Solution:

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

To maximize the sum of three distinct positive integers with this product, take $I = 1,$ $M = 3,$ and $O = 23 \cdot 29 = 667.$

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

Thus, the correct answer is E.

Solution:

Writing $2000 = 2000^1,$ we get $$2000^1 \cdot 2000^{2000} = 2000^{1 + 2000} = 2000^{2001}.$$

All of the other options are larger than this.

Thus, the correct answer is A.

Solution:

Since $20\%$ is eaten each day, $80\%$ remains at the end of each day. If $x$ is the original number, then $$(0.8)^2 x = 32.$$

Solving gives $0.64x = 32,$ so $x = 50.$

Thus, the correct answer is B.

Solution:

The sequence of units digits begins $$1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots$$

Scanning this list, the digit $6$ is the last of the ten digits to appear.

Thus, the correct answer is C.

Solution:

Since $x \lt 2,$ 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:

The primes between $4$ and $18$ are $5, 7, 11, 13,$ and $17.$

For two such primes, $xy - (x + y) = (x - 1)(y - 1) - 1$ is a product of two even numbers minus $1,$ hence odd. Among the choices, only $119$ is odd.

Indeed, $11 \cdot 13 - (11 + 13) = 143 - 24 = 119.$

Thus, the correct answer is C.

Solution:

If $\log_b 729 = n,$ then $b^n = 729 = 3^6,$ so $n$ must be a positive divisor of $6.$

The possibilities $n = 1, 2, 3, 6$ give $b = 3^6, 3^3, 3^2, 3^1,$ that is, $729, 27, 9,$ and $3.$

There are $4$ such values of $b.$

Thus, the correct answer is E.

Solution:

Figure $n$ is a diamond whose row lengths increase through the odd numbers and back down, giving a total of $n^2 + (n + 1)^2$ unit squares. This matches $1, 5, 13, 25$ for $n = 0, 1, 2, 3.$

Therefore figure $100$ has $$100^2 + 101^2 = 10000 + 10201 = 20201$$ unit squares.

Thus, the correct answer is C.

Solution:

The total is $71 + 76 + 80 + 82 + 91 = 400,$ which is divisible by $5.$ The sum of the first three scores must be divisible by $3.$

Modulo $3,$ the scores are $2, 1, 2, 1, 1.$ The only triple summing to a multiple of $3$ is $76 + 82 + 91 = 249,$ so these are the first three (with $91$ third, since the first two, $76$ and $82,$ must have equal parity).

Since $249 \equiv 1 \pmod 4,$ the fourth score must be $\equiv 3 \pmod 4,$ which is $71.$ That leaves $80$ as the last score entered.

Thus, the correct answer is C.

Solution:

Reflecting $(1, 2, 3)$ in the $xy$-plane gives $Q = (1, 2, -3).$

Rotating $180^\circ$ about the $x$-axis negates $y$ and $z,$ giving $R = (1, -2, 3).$

Translating $5$ units in the positive $y$ direction gives $S = (1, 3, 3).$

Thus, the correct answer is E.

Solution:

Combining over a common denominator, $$\frac{a}{b} + \frac{b}{a} - ab = \frac{a^2 + b^2 - (ab)^2}{ab}.$$

Replacing $ab$ with $a - b$ in the numerator, $$a^2 + b^2 - (a - b)^2 = a^2 + b^2 - (a^2 - 2ab + b^2) = 2ab.$$

Therefore the expression equals $\dfrac{2ab}{ab} = 2.$

Thus, the correct answer is E.

Solution:

Observe that $$AMC + AM + MC + CA = (A + 1)(M + 1)(C + 1) - (A + M + C) - 1.$$

Since $A + M + C = 12,$ this equals $(A + 1)(M + 1)(C + 1) - 13.$ The three factors sum to $15,$ so their product is maximized when each equals $5,$ giving $5^3 = 125.$

The maximum value is $125 - 13 = 112.$

Thus, the correct answer is E.

Solution:

Measure amounts in $8$-ounce cups, so Angela's cup holds $c$ coffee and $m$ milk with $c + m = 1.$

Since Angela drank a sixth of the coffee, the total coffee is $6c$; since she drank a quarter of the milk, the total milk is $4m.$ The number of people equals the total number of cups, $$6c + 4m = 6c + 4(1 - c) = 4 + 2c.$$

This is an integer only when $2c$ is an integer, and since $0 \lt c \lt 1$ this forces $c = \tfrac12,$ giving $4 + 1 = 5$ people.

Thus, the correct answer is C.

Solution:

The six fixed numbers sum to $25,$ so the mean is $\dfrac{25 + x}{7},$ and the mode is $2.$ If $x \le 2,$ then $2$ is both median and mode, forcing a constant progression, so $x \gt 2.$

Case $2 \lt x \lt 4$: the median is $x.$ Requiring $2, x, \dfrac{25 + x}{7}$ to form an arithmetic progression yields $x = 3$ as the only value in this range.

Case $x \ge 4$: the median is $4,$ and the progression $2, 4, 6$ forces the mean to be $6,$ so $\dfrac{25 + x}{7} = 6,$ giving $x = 17.$

The sum of all possible values is $3 + 17 = 20.$

Thus, the correct answer is E.

Solution:

Setting $\dfrac{x}{3} = 3z$ gives $x = 9z,$ so $$f(3z) = (9z)^2 + 9z + 1 = 81z^2 + 9z + 1 = 7.$$

This rearranges to $81z^2 + 9z - 6 = 0.$

By the sum-of-roots formula, the sum of the values of $z$ is $-\dfrac{9}{81} = -\dfrac{1}{9}.$

Thus, the correct answer is B.

Solution:

The square $(m, n)$ is numbered $17(m - 1) + n$ originally and $13(n - 1) + m$ after renumbering. Setting these equal gives $$4m - 3n = 1.$$

The solutions with $1 \le m \le 13$ and $1 \le n \le 17$ are $(1, 1),$ $(4, 5),$ $(7, 9),$ $(10, 13),$ and $(13, 17).$

These squares hold the numbers $1, 56, 111, 166,$ and $221,$ whose sum is $555.$

Thus, the correct answer is D.

Solution:

Because $OA = 1$ and $AB$ is tangent at $A,$ angle $OAB$ is right, so $$BA = \tan\theta, \qquad OB = \sec\theta.$$

Since $BC$ bisects $\angle ABO,$ the angle bisector theorem gives $\dfrac{OC}{CA} = \dfrac{OB}{BA}.$ Using $OC + CA = OA = 1,$ $$OC = \frac{OB}{OB + BA} = \frac{\sec\theta}{\sec\theta + \tan\theta}.$$

Multiplying numerator and denominator by $\cos\theta$ gives $OC = \dfrac{1}{1 + \sin\theta}.$

Thus, the correct answer is D.

Solution:

From day $300$ of year $N$ to day $200$ of year $N + 1,$ the number of days is $365 - 300 + 200 = 265$ if $N$ is not a leap year. But $265 = 7 \cdot 37 + 6,$ which would land on a Monday, not a Tuesday.

So year $N$ is a leap year, and the gap is $266 = 7 \cdot 38$ days, giving a Tuesday as stated. It follows that year $N - 1$ is not a leap year.

The $100$th day of year $N - 1$ precedes the Tuesday on day $300$ of year $N$ by $365 - 100 + 300 = 565$ days. Since $565 = 7 \cdot 80 + 5,$ that day is $5$ weekdays before Tuesday, namely Thursday.

Thus, the correct answer is A.

Solution:

By Heron's formula, the area of $\triangle ABC$ is $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,$ so the altitude from $A$ to $BC$ is $\dfrac{2 \cdot 84}{14} = 12.$

The midpoint $D$ is $7$ from $B.$ The bisector from $A$ meets $BC$ at $E$ with $BE : EC = AB : AC = 13 : 15,$ so $BE = 14 \cdot \dfrac{13}{28} = 6.5.$

Both $D$ and $E$ lie on $BC,$ so $\triangle ADE$ has base $DE = 7 - 6.5 = 0.5$ and altitude $12,$ giving area $$\tfrac12 \cdot 0.5 \cdot 12 = 3.$$

Thus, the correct answer is C.

