2005 AMC 12A Solutions | LIVE by Po-Shen Loh

1.

Two is $10\%$ of $x$ and $20\%$ of $y.$ What is $x - y?$

$1$

$2$

$5$

$10$

$20$

Concepts:percentage linear equation

Difficulty rating: 770

Small Hint:

$10\%$ of $x$ means $0.1x$

Big Hint:

Solve $0.1x = 2$ and $0.2y = 2$ separately

Solution:

From $0.1x = 2$ we get $x = 20,$ and from $0.2y = 2$ we get $y = 10.$

Therefore $x - y = 20 - 10 = 10.$

Thus, the correct answer is D.

2.

The equations $2x + 7 = 3$ and $bx - 10 = -2$ have the same solution for $x.$ What is the value of $b?$

$-8$

$-4$

$-2$

$4$

$8$

Concepts:linear equation substitution

Difficulty rating: 910

Small Hint:

First solve $2x + 7 = 3$ for $x$

Big Hint:

Substitute that value of $x$ into $bx - 10 = -2$

Solution:

Solving $2x + 7 = 3$ gives $x = -2.$

Substituting into the second equation, $-2b - 10 = -2,$ so $-2b = 8$ and $b = -4.$

Thus, the correct answer is B.

3.

A rectangle with a diagonal of length $x$ is twice as long as it is wide. What is the area of the rectangle?

$\dfrac{1}{4}x^2$

$\dfrac{2}{5}x^2$

$\dfrac{1}{2}x^2$

$x^2$

$\dfrac{3}{2}x^2$

Concepts:rectangle Pythagorean Theorem area

Difficulty rating: 1100

Small Hint:

Let the width be $w,$ so the length is $2w$

Big Hint:

The diagonal satisfies $x^2 = w^2 + (2w)^2$

Solution:

Let the width be $w.$ Then the length is $2w,$ and the diagonal gives $x^2 = w^2 + (2w)^2 = 5w^2.$

The area is $w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2.$

Thus, the correct answer is B.

4.

A store normally sells windows at $\$100$ each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?

$100$

$200$

$300$

$400$

$500$

Concepts:basic counting

Difficulty rating: 1200

Small Hint:

For every $4$ windows paid for, a $5$ th comes free

Big Hint:

Compare the number of paid windows separately versus for $15$ windows together

Solution:

Buying separately, Dave gets $7$ windows by paying for $6$ ( $\$600$ ), and Doug gets $8$ by paying for $7$ ( $\$700$ ), for a total of $\$1300.$

Buying together, they need $15$ windows: paying for $12$ yields $3$ free, for a cost of $\$1200.$

The savings are $\$1300 - \$1200 = \$100.$

Thus, the correct answer is A.

5.

The average (mean) of $20$ numbers is $30,$ and the average of $30$ other numbers is $20.$ What is the average of all $50$ numbers?

$23$

$24$

$25$

$26$

$27$

Concepts:mean

Difficulty rating: 1020

Small Hint:

Find the total sum of all $50$ numbers first

Big Hint:

The combined sum is $20 \cdot 30 + 30 \cdot 20$

Solution:

The total of all $50$ numbers is $20 \cdot 30 + 30 \cdot 20$ $= 600 + 600$ $= 1200.$

The average is $1200 \div 50 = 24.$

Thus, the correct answer is B.

6.

Josh and Mike live $13$ miles apart. Yesterday Josh started to ride his bicycle toward Mike’s house. A little later Mike started to ride his bicycle toward Josh’s house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike’s rate. How many miles had Mike ridden when they met?

$4$

$5$

$6$

$7$

$8$

Concepts:distance rate and time ratio and proportion

Difficulty rating: 1270

Small Hint:

Distance equals rate times time for each rider

Big Hint:

Josh’s distance is $\dfrac{4}{5} \cdot 2 = \dfrac{8}{5}$ of Mike’s distance

Solution:

Since distance is rate times time, Josh rode $\dfrac{4}{5} \cdot 2 = \dfrac{8}{5}$ as far as Mike.

Let $m$ be the miles Mike rode. Then $13 = m + \dfrac{8}{5}m = \dfrac{13}{5}m,$ so $m = 5.$

Thus, the correct answer is B.

7.

Square $EFGH$ is inside square $ABCD$ so that each side of $EFGH$ can be extended to pass through a vertex of $ABCD.$ Square $ABCD$ has side length $\sqrt{50},$ $E$ is between $B$ and $H,$ and $BE = 1.$ What is the area of the inner square $EFGH?$

$25$

$32$

$36$

$40$

$42$

Concepts:square (geometry)Pythagorean Theorem congruence (geometry)

Difficulty rating: 1460

Small Hint:

By symmetry the four corner triangles are congruent right triangles

Big Hint:

In right triangle $BCE,$ the leg $CE = BH = \sqrt{BC^2 - BE^2}$

Solution:

By the symmetry of the figure, triangles $ABH,$ $BCE,$ $CDF,$ and $DAG$ are congruent right triangles. Hence $\begin{aligned} &BH = CE = \sqrt{BC^2 - BE^2} \\ &= \sqrt{50 - 1} = 7. \end{aligned}$

Since $E$ lies between $B$ and $H,$ the side of the inner square is $EH = BH - BE = 7 - 1 = 6.$

Therefore the area of $EFGH$ is $6^2 = 36.$

Thus, the correct answer is C.

8.

Let $A,$ $M,$ and $C$ be digits with $\begin{aligned} &(100A + 10M + C) \\ &\quad {}\cdot (A + M + C) = 2005. \end{aligned}$ What is $A?$

$1$

$2$

$3$

$4$

$5$

Concepts:prime factorization digits bounding to limit cases

Difficulty rating: 1350

Small Hint:

Factor $2005$ into primes

Big Hint:

The digit sum $A + M + C$ is at most $27,$ which pins down which factor it equals

Solution:

Since $A + M + C \le 9 + 9 + 9 = 27,$ and $2005 = 5 \cdot 401,$ the digit sum can only be $1$ or $5.$ It cannot be $1,$ because then $100A+10M+C=2005>999.$ Thus it must be the smaller nontrivial factor: $\begin{aligned} &100A + 10M + C = 401, \\ &\quad A + M + C = 5. \end{aligned}$

Reading off the digits, $A = 4,$ $M = 0,$ and $C = 1.$

Thus, the correct answer is D.

9.

There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x.$ What is the sum of those values of $a?$

$-16$

$-8$

$0$

$8$

$20$

Concepts:quadratic Vieta’s Formulas

Difficulty rating: 1380

Small Hint:

A quadratic has one solution exactly when its discriminant is $0$

Big Hint:

Write the linear coefficient as $a + 8$ and set $(a+8)^2 - 4 \cdot 4 \cdot 9 = 0$

Solution:

The equation is $4x^2 + (a+8)x + 9 = 0.$ It has one solution when the discriminant vanishes: $(a+8)^2 - 4 \cdot 4 \cdot 9 = 0,$ so $(a+8)^2 = 144$ and $a + 8 = \pm 12.$

Thus $a = 4$ or $a = -20,$ and their sum is $-16.$

Thus, the correct answer is A.

10.

A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n?$

$3$

$4$

$5$

$6$

$7$

Concepts:cube geometry surface area ratio and proportion

Difficulty rating: 1430

Small Hint:

Count the total number of small-cube faces and the number that are red

Big Hint:

There are $6n^3$ faces in all and $6n^2$ red ones

Solution:

The $n^3$ unit cubes have $6n^3$ faces total. The red faces are exactly the surface of the original cube, $6n^2$ of them.

Setting the red fraction to one-fourth, $\dfrac{6n^2}{6n^3} = \dfrac{1}{n} = \dfrac{1}{4},$ so $n = 4.$

Thus, the correct answer is B.

11.

How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

$41$

$42$

$43$

$44$

$45$

Concepts:digits parity basic counting

Difficulty rating: 1620

Small Hint:

The middle digit is an integer only when the first and last digits have the same parity

Big Hint:

Count the odd-odd pairs and the even-even pairs separately, remembering the first digit is nonzero

Solution:

The middle digit is an integer only when the first and last digits are both odd or both even. Each such pair determines the middle digit uniquely.

There are $5 \cdot 5 = 25$ odd-odd choices for the first and last digits. For even-even, the first digit cannot be $0,$ giving $4 \cdot 5 = 20$ choices.

The total is $25 + 20 = 45.$

Thus, the correct answer is E.

12.

