The function f has the property that for each real number x in its domain, (1)/(x) is also in its domain and f(x) + f ((1)/(x)) = x. What is the largest set of real numbers that can be in the domain of f?

(D) \x \mid x ≠ -1, x ≠ 0, x ≠ 1\

2006 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

Sandwiches at Joe’s Fast Food cost $\$3$ each and sodas cost $\$2$ each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?

$31$

$32$

$33$

$34$

$35$

Answer: A

Concepts:money

Difficulty rating: 770

Small Hint:

Multiply each price by its quantity

Big Hint:

The total is $5\cdot 3 + 8\cdot 2$

Solution:

Five sandwiches cost $5\cdot 3 = 15$ dollars and eight sodas cost $8\cdot 2 = 16$ dollars. Together they cost $15 + 16 = 31$ dollars.

Thus, the correct answer is A.

2.

Define $x \otimes y = x^3 - y.$ What is $h \otimes (h \otimes h)?$

$-h$

$0$

$h$

$2h$

$h^3$

Answer: C

Concepts:custom operation substitution

Difficulty rating: 920

Small Hint:

Evaluate the inner $h \otimes h$ first

Big Hint:

$h \otimes h = h^3 - h,$ then apply $\otimes$ once more

Solution:

By the definition, $h \otimes h = h^3 - h.$ Then $\begin{gathered} h \otimes (h^3 - h) \\ = h^3 - (h^3 - h) \\ = h. \end{gathered}$

Thus, the correct answer is C.

3.

The ratio of Mary’s age to Alice’s age is $3 : 5.$ Alice is $30$ years old. How old is Mary?

$15$

$18$

$20$

$24$

$50$

Answer: B

Concepts:ratio and proportion ages

Difficulty rating: 800

Small Hint:

Mary’s age is $\tfrac{3}{5}$ of Alice’s age

Big Hint:

Compute $\tfrac{3}{5}\cdot 30$

Solution:

Mary’s age is $\tfrac{3}{5}$ of Alice’s, so Mary is $\tfrac{3}{5}\cdot 30 = 18$ years old.

Thus, the correct answer is B.

4.

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?

$17$

$19$

$21$

$22$

$23$

Answer: E

Concepts:digits clock

Difficulty rating: 1050

Small Hint:

Maximize the minutes digits and the hour digits separately

Big Hint:

The two minutes digits sum to at most $5+9;$ the best hour is the single digit $9$

Solution:

The two minutes digits sum to at most $5 + 9 = 14,$ at $59$ minutes past the hour. For the hour, a single digit $9$ gives digit sum $9,$ which beats any two-digit hour $(10, 11, 12$ give at most $1 + 2 = 3).$

The largest total is $14 + 9 = 23,$ occurring at $9\!:\!59.$

Thus, the correct answer is E.

5.

Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half of the pizza. The cost of a plain pizza was $\$8,$ and there was an additional cost of $\$2$ for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each then paid for what he had eaten. How many more dollars did Dave pay than Doug?

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:fraction money

Difficulty rating: 1190

Small Hint:

The plain pizza is $\$8$ for $8$ slices, so each plain slice costs $\$1$

Big Hint:

Spread the $\$2$ anchovy surcharge over the $4$ anchovy slices, then total each person’s cost

Solution:

Each plain slice costs $\$1.$ The $\$2$ anchovy charge is spread over the $4$ anchovy slices, adding $\$0.50$ each, so an anchovy slice costs $\$1.50.$

Dave ate $4$ anchovy slices and $1$ plain slice: $4\cdot 1.5 + 1 = \$7.$ Doug ate the $3$ remaining plain slices: $\$3.$ Dave paid $7 - 3 = \$4$ more.

Thus, the correct answer is D.

6.

The $8 \times 18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y?$

$6$

$7$

$8$

$9$

$10$

Answer: A

Concepts:area square (geometry)

Difficulty rating: 1310

Small Hint:

The two hexagons reassemble into a square of area $8\cdot 18$

Big Hint:

The staircase splits the width into three equal horizontal pieces that sum to $18$

Solution:

The two hexagons form a square of area $8 \cdot 18 = 144,$ so the square has side $12.$

The staircase cut splits the width into three equal horizontal pieces of length $y,$ which together span the full width: $y + y + y = 18,$ so $y = 6.$ (The two vertical steps each rise $12 - 8 = 4,$ building the extra height of the square.)

Thus, the correct answer is A.

7.

Mary is $20\%$ older than Sally, and Sally is $40\%$ younger than Danielle. The sum of their ages is $23.2$ years. How old will Mary be on her next birthday?

$7$

$8$

$9$

$10$

$11$

Answer: B

Concepts:percentage linear equation

Difficulty rating: 1240

Small Hint:

Let Danielle’s age be $x,$ then express Sally’s and Mary’s ages in terms of $x$

Big Hint:

Sally $= 0.6x$ and Mary $= 1.2(0.6x) = 0.72x;$ the ages sum to $23.2$

Solution:

Let Danielle be $x$ years old. Then Sally is $0.6x$ and Mary is $1.2(0.6x) = 0.72x.$

The sum $x + 0.6x + 0.72x = 2.32x = 23.2$ gives $x = 10.$ So Mary is $0.72(10) = 7.2$ years old, and on her next birthday she will be $8.$

Thus, the correct answer is B.

8.

How many sets of two or more consecutive positive integers have a sum of $15?$

$1$

$2$

$3$

$4$

$5$

Answer: C

Concepts:arithmetic sequence casework

Difficulty rating: 1330

Small Hint:

A run of $n$ consecutive integers sums to $n$ times its middle value

Big Hint:

Test $n = 2, 3, 4, 5;$ beyond $5$ the smallest possible sum already exceeds $15$

Solution:

The sum of $n$ consecutive integers equals $n$ times their median. For a sum of $15:$ $n = 2$ gives $7 + 8,$ $n = 3$ gives $4 + 5 + 6,$ and $n = 5$ gives $1 + 2 + 3 + 4 + 5.$

A run of four consecutive integers sums to an even number, and more than five terms already exceed $1 + 2 + 3 + 4 + 5 = 15.$ So there are $3$ sets.

Thus, the correct answer is C.

9.

Oscar buys $13$ pencils and $3$ erasers for $\$1.00.$ A pencil costs more than an eraser, and both items cost a whole number of cents. What is the total cost, in cents, of one pencil and one eraser?

$10$

$12$

$15$

$18$

$20$

Answer: A

Concepts:Diophantine Equation divisibility

Difficulty rating: 1430

Small Hint:

Group as $3$ sets of one pencil and one eraser, plus $10$ extra pencils

Big Hint:

If $s$ is the cost of one pencil plus one eraser, then $3s + 10p = 100,$ so $3s$ is a multiple of $10$

Solution:

Let $p$ be a pencil’s cost and $s$ the cost of one pencil plus one eraser, in cents. Then $13p + 3e = 3s + 10p = 100,$ so $3s$ is a multiple of $10$ less than $100.$ Hence $s \in \{10, 20, 30\},$ with $p = 7, 4, 1$ respectively.

Since a pencil costs more than an eraser, $p \gt \tfrac{s}{2},$ which holds only for $s = 10$ (pencil $7,$ eraser $3$ ). So one pencil and one eraser cost $10$ cents.

Thus, the correct answer is A.

10.

For how many real values of $x$ is $\sqrt{120 - \sqrt{x}}$ an integer?

$3$

$6$

$9$

$10$

$11$

Answer: E

Concepts:radical counting integers in a range

Difficulty rating: 1490

Small Hint:

Set $\sqrt{120 - \sqrt{x}} = k$ for a nonnegative integer $k$

Big Hint:

Need $k^2 \le 120,$ and each valid $k$ gives a distinct $x = (120 - k^2)^2$

Solution:

Let $k = \sqrt{120 - \sqrt{x}}$ be an integer. Then $k \ge 0$ and $k^2 = 120 - \sqrt{x} \le 120,$ so $0 \le k \le 10.$

Each such $k$ gives $\sqrt{x} = 120 - k^2 \ge 0,$ hence a distinct value $x = (120 - k^2)^2.$ That is $11$ values.

Thus, the correct answer is E.

11.

Which of the following describes the graph of the equation $(x + y)^2 = x^2 + y^2?$

the empty set

one point

two lines

a circle

the entire plane

Answer: C

Concepts:algebraic manipulation factoring

Difficulty rating: 1390

Small Hint:

Expand $(x + y)^2$ and simplify

Big Hint:

The equation reduces to $2xy = 0$

Solution:

Expanding, $x^2 + 2xy + y^2 = x^2 + y^2,$ so $2xy = 0,$ i.e. $xy = 0.$

This is the union of the two coordinate axes, a pair of lines.

Thus, the correct answer is C.

12.

A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the other rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?

$171$

$173$

$182$

$188$

$210$

Answer: B

Concepts:arithmetic sequence summation

Difficulty rating: 1370

Small Hint:

The top ring spans $20$ cm; each lower ring adds its outside diameter minus $2$ for the overlap

Big Hint:

Total $= 20 + (17 + 16 + \cdots + 1)$

Solution:

The top ring spans $20$ cm. Each ring below overlaps the ring above by $2$ cm (twice the $1$ -cm thickness), so it adds its outside diameter minus $2.$

The lower rings have outside diameters $19, 18, \ldots, 3,$ contributing $17, 16, \ldots, 1.$ Thus the total distance is $\begin{gathered} 20 + (17 + 16 + \cdots + 1) \\ = 20 + \frac{17 \cdot 18}{2} \\ = 20 + 153 \\ = 173 \text{ cm}. \end{gathered}$

Thus, the correct answer is B.

13.

The vertices of a $3$ – $4$ – $5$ right triangle are the centers of three mutually externally tