Solution:

Factoring gives $2001 = 3 \cdot 23 \cdot 29.$

To maximize the sum with the product fixed, spread the factors as much as possible: take $I = 1$ and combine the two largest primes, $M = 3$ and $O = 23 \cdot 29 = 667.$

The sum is $1 + 3 + 667 = 671.$

Thus, the correct answer is E.

Solution:

Write the factor $2000$ as $2000^1.$ Then $$2000^1 \cdot 2000^{2000} = 2000^{1 + 2000} = 2000^{2001}.$$

Each of the other options is larger than $2000^{2001}.$

Thus, the correct answer is A.

Solution:

Since Jenny eats $20\%$ each day, $80\%$ remain at the end of each day.

If $x$ is the original number, then $$(0.8)^2 x = 0.64x = 32,$$ so $x = 50.$

Thus, the correct answer is B.

Solution:

January doubled only the connect time, so the increase $\$17.54 - \$12.48 = \$5.06$ equals December's connect-time cost.

The fixed monthly fee is therefore $\$12.48 - \$5.06 = \$7.42.$

Thus, the correct answer is D.

Solution:

Since $MN$ is a midsegment, $MN = \tfrac12 AB,$ which is fixed.

The base $AB$ and the height from $P$ to $AB$ are both constant as $P$ slides along the parallel line, so the area of $\triangle PAB$ does not change. The trapezoid $ABNM$ is the triangle minus $\triangle PMN,$ both of whose areas are constant, so its area does not change either.

Only the perimeter changes, since $PA$ and $PB$ vary as $P$ moves. So exactly one quantity changes.

Thus, the correct answer is B.

Solution:

Recording only the units digits gives the sequence $$1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots$$

Scanning for the first appearance of each digit, the digit $6$ is the last of the ten digits to show up.

Thus, the correct answer is C.

Solution:

The right angle $\angle ADC = 90^\circ$ is trisected into three $30^\circ$ angles, so $\angle ADP = 30^\circ$ and $\angle ADB = 60^\circ.$

In right triangle $ADP,$ with $AD = 1,$ we get $DP = \dfrac{1}{\cos 30^\circ} = \dfrac{2\sqrt3}{3}$ and $AP = \tan 30^\circ = \dfrac{\sqrt3}{3}.$

In right triangle $ADB,$ with $AD = 1,$ we get $DB = \dfrac{1}{\cos 60^\circ} = 2$ and $AB = \tan 60^\circ = \sqrt3.$

Then $PB = AB - AP = \sqrt3 - \dfrac{\sqrt3}{3} = \dfrac{2\sqrt3}{3}.$

The perimeter of $\triangle BDP$ is $$DP + PB + DB = \dfrac{2\sqrt3}{3} + \dfrac{2\sqrt3}{3} + 2 = 2 + \dfrac{4\sqrt3}{3}.$$

Thus, the correct answer is B.

Solution:

Let $f$ and $s$ be the numbers of freshmen and sophomores. The contestant counts are equal, so $$\tfrac25 f = \tfrac45 s.$$

Multiplying by $5$ gives $2f = 4s,$ so $f = 2s.$ There are twice as many freshmen as sophomores.

Thus, the correct answer is D.

Solution:

Because $x \lt 2,$ we have $|x - 2| = 2 - x = p,$ so $x = 2 - p.$

Then $$x - p = (2 - p) - p = 2 - 2p.$$

Thus, the correct answer is C.

Solution:

By the triangle inequality, each of $x$ and $y$ can be any number strictly between $6 - 4 = 2$ and $6 + 4 = 10.$

Then $|x - y|$ can take any value with $0 \le |x - y| \lt 8.$

The smallest positive number not attainable is $10 - 2 = 8.$

Thus, the correct answer is D.

Solution:

The primes between $4$ and $18$ are $5, 7, 11, 13,$ and $17.$ The product of two of them is odd and the sum is even, so $xy - (x + y)$ is odd.

Since $xy - (x + y) = (x - 1)(y - 1) - 1$ increases as either prime increases, the result ranges from $5 \cdot 7 - 12 = 23$ up to $13 \cdot 17 - 30 = 191.$

The only odd option in $[23, 191]$ is $119 = 11 \cdot 13 - (11 + 13).$

Thus, the correct answer is C.

Solution:

Figure $n$ can be split into the sum of the first $n$ odd numbers and the first $n+1$ odd numbers, giving $n^2 + (n+1)^2$ unit squares.

For figure $100,$ this is $$100^2 + 101^2 = 10000 + 10201 = 20201.$$

Thus, the correct answer is C.

Solution:

The board has five rows and five columns. To avoid two yellow pegs in a row or column, there must be exactly one yellow peg in each row, forcing the yellow pegs onto the long diagonal.

The four red pegs must then each go in rows $2$ through $5,$ and the only positions left force them into a single diagonal as well. Continuing with green, blue, and orange, every color is forced into a unique position.

Hence there is exactly one valid arrangement.

Thus, the correct answer is B.

Solution:

The residues of $71, 76, 80, 82, 91$ modulo $3$ are $2, 1, 2, 1, 1.$ The sum of the first three scores must be divisible by $3,$ and the only such triple is $76 + 82 + 91 = 249,$ so the third score entered is $91$ and the first two are $76$ and $82.$

Since $249$ is one more than a multiple of $4,$ the fourth score must be three more than a multiple of $4,$ which only $71$ satisfies. That leaves $80$ as the fifth and last score.

Indeed $76, 158, 249, 320, 400$ are divisible by $1, 2, 3, 4, 5.$

Thus, the correct answer is C.

Solution:

Over the common denominator $ab,$ $$\dfrac{a}{b} + \dfrac{b}{a} - ab = \dfrac{a^2 + b^2 - (ab)^2}{ab}.$$

Substituting $ab = a - b$ gives $$\dfrac{a^2 + b^2 - (a - b)^2}{ab} = \dfrac{2ab}{ab} = 2.$$

Thus, the correct answer is E.

Solution:

Place the points at $A = (0, 3),$ $B = (6, 0),$ $C = (4, 2),$ $D = (2, 0).$

Line $AB$ is $x + 2y = 6$ and line $CD$ is $x - y = 2.$ Solving simultaneously gives $E = \left(\dfrac{10}{3}, \dfrac{4}{3}\right).$

Then $$AE = \sqrt{\left(\dfrac{10}{3}\right)^2 + \left(\dfrac{4}{3} - 3\right)^2} = \sqrt{\dfrac{100}{9} + \dfrac{25}{9}} = \dfrac{5\sqrt5}{3}.$$

Thus, the correct answer is B.

Solution:

Trading a quarter for five nickels or a nickel for five pennies does not change the total value. Only trading a penny for five quarters changes it, adding $5 \cdot 25 - 1 = 124$ cents.

Starting from $1$ cent, Boris always has $1 + 124n$ cents for some nonnegative integer $n.$

Only $\$7.45$ has this form, since $745 = 1 + 124 \cdot 6.$

Thus, the correct answer is D.

Solution:

Inside the square, Charlyn sees everything except a central square of side $5 - 2 = 3,$ an area of $25 - 9 = 16$ km$^2.$

Outside the square, the region is four rectangles each $5 \times 1$ plus four quarter circles of radius $1,$ an area of $4 \cdot 5 + \pi = 20 + \pi$ km$^2.$

The total area is $36 + \pi \approx 39$ km$^2.$

Thus, the correct answer is C.

Solution:

Let the square have side $1.$ One small triangle has legs $1$ and $r,$ with area $\tfrac12 r = m,$ so $r = 2m.$

The two small triangles are similar, so the other has legs $1$ and $\tfrac1r,$ with area $\tfrac12 \cdot \tfrac1r = \tfrac{1}{4m}.$

Since the square has area $1,$ the desired ratio is $\dfrac{1}{4m}.$

Thus, the correct answer is D.

Solution:

Notice that $$A \cdot M \cdot C + AM + MC + CA = (A+1)(M+1)(C+1) - (A + M + C) - 1 = (A+1)(M+1)(C+1) - 11.$$

We maximize a product of three positive integers summing to $13.$ The most balanced split is $4, 4, 5,$ giving $4 \cdot 4 \cdot 5 = 80.$

The maximum is $80 - 11 = 69.$

Thus, the correct answer is C.

Solution:

Some creepy crawlers are alligators, and all alligators are ferocious, so those creatures are both creepy crawlers and ferocious. Hence some ferocious creatures are creepy crawlers, making II true.

Statement I fails because not every alligator need be a creepy crawler, and III fails because it is possible that all alligators are creepy crawlers. Only II must hold.

Thus, the correct answer is B.

Solution:

Let there be $n$ people, drinking $8n$ ounces total, split into milk $M$ and coffee $C.$ Angela drank one cup, so $$\tfrac14 M + \tfrac16 C = \tfrac1n (M + C).$$

The left side is a weighted average of $\tfrac14$ and $\tfrac16,$ so $\tfrac1n$ lies strictly between $\tfrac16$ and $\tfrac14.$ That forces $4 \lt n \lt 6,$ so $n = 5.$

Thus, the correct answer is C.

Solution:

The mode is always $2,$ and the mean is $\dfrac{25 + x}{7}.$ For the values to form a non-constant arithmetic progression we examine the median.

If $x = 3,$ the sorted list is $2, 2, 2, 3, 4, 5, 10,$ with median $3$ and mean $4,$ giving the progression $2, 3, 4.$

If $x = 17,$ the sorted list is $2, 2, 2, 4, 5, 10, 17,$ with median $4$ and mean $6,$ giving the progression $2, 4, 6.$

No other value of $x$ works, so the sum of all possible values is $3 + 17 = 20.$

Thus, the correct answer is E.

Solution:

To evaluate $f(3z),$ set $\dfrac{x}{3} = 3z,$ so $x = 9z.$ Then $$f(3z) = (9z)^2 + 9z + 1 = 81z^2 + 9z + 1.$$

Setting this equal to $7$ gives $81z^2 + 9z - 6 = 0.$

By the sum-of-roots formula, the sum of the values of $z$ is $-\dfrac{9}{81} = -\dfrac19.$

Thus, the correct answer is B.

Solution:

From day $300$ of year $N$ to day $200$ of year $N + 1$ is $(L - 300) + 200$ days, where $L$ is the length of year $N.$ If $N$ were not a leap year, this is $265 \equiv 6 \pmod 7,$ giving a Monday, not a Tuesday. So year $N$ is a leap year, and the count is $266 = 7 \cdot 38,$ consistent with Tuesday.

Then years $N - 1$ and $N + 1$ are not leap years.

The $100$th day of year $N - 1$ precedes the Tuesday (day $300$ of year $N$) by $(365 - 100) + 300 = 565$ days. Since $565 = 7 \cdot 80 + 5,$ that day is $5$ days earlier in the week than Tuesday, which is a Thursday.

Thus, the correct answer is A.