Solution:

The positive factors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18, 36$. Of these, only $4, 12,$ and $36$ are multiples of $4$.

Thus, the correct answer is B .

Solution:

Starting from $10$: José computes $9, 18, 20$; Thuy computes $20, 19, 21$; Kareem computes $9, 11, 22$.

Kareem doubles last, so the $2$ he adds is doubled too, giving the largest result.

Thus, the correct answer is C .

Solution:

The first row is $1, 2, \ldots, 8$ and the last row is $57, 58, \ldots, 64$. The four corners are $1, 8, 57,$ and $64$.

Their sum is $1 + 8 + 57 + 64 = 130$.

Thus, the correct answer is A .

Solution:

The numerator is $2(1 + 2 + \cdots + 17)$ and the denominator is $3(1 + 2 + \cdots + 17)$.

The common factor cancels, leaving $\dfrac{2}{3}$.

Thus, the correct answer is B .

Solution:

From the number line, $P$ and $Q$ are negative and $R, S, T$ are positive.

Then $P \cdot Q$ is positive; $\dfrac{S}{Q} \cdot P$ has two negative factors, so it is positive; $\dfrac{R}{P \cdot Q}$ is positive; and $\dfrac{S + T}{R}$ is positive. Since $P$ is to the left of $Q$, $P - Q$ is negative.

Thus, the correct answer is A .

Solution:

Use the three smallest numbers $3, 5, 7$. The choices are $3(5 + 7) = 36$, $5(3 + 7) = 50$, and $7(3 + 5) = 56$.

Making the smallest number the multiplier gives the least result, $36$.

Thus, the correct answer is C .

Solution:

Brent's counts are $4, 16, 64, 256, 1024, 4096$, and Gretel's are $128, 256, 512, 1024, 2048, 4096$.

They are equal after $5$ months, when both have $4096$.

Thus, the correct answer is B .

Solution:

The distance is smallest when the points are collinear with $C$ and $D$ toward $A$: take $A = 0, B = 10, C = 6, D = 3$.

Then $AD = 10 - 4 - 3 = 3$.

Thus, the correct answer is B .

Solution:

The number is $\dfrac{2}{5}$, whose reciprocal is $\dfrac{5}{2}$.

Then $100 \cdot \dfrac{5}{2} = 250$.

Thus, the correct answer is D .

Solution:

The increase is $\dfrac58 - \dfrac18 = \dfrac12$ of a tank, which equals $7.5$ gallons.

So a full tank holds $2 \cdot 7.5 = 15$ gallons.

Thus, the correct answer is D .

Solution:

Since $x$ is a very small positive number, $3 + x$ and $3 - x$ are near $3$, while $3 \cdot x$ and $x / 3$ are near $0$.

But $3 / x$ is $3$ followed by $1997$ zeros, far larger than any other choice.

Thus, the correct answer is D .

Solution:

The sum of $1$ through $11$ is $66$. For ten numbers to average $6.1$, their sum must be $10 \cdot 6.1 = 61$.

So the removed number is $66 - 61 = 5$.

Thus, the correct answer is B .

Solution:

Each year multiplies the count by $1.5$: $800 \to 1200 \to 1800 \to 2700$.

So $800 \cdot 1.5^3 = 2700$ participants are expected in $1999$.

Thus, the correct answer is E .

Solution:

Three distinct digits from $1$ through $9$ summing to $23$ must be $6, 8, 9$. The row's other three digits are at least $1 + 2 + 3 = 6$, so the shared square (belonging to both the column and the row) is at most $12 - 6 = 6$. Hence the shared digit is $6$.

The six digits are then $6, 8, 9$ and $1, 2, 3$, whose sum is $29$. Equivalently, $23 + 12 - 6 = 29$.

Thus, the correct answer is B .

Solution:

The units digit of the product equals the units digit of $2 \cdot 6 \cdot 2 \cdot 6 = 144$, which is $4$.

A number ending in $4$ leaves remainder $4$ when divided by $5$.

Thus, the correct answer is E .

Solution:

Grouping in blocks of four gives $1 - 2 - 3 + 4 = 0$, $5 - 6 - 7 + 8 = 0$, and so on.

There are $1996 / 4 = 499$ such blocks, each equal to $0$, so the total is $0$.

Thus, the correct answer is C .

Solution:

Since $OPQR$ is a square with $O = (0, 0)$ and $Q = (2, 2)$, we have $P = (2, 0)$ and $R = (0, 2)$, so the area is $2^2 = 4$.

Triangle $PQT$ has vertical base $PQ$ of length $2$, and $T = (t, 0)$ lies on the $x$-axis. Its area is $\tfrac12 \cdot 2 \cdot (2 - t) = 2 - t$. Setting $2 - t = 4$ gives $t = -2$, so $T = (-2, 0)$.

Thus, the correct answer is C .

Solution:

After the raise, the salary is $2000 \cdot 1.2 = 2400$.

After the cut, it is $2400 \cdot 0.8 = 1920$.

Thus, the correct answer is A .

Solution:

East has $22\%$ of $2000 = 440$ tennis fans, and West has $40\%$ of $2500 = 1000$.

Together $1440$ of the $4500$ students prefer tennis, which is $\dfrac{1440}{4500} = 32\%$.

Thus, the correct answer is C .

Solution:

Starting from $5$: $1/(1 - 5) = -0.25$, then $1/(1 + 0.25) = 0.8$, then $1/(1 - 0.8) = 5$. The values repeat with period $3$.

Since $100 = 3 \cdot 33 + 1$, the $100$th press gives the same result as the first press, $-0.25$.

Thus, the correct answer is A .

Solution:

The set has $4$ odd numbers ($89, 95, 99, 173$) and $2$ even numbers ($132, 166$). A sum of three is even only with two odds and one even, since three evens is impossible with just two available.

The count is $\binom{4}{2} \cdot \binom{2}{1} = 6 \cdot 2 = 12$.

Thus, the correct answer is D .

Solution:

Taking $A = (0, 0)$, $B = (3, 2)$, and $C = (4, 3)$, the enclosing $4 \times 3$ rectangle has area $12$; subtracting the surrounding regions of areas $6, 3, 2,$ and $\tfrac12$ leaves $\tfrac12$.

Equivalently, by Pick's theorem with no interior lattice points and $3$ boundary points, the area is $0 + \tfrac32 - 1 = \tfrac12$.

Thus, the correct answer is B .

Solution:

Let $n$ be the number of employees. The fund is $50n - 5$ (five dollars short of $50$ each) and also $45n + 95$.

Setting $50n - 5 = 45n + 95$ gives $5n = 100$, so $n = 20$. The fund is $45 \cdot 20 + 95 = 995$.

Thus, the correct answer is E .

Solution:

In triangle $ABC$, $\angle BAC + \angle BCA = 180^\circ - 50^\circ = 130^\circ$.

The bisectors give $\angle DAC + \angle DCA = \tfrac{130^\circ}{2} = 65^\circ$. In triangle $ADC$, $\angle ADC = 180^\circ - 65^\circ = 115^\circ$.

Thus, the correct answer is C .

Solution:

Take the radius to be $1$. A point at distance $r$ from the center is closer to the center than to the boundary when $r \lt 1 - r$, i.e. $r \lt \tfrac12$.

The favorable region is a circle of radius $\tfrac12$, with area $\pi(\tfrac12)^2 = \tfrac{\pi}{4}$, out of the total area $\pi$. The probability is $\dfrac{\pi/4}{\pi} = \dfrac14$.

Thus, the correct answer is A .