Answer: E

Small Hint:

Write $6^{2002}$ as $2^{2002}\cdot 3^{2002}$

Big Hint:

Cancel the common powers of $2$ and $3$ from top and bottom

Solution:

Since $6^{2002} = 2^{2002}\cdot 3^{2002},$ the ratio becomes $$\dfrac{2^{2001}\cdot 3^{2003}}{2^{2002}\cdot 3^{2002}} = \dfrac{3}{2}.$$

Thus, the correct answer is E.

Answer: C

Small Hint:

Substitute $a=2,$ $b=4,$ $c=6$ into the definition

Big Hint:

The numerator is $2\cdot4\cdot6$ and the denominator is $2+4+6$

Solution:

Substituting directly, $$(2, 4, 6) = \dfrac{2\cdot4\cdot6}{2+4+6} = \dfrac{48}{12} = 4.$$

Thus, the correct answer is C.

Answer: A

Small Hint:

Each term is $9$ times a repunit, so divide the sum by $9$ first

Big Hint:

$\dfrac19(9 + 99 + \cdots)$ $= 1 + 11 + 111 + \cdots$ $+ 111111111$

Solution:

The mean is $$\begin{aligned} &\dfrac{1}{9} \\ &\quad {}\cdot \small \left(9 + 99 + 999 + \cdots + 999999999\right) \\ &= 1 + 11 + 111 + \cdots \\ &\quad {}+ 111111111. \end{aligned}$$

Adding these nine repunits column by column gives $M = 123456789.$

The only digit missing from $M$ is $0.$

Thus, the correct answer is A.

Answer: D

Small Hint:

Factor $3x - 2$ out of the first two terms before plugging in

Big Hint:

The bracket $(4x + 1) - 4x$ collapses to $1$

Solution:

Factoring $3x-2$ from the first two terms, $$\begin{aligned} &(3x - 2)\big[(4x + 1) - 4x\big] + 1 \\ &= (3x - 2)(1) + 1 \\ &= 3x - 1. \end{aligned}$$

At $x = 4,$ this equals $3\cdot4 - 1 = 11.$

Thus, the correct answer is D.

Answer: E

Small Hint:

The two small circles lie along a diameter of the large circle

Big Hint:

The shaded area is the large disk minus the two smaller disks

Solution:

The two small circles line up along a diameter of the big circle, so that diameter is $2\cdot3 + 2\cdot2 = 10$ and the large radius is $5.$

The shaded region is the large disk with the two small disks removed: $$\begin{aligned} &\pi(5^2) - \pi(3^2) - \pi(2^2) \\ &= \pi(25 - 9 - 4) \\ &= 12\pi. \end{aligned}$$

Thus, the correct answer is E.

Answer: B

Small Hint:

Factor $n^2 - 3n + 2$

Big Hint:

For $(n-1)(n-2)$ to be prime, one factor must equal $1$

Solution:

Factor as $n^2 - 3n + 2 = (n-1)(n-2).$

For $n \ge 4,$ both factors exceed $1,$ so the product is composite. For $n = 1$ and $n = 2$ the value is $0,$ and for $n = 3$ the value is $(2)(1) = 2,$ which is prime.

So exactly one value of $n$ works.

Thus, the correct answer is B.

Answer: E

Small Hint:

The sum is positive and less than $2,$ so it must equal exactly $1$

Big Hint:

Solve $\dfrac1n = 1 - \dfrac12 - \dfrac13 - \dfrac17$ for $n$

Solution:

The sum $\dfrac12 + \dfrac13 + \dfrac17 + \dfrac1n$ is greater than $0$ and less than $\dfrac12 + \dfrac13 + \dfrac17 + 1 \lt 2,$ so as an integer it must equal $1.$

Since $\dfrac12 + \dfrac13 + \dfrac17 = \dfrac{41}{42},$ we need $\dfrac1n = \dfrac{1}{42},$ so $n = 42.$

Then $2,$ $3,$ $6,$ and $7$ all divide $42,$ but $n = 42$ is not greater than $84.$ So the false statement is $n \gt 84.$

Thus, the correct answer is E.

Answer: D

Small Hint:

A $31$-day month has exactly three weekdays that occur five times

Big Hint:

Those are the weekdays of days $1,$ $2,$ and $3$; find where a five-Monday July forces them

Solution:

A $31$-day month is $4$ weeks plus $3$ extra days, so exactly the weekdays of the $1$st, $2$nd, and $3$rd of the month occur five times.

For July to have five Mondays, Monday must be one of July $1,$ $2,$ or $3.$ In all three cases August $1$ lands on a Tuesday, Wednesday, or Thursday, and the common weekday among the resulting five-time days is Thursday.

More directly, since July has $31$ days, August $1$ is the same weekday as July $4.$ With Monday on July $1,$ $2,$ or $3,$ the three five-time weekdays of August always include Thursday.

Thus, the correct answer is D.

Answer: D

Small Hint:

Words starting with $A,$ $M,$ $O,$ or $S$ come first; count them

Big Hint:

Among the words starting with $U,$ find how many precede $USAMO$

Solution:

The alphabetical order of the letters is $A, M, O, S, U.$ Words beginning with $A,$ $M,$ $O,$ or $S$ fill positions $1$ through $96$ (four choices of first letter, $24$ each).

Words beginning with $U$ occupy positions $97$–$120.$ Listing them alphabetically, $USAMO$ is the $19$th such word, so it occupies position $96 + 19 = 115.$

Thus, the correct answer is D.

Answer: C

Small Hint:

By Vieta’s formulas, the sum of the roots is $-a$ and the product is $b$

Big Hint:

So $a + b = -a$ and $ab = b$; use $b \ne 0$

Solution:

Since the roots are $a$ and $b,$ Vieta’s formulas give $a + b = -a$ and $ab = b.$

From $ab = b$ with $b \ne 0,$ we get $a = 1.$ Then $a + b = -a$ gives $1 + b = -1,$ so $b = -2.$

Thus $(a, b) = (1, -2),$ and the correct answer is C.

Answer: B

Small Hint:

Call the integers $n - 1,$ $n,$ $n + 1$; their sum is $3n$

Big Hint:

The product is $(n-1)n(n+1) = n(n^2 - 1),$ which equals $8\cdot 3n$

Solution:

Let the integers be $n - 1,$ $n,$ $n + 1.$ Their product is $n(n^2 - 1)$ and their sum is $3n,$ so $$n(n^2 - 1) = 8(3n) = 24n.$$

Since $n \ne 0,$ we get $n^2 - 1 = 24,$ so $n^2 = 25$ and $n = 5.$

The three integers are $4,$ $5,$ and $6,$ and $4^2 + 5^2 + 6^2 = 16 + 25 + 36$ $= 77.$

Thus, the correct answer is B.

Answer: E

Small Hint:

Cross multiply and expand both sides

Big Hint:

The equation becomes linear; it has no solution when the coefficient of $x$ vanishes

Solution:

Cross multiplying gives $(x - 1)(x - 6) = (x - 2)(x - k),$ which expands to $$\begin{aligned} x^2 - 7x + 6 &= x^2 - (2 + k)x \\ &\quad {}+ 2k. \end{aligned}$$

Cancelling $x^2$ leaves $(k - 5)x = 2k - 6.$ For $k$ equal to $1,$ $2,$ $3,$ or $4,$ this gives a valid value of $x$ that is neither excluded denominator value $2$ nor $6.$ For $k=5,$ the equation instead becomes $0\cdot x=4,$ which has no solution.

Thus, the correct answer is E.

Answer: D

Small Hint:

Group terms to factor the left side

Big Hint:

$8xy - 12y + 2x - 3$ $= (4y + 1)(2x - 3)$; it must vanish for every $y$

Solution:

Grouping and factoring, $$\begin{aligned} &8xy - 12y + 2x - 3 \\ &= 4y(2x - 3) + (2x - 3) \\ &= (4y + 1)(2x - 3). \end{aligned}$$

For this to equal $0$ for all $y,$ the factor that depends on $y$ cannot be forced to zero, so we need $2x - 3 = 0,$ giving $x = \dfrac32.$

Thus, the correct answer is D.

Answer: B

Small Hint:

Take the square root by writing everything as powers of $2$ and $5$

Big Hint:

$N = 5^{64}\cdot 2^{75} = 10^{64}\cdot 2^{11}$; only the digits of $2^{11}$ matter

Solution:

Since $25 = 5^2$ and $64 = 2^6,$ we have $25^{64}\cdot 64^{25} = 5^{128}\cdot 2^{150},$ so $$N = \sqrt{5^{128}\cdot 2^{150}} = 5^{64}\cdot 2^{75}.$$

Writing $2^{75} = 2^{64}\cdot 2^{11},$ we get $N = (5\cdot 2)^{64}\cdot 2^{11} = 10^{64}\cdot 2048.$

So $N$ is $2048$ followed by $64$ zeros, and its digit sum is $2 + 0 + 4 + 8 = 14.$

Thus, the correct answer is B.

Answer: E

Small Hint:

$A - B$ and $A + B$ have the same parity, and both are prime

Big Hint:

This forces $B = 2,$ so $A - 2,$ $A,$ and $A + 2$ are three primes in a row

Solution:

The numbers $A - B$ and $A + B$ differ by $2B,$ so they have the same parity. Being prime, they must both be odd, which forces $A$ and $B$ to have opposite parity.

Since $2$ is the only even prime, either $A=2$ or $B=2.$ The first case is impossible because the positive prime $B$ would make $A-B<0.$ Hence $B=2.$

Now $A - 2,$ $A,$ and $A + 2$ are three primes. One of any three integers spaced $2$ apart is divisible by $3,$ so that member must itself be $3.$ The only positive possibility is $3,$ $5,$ $7.$

The four primes are $2,$ $3,$ $5,$ $7,$ and their sum is $17,$ which is prime.

Thus, the correct answer is E.

Answer: D

Small Hint:

Set $\dfrac{n}{20 - n} = k^2$ and solve for $n$ in terms of $k$

Big Hint:

$n = \dfrac{20k^2}{k^2 + 1}$; since $k^2$ and $k^2 + 1$ are coprime, $20$ is divisible by $k^2 + 1$

Solution:

Suppose $\dfrac{n}{20 - n} = k^2$ for some integer $k \ge 0.$ Solving, $$n = \dfrac{20k^2}{k^2 + 1}.$$

Since $k^2$ and $k^2 + 1$ share no common factor, $k^2 + 1$ must divide $20.$ This happens only for $k = 0,$ $1,$ $2,$ $3,$ giving $k^2 + 1 = 1,$ $2,$ $5,$ $10.$

The corresponding values $n = 0,$ $10,$ $16,$ $18$ are all integers, so there are $4$ such $n.$

Thus, the correct answer is D.

Answer: C

Small Hint:

Place the octagon on a grid; each slanted side moves $\sqrt2$ horizontally and vertically

Big Hint:

$DG$ is a horizontal chord; use it as the base and the vertical distance from $A$ as the height

Solution:

Set the octagon on coordinate axes with the axis-aligned sides of length $2$ and each slanted side spanning $\sqrt2$ horizontally and $\sqrt2$ vertically. Then $$\begin{aligned} A &= (\sqrt2, 0), \\ D &= (2 + 2\sqrt2, 2 + \sqrt2), \\ G &= (0, 2 + \sqrt2). \end{aligned}$$

Since $D$ and $G$ share the height $2 + \sqrt2,$ segment $DG$ is horizontal with length $2 + 2\sqrt2,$ and the height from $A$ up to that level is $2 + \sqrt2.$

Therefore $$\begin{aligned} [\triangle ADG] &= \tfrac12(2 + 2\sqrt2)(2 + \sqrt2) \\ &= (1 + \sqrt2)(2 + \sqrt2) \\ &= 4 + 3\sqrt2. \end{aligned}$$

Thus, the correct answer is C.

Answer: D

Small Hint:

Two distinct circles meet in at most $2$ points

Big Hint:

Count the pairs of circles, then multiply by $2$

Solution:

Any two distinct circles intersect in at most $2$ points. There are $\binom{4}{2} = 6$ pairs of circles, giving at most $6\cdot 2 = 12$ intersection points.

This maximum is achievable by a configuration where every pair of circles crosses twice, so the answer is $12.$

Thus, the correct answer is D.