A line passes through $A(1, 1)$ and $B(100, 1000).$ How many other points with integer coordinates are on the line and strictly between $A$ and $B?$

$0$

$2$

$3$

$8$

$9$

Concepts:lattice point greatest common divisor slope

Difficulty rating: 1660

Small Hint:

Reduce the slope $\dfrac{1000 - 1}{100 - 1}$ to lowest terms

Big Hint:

Points on the line take the form $(1 + 11t,\ 1 + 111t);$ count integer $t$ strictly between the endpoints

Solution:

The slope is $\dfrac{1000 - 1}{100 - 1} = \dfrac{999}{99} = \dfrac{111}{11}.$

So every point on the line has the form $(1 + 11t,\ 1 + 111t),$ which is a lattice point exactly when $t$ is an integer. The point is strictly between $A$ and $B$ when $0 \lt t \lt 9.$

There are $8$ such integers $t,$ giving $8$ lattice points.

Thus, the correct answer is D.

13.

In the five-sided star shown, the letters $A,$ $B,$ $C,$ $D,$ and $E$ are replaced by the numbers $3,$ $5,$ $6,$ $7,$ and $9,$ although not necessarily in that order. The sums of the numbers at the ends of the line segments $AB,$ $BC,$ $CD,$ $DE,$ and $EA$ form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?

$9$

$10$

$11$

$12$

$13$

Concepts:arithmetic sequence double counting

Difficulty rating: 1620

Small Hint:

Each of the five numbers is an endpoint of exactly two segments

Big Hint:

The middle term of a five-term arithmetic sequence equals the mean of all five terms

Solution:

Every number appears as an endpoint of exactly two of the five segments, so the total of the five sums is $2(3 + 5 + 6 + 7 + 9) = 60.$

The middle term of a five-term arithmetic sequence is its mean, namely $60 \div 5 = 12.$

Thus, the correct answer is D.

14.

On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

$\dfrac{5}{11}$

$\dfrac{10}{21}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

$\dfrac{6}{11}$

Concepts:dice (probability)conditional probability parity

Difficulty rating: 1870

Small Hint:

A standard die has $1 + 2 + \cdots + 6 = 21$ dots, so the removed dot comes from face $n$ with probability $\dfrac{n}{21}$

Big Hint:

Removing a dot from an even face makes it odd; removing from an odd face makes it even

Solution:

The die has $21$ dots, so a dot is removed from the face with $n$ dots with probability $\dfrac{n}{21}.$

If a dot is removed from an odd face, that face becomes even, leaving two odd faces and hence probability $\dfrac{2}{6}=\dfrac{1}{3}$ of an odd top. If a dot is removed from an even face, that face becomes odd, leaving four odd faces and hence probability $\dfrac{4}{6}=\dfrac{2}{3}.$ The removed dot lies on an odd face with probability $\dfrac{1 + 3 + 5}{21}$ and on an even face with probability $\dfrac{2 + 4 + 6}{21}.$

Hence the answer is $\dfrac{1}{3} \cdot \dfrac{9}{21} + \dfrac{2}{3} \cdot \dfrac{12}{21} = \dfrac{33}{63} = \dfrac{11}{21}.$

Thus, the correct answer is D.

15.

Let $AB$ be a diameter of a circle and $C$ be a point on $AB$ with $2 \cdot AC = BC.$ Let $D$ and $E$ be points on the circle such that $DC \perp AB$ and $DE$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD?$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Concepts:circle area ratio midpoint

Difficulty rating: 1770

Small Hint:

Let $O$ be the center. Since $AC = \dfrac{AB}{3}$ and $AO = \dfrac{AB}{2},$ find $CO$

Big Hint:

$\triangle DCO$ and $\triangle ECO$ have equal areas because $O$ is the midpoint of $DE$

Solution:

Let $O$ be the center. Since $2 \cdot AC = BC,$ we have $AC = \dfrac{AB}{3},$ and $AO = \dfrac{AB}{2},$ so $\begin{aligned} &CO = AO - AC \\ &= \dfrac{AB}{2} - \dfrac{AB}{3} \\ &= \dfrac{AB}{6}. \end{aligned}$

Triangles $DCO$ and $DAB$ share the same altitude from $D$ to line $AB,$ so $\dfrac{[DCO]}{[DAB]} = \dfrac{CO}{AB} = \dfrac{1}{6}.$

Because $O$ is the midpoint of $DE,$ triangles $DCO$ and $ECO$ have equal areas, so $\begin{aligned} &[DCE] = 2\,[DCO] \\ &= \dfrac{2}{6}[DAB] = \dfrac{1}{3}[DAB]. \end{aligned}$

Thus, the correct answer is C.

16.

Three circles of radius $s$ are drawn in the first quadrant of the $xy$ -plane. The first circle is tangent to both axes, the second is tangent to the first circle and the $x$ -axis, and the third is tangent to the first circle and the $y$ -axis. A circle of radius $r \gt s$ is tangent to both axes and to the second and third circles. What is $\frac{r}{s}?$

$5$

$6$

$8$

$9$

$10$

Concepts:tangent circles coordinate geometry Pythagorean Theorem

Difficulty rating: 2000

Small Hint:

Place the centers by their radii: the big circle is at $(r, r)$ and the second small circle at $(3s, s)$

Big Hint:

The distance between the big center and a tangent small center equals $r + s;$ apply the Pythagorean theorem to the horizontal and vertical gaps

Solution:

Put the big circle’s center at $(r, r)$ and the second small circle’s center at $(3s, s).$ They are externally tangent, so the distance between centers is $r + s.$

The horizontal and vertical gaps are $r - 3s$ and $r - s,$ so $(r + s)^2 = (r - 3s)^2 + (r - s)^2.$

Expanding gives $0 = r^2 - 10rs + 9s^2$ $= (r - 9s)(r - s).$ Since $r \ne s,$ we get $r = 9s,$ so $\frac{r}{s} = 9.$

Thus, the correct answer is D.

17.

A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure $1.$ The cube is then cut in the same manner along the dashed lines shown in Figure $2.$ This creates nine pieces. What is the volume of the piece that contains vertex $W?$

$\dfrac{1}{12}$

$\dfrac{1}{9}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

Concepts:pyramid volume 3D geometry

Difficulty rating: 1910

Small Hint:

The piece at $W$ is a pyramid; identify its base and apex

Big Hint:

If its base area is $B$ and height is $h,$ use $V = \dfrac{1}{3}Bh$

Solution:

The two perpendicular sets of cuts each run from top edges to midlines of the bottom face. Near $W$ they carve out a square pyramid. Its base is the quarter of the bottom face adjacent to $W,$ and its apex lies one unit above the center of the bottom face.

Thus its base is a square of side $\dfrac{1}{2}$ and its altitude is the full height $1.$ Therefore the volume is $\dfrac{1}{3} \left(\dfrac{1}{2}\right)^2 (1) = \dfrac{1}{12}.$

Thus, the correct answer is A.

18.

Call a number “prime-looking” if it is composite but not divisible by $2,$ $3,$ or $5.$ The three smallest prime-looking numbers are $49,$ $77,$ and $91.$ There are $168$ prime numbers less than $1000.$ How many prime-looking numbers are there less than $1000?$

$100$

$102$

$104$

$106$

$108$

Concepts:inclusion-exclusion prime counting integers in a range

Difficulty rating: 1950

Small Hint:

Use inclusion-exclusion to count the numbers below $1000$ divisible by $2, 3,$ or $5$

Big Hint:

From the numbers coprime to $2, 3, 5,$ remove the primes and the number $1$

Solution:

Among the $999$ numbers from $1$ to $999,$ inclusion-exclusion gives $\begin{aligned} &499 + 333 + 199 - 166 \\ &\quad {}- 99 - 66 + 33 = 733 \end{aligned}$ that are divisible by $2, 3,$ or $5.$

That leaves $999 - 733 = 266$ numbers coprime to $2, 3, 5.$ Of these, $165$ are primes (the $168$ primes minus $2, 3, 5$ ), and $1$ is neither prime nor composite.

The remaining $266 - 165 - 1 = 100$ numbers are prime-looking.

Thus, the correct answer is A.

19.

A faulty car odometer proceeds from digit $3$ to digit $5,$ always skipping the digit $4,$ regardless of position. For example, after traveling one mile the odometer changed from $000039$ to $000050.$ If the odometer now reads $002005,$ how many miles has the car actually traveled?

$1404$

$1462$

$1604$

$1605$

$1804$

Concepts:number base

Difficulty rating: 1950

Small Hint:

The odometer only ever shows $9$ distinct digits, so it is really counting in base $9$

Big Hint:

Map each shown digit to its base- $9$ value: $5, 6, 7, 8, 9$ stand for $4, 5, 6, 7, 8$

Solution:

Because the odometer never displays a $4,$ it uses only $9$ symbols and counts in base $9,$ where its digits $5, 6, 7, 8, 9$ represent the base- $9$ digits $4, 5, 6, 7, 8.$

The reading $002005$ therefore corresponds to $2004$ in base $9,$ which equals $2 \cdot 9^3 + 4 = 2 \cdot 729 + 4 = 1462.$

Thus, the correct answer is B.

20.

For each $x$ in $[0, 1],$ define $f(x) = \begin{cases} 2x, & 0 \le x \le \tfrac{1}{2},\\ 2 - 2x, & \tfrac{1}{2} \lt x \le 1. \end{cases}$ Let $f^{[2]}(x) = f(f(x)),$ and $f^{[n+1]}(x) = f^{[n]}(f(x))$ for each integer $n \ge 2.$ For how many values of $x$ in $[0, 1]$ is $f^{[2005]}(x) = \tfrac{1}{2}?$

$0$

$2005$

$4010$

$2005^2$

$2^{2005}$

Concepts:function recursion

Difficulty rating: 2330

Small Hint:

Let $g(n)$ be the number of solutions of $f^{[n]}(x) = \tfrac{1}{2};$ find a recursion for $g(n)$

Big Hint:

Because $f$ maps each of $[0, \tfrac12]$ and $[\tfrac12, 1]$ onto all of $[0, 1],$ each solution splits into two, giving $g(n) = 2\,g(n-1)$

Solution:

Let $g(n)$ count the solutions of $f^{[n]}(x) = \tfrac{1}{2}$ in $[0, 1].$ Since $f$ maps each of the two halves $[0, \tfrac12]$ and $[\tfrac12, 1]$ onto all of $[0, 1],$ every solution of $f^{[n-1]}(y) = \tfrac12$ comes from two values of $x$ (one in each half).

The boundary value $x = \tfrac12$ satisfies $f^{[n]}(\tfrac12) = f^{[n-1]}(1) = 0 \ne \tfrac12,$ so no solutions are lost, giving $g(n) = 2\,g(n-1).$

Since $g(1) = 2,$ we conclude $g(2005) = 2^{2005}.$

Thus, the correct answer is E.

21.

How many ordered triples of integers $(a, b, c),$ with $a \ge 2,$ $b \ge 1,$ and $c \ge 0,$ satisfy both $\log_a b = c^{2005}$ and $a + b + c = 2005?$

$0$

$1$

$2$

$3$

$4$

Concepts:logarithm Diophantine Equation bounding to limit cases

Difficulty rating: 2440

Small Hint:

Rewrite the first condition as $b = a^{\left(c^{2005}\right)}$

Big Hint:

If $c \ge 2,$ then $b$ is astronomically larger than $2005;$ only $c = 0$ and $c = 1$ can work

Solution:

The condition $\log_a b = c^{2005}$ means $b = a^{\left(c^{2005}\right)}.$

If $c \ge 2,$ then $b = a^{\left(c^{2005}\right)} \ge 2^{\left(2^{2005}\right)},$ which vastly exceeds $2005,$ so $a + b + c = 2005$ is impossible.

For $c = 0:$ $b = a^0 = 1,$ so $a + 1 + 0 = 2005$ gives $(a, b, c) = (2004, 1, 0).$ For $c = 1:$ $b = a^1 = a,$ so $2a + 1 = 2005$ gives $(a, b, c) = (1002, 1002, 1).$

There are $2$ such triples.

Thus, the correct answer is C.

22.

A rectangular box $P$ is inscribed in a sphere of radius $r.$ The surface area of $P$ is $384,$ and the sum of the lengths of its $12$ edges is $112.$ What is $r?$

$8$

$10$

$12$

$14$

$16$

Concepts:rectangular prism sphere algebraic manipulation

Difficulty rating: 1990

Small Hint:

Let the dimensions be $x, y, z.$ The edge sum gives $x + y + z$ and the surface area gives $2xy + 2yz + 2xz$

Big Hint:

The sphere’s diameter is the space diagonal: $(2r)^2$ $= x^2 + y^2 + z^2$ $= (x+y+z)^2$ $- (2xy + 2yz + 2xz)$

Solution:

Let the dimensions be $x, y, z.$ The $12$ edges give $4(x + y + z) = 112,$ so $x + y + z = 28,$ and the surface area gives $2xy + 2yz + 2xz = 384.$

The space diagonal is a diameter of the sphere, so $\begin{aligned} &(2r)^2 = x^2 + y^2 + z^2 \\ &= (x + y + z)^2 \\ &\quad {}- (2xy + 2yz + 2xz) \\ &= 28^2 - 384 = 400. \end{aligned}$

Thus $2r = 20$ and $r = 10.$

Thus, the correct answer is B.

23.

Two distinct numbers $a$ and $b$ are chosen randomly from the set $\{2, 2^2, 2^3, \ldots, 2^{25}\}.$ What is the probability that $\log_a b$ is an integer?

$\dfrac{2}{25}$

$\dfrac{31}{300}$

$\dfrac{13}{100}$

$\dfrac{7}{50}$

$\dfrac{1}{2}$

Concepts:divisibility floor and ceiling functions basic probability

Difficulty rating: 2330

Small Hint:

Write $a = 2^j$ and $b = 2^k,$ so $\log_a b = \dfrac{k}{j}$

Big Hint:

For each $j,$ count the multiples of $j$ among $1, \ldots, 25$ other than $j$ itself, namely $\left\lfloor \tfrac{25}{j} \right\rfloor - 1$

Solution:

Let $a = 2^j$ and $b = 2^k.$ Then $\log_a b = \dfrac{k}{j},$ which is an integer exactly when $k$ is a multiple of $j.$

For each $j,$ the number of valid $k \ne j$ in $\{1, \ldots, 25\}$ is $\left\lfloor \tfrac{25}{j} \right\rfloor - 1.$ Summing over $j$ gives $\begin{aligned} &24 + 11 + 7 + 5 + 4 + 3 + 2 \\ &\quad {}+ 2 + 4 \cdot 1 = 62 \end{aligned}$ ordered pairs $(a, b).$

Since there are $25 \cdot 24 = 600$ ordered pairs of distinct elements, the probability is $\dfrac{62}{600} = \dfrac{31}{300}.$

Thus, the correct answer is B.

24.

Let $P(x) = (x - 1)(x - 2)(x - 3).$ For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree $3$ such that $P(Q(x)) = P(x) \cdot R(x)?$

$19$

$22$

$24$

$27$

$32$

Concepts:polynomial casework

Difficulty rating: 2520

Small Hint:

Compare degrees: $P(Q(x))$ has degree $3\deg Q$ and $P(x)R(x)$ has degree $6,$ so $\deg Q = 2$

Big Hint:

For $x = 1, 2, 3,$ the right side is $0,$ so each of $Q(1), Q(2), Q(3)$ lies in $\{1, 2, 3\};$ discard the triples giving degree less than $2$

Solution:

Since $P(x)R(x)$ has degree $6$ and $P(Q(x))$ has degree $3\deg Q,$ we need $\deg Q = 2.$ A quadratic $Q$ is determined by the ordered triple $(Q(1), Q(2), Q(3)).$

At $x = 1, 2, 3$ the right side vanishes, so $P(Q(x)) = 0,$ forcing each of $Q(1), Q(2), Q(3)$ into $\{1, 2, 3\}.$ That gives $27$ triples.

Five of them give a polynomial of degree less than $2:$ the constants from $(1,1,1), (2,2,2), (3,3,3)$ and the linear $Q(x) = x$ from $(1,2,3)$ and $Q(x) = 4 - x$ from $(3,2,1).$ The other $27 - 5 = 22$ triples are non-collinear and yield genuine quadratics.

Thus, the correct answer is B.

25.

Let $S$ be the set of all points with coordinates $(x, y, z),$ where $x,$ $y,$ and $z$ are each chosen from the set $\{0, 1, 2\}.$ How many equilateral triangles have all their vertices in $S?$

$72$

$76$

$80$

$84$

$88$

Concepts:equilateral triangle 3D geometry casework

Difficulty rating: 2640

Small Hint:

An equilateral triangle in this grid must have all three sides equal, so each side is a face diagonal or a specific slanted segment

Big Hint:

Count separately the triangles whose sides are unit-cube face diagonals, big-cube face diagonals, and edge-midpoint-to-edge-midpoint segments of length $\sqrt{6}$

Solution:

The three equal sides of such a triangle must all have the same length. Checking the possible squared lengths in the $3 \times 3 \times 3$ grid, only three families of side occur.

Face diagonals of a unit cube (length $\sqrt2$ ): each of the $8$ unit cubes contributes $8$ triangles, one at each corner, for $8 \cdot 8 = 64.$

Face diagonals of the $2 \times 2 \times 2$ cube (length $2\sqrt2$ ): the three faces meeting at a vertex form one triangle, giving $8$ triangles.

Edge-midpoint segments (length $\sqrt6,$ joining midpoints of two edges): each of the $12$ edge midpoints is a vertex of two such triangles, for $\dfrac{12 \cdot 2}{3} = 8.$

The total is $64 + 8 + 8 = 80.$

Thus, the correct answer is C.