Solution:

Adding the three equations gives $$\left(x + \tfrac1y\right) + \left(y + \tfrac1z\right) + \left(z + \tfrac1x\right) = 4 + 1 + \tfrac73 = \tfrac{22}{3}.$$

Multiplying them gives $$4 \cdot 1 \cdot \tfrac73 = \tfrac{28}{3}.$$

Expanding the product, $$\left(x + \tfrac1y\right)\left(y + \tfrac1z\right)\left(z + \tfrac1x\right) = xyz + \left(x + y + z + \tfrac1x + \tfrac1y + \tfrac1z\right) + \frac{1}{xyz}.$$ The middle group is the sum $\tfrac{22}{3},$ so $xyz + \dfrac{1}{xyz} = \tfrac{28}{3} - \tfrac{22}{3} = 2.$

Hence $(xyz - 1)^2 = 0,$ so $xyz = 1.$

Thus, the correct answer is B.

Solution:

Let the square have side $1.$ The small triangle sharing one side of the square has a perpendicular leg $r,$ so its area is $\tfrac12 \cdot 1 \cdot r = m,$ giving $r = 2m.$

The two small triangles are similar, so the other triangle's leg along the square is $\dfrac1r,$ and its area is $$\frac12 \cdot 1 \cdot \frac1r = \frac{1}{2r} = \frac{1}{4m}.$$

Thus, the correct answer is D.

Solution:

The graph crosses the $x$-axis exactly twice, both times at positive values, so $P$ has two real zeros and two non-real (complex conjugate) zeros.

Reading off the graph: the sum of the coefficients is $P(1) \gt 3;$ $P(-1) \gt 4;$ the sum of the real zeros is greater than $4.5;$ and the product of all zeros is $d,$ the $y$-intercept, which is less than $6.$

The product of the real zeros is greater than $4.5,$ so the product of the non-real zeros, equal to $\dfrac{\text{product of all zeros}}{\text{product of real zeros}},$ is less than $\dfrac{6}{4.5} \lt 2.$

This is smaller than every other listed quantity.

Thus, the correct answer is C.

Solution:

The sum of the logarithms is an integer $k$ exactly when the product of the six numbers is $10^k.$ Since $10 = 2 \cdot 5,$ each chosen number must be of the form $2^a 5^b,$ so it comes from $$1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40.$$

For each, record the excess of factors of $2$ over factors of $5$: $0, 1, 2, -1, 3, 0, 4, 1, -2, 5, 2.$ The product is a power of $10$ only if the six chosen values have equal totals of $2$s and $5$s, i.e. their excesses sum to $0.$

Working through the possibilities, exactly four valid tickets exist: $\{1, 5, 10, 20, 25, 40\},$ $\{1, 2, 5, 10, 25, 40\},$ $\{1, 2, 4, 5, 10, 25\},$ and $\{1, 4, 5, 10, 20, 25\}.$

Professor Gamble holds one of these four, and only one matches the winning ticket, so the probability is $\dfrac14.$

Thus, the correct answer is B.

Solution:

Each arc has radius $AB,$ and $C$ is at distance $AB$ from both $A$ and $B,$ so $\triangle ABC$ is equilateral. Thus arc $BC$ subtends $60^\circ$ of a circle of radius $AB,$ whose full circumference is $6 \cdot 12 = 72.$

Let the small circle have radius $r$ and touch $\overline{AB}$ at its midpoint $D,$ where $AD = \tfrac12 AB.$ By Power of a Point, $AD^2 = AB(AB - 2r),$ so $$\frac{AB^2}{4} = AB^2 - 2r\,AB,$$ giving $2r = \tfrac34 AB,$ hence $r = \tfrac38 AB.$

The circumferences are in the ratio of the radii, so the small circle's circumference is $\tfrac38 \cdot 72 = 27.$

Thus, the correct answer is D.

Solution:

There are $8!$ ways to assign the eight distinct colors to the eight faces. Two assignments give the same octahedron exactly when one is a rotation of the other.

The rotation group of a regular octahedron has $24$ elements. Because all eight colors are different, no nontrivial rotation fixes a coloring, so each distinguishable octahedron corresponds to exactly $24$ assignments.

Therefore the number of distinguishable octahedrons is $$\frac{8!}{24} = \frac{40320}{24} = 1680.$$

Thus, the correct answer is E.