34 \cdot \dfrac43 \pi r^3 = \dfrac13 \pi r^2 h.$$

Simplifying gives $\pi r^3 = \dfrac13 \pi r^2 h,$ so $h=3r.$ The ratio of height to radius is $3:1.$

Thus, the correct answer is B.

Answer: D

Small Hint:

When $n$ is even, the five factors are consecutive odd numbers

Big Hint:

Among five consecutive odd numbers one is a multiple of $3$ and one of $5;$ check no larger factor always works

Solution:

When $n$ is even, the factors are five 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 $15.$

To prove that no larger fixed divisor is forced, compare three cases: $$\begin{aligned} n=2 &: 3\cdot5\cdot7\cdot9\cdot11,\\ n=10 &: 11\cdot13\cdot15\cdot17\cdot19,\\ n=12 &: 13\cdot15\cdot17\cdot19\cdot21. \end{aligned}$$ The greatest common divisor of these three products is exactly $15,$ so a divisor common to every case cannot be any larger.

Thus, the correct answer is D.

Answer: E

Small Hint:

Start with the large semicircle’s area $\tfrac12 \pi (2)^2 = 2\pi$

Big Hint:

The overlaps of adjacent small semicircles are $60^\circ$ sectors plus equilateral triangles

Solution:

The large semicircle has area $\dfrac12 \pi (2)^2 = 2\pi.$

Removing the small semicircles deletes a region equal to five congruent $60^\circ$ sectors of radius $1$ plus two equilateral triangles of side $1.$ Each sector has area $\dfrac{\pi}{6}$ and each triangle has area $\dfrac{\sqrt3}{4}.$

The shaded area is $$\begin{gathered} 2\pi - 5 \cdot \dfrac{\pi}{6} - 2 \cdot \dfrac{\sqrt3}{4} \\ = \dfrac{7}{6}\pi - \dfrac{\sqrt3}{2}. \end{gathered}$$

Thus, the correct answer is E.

Answer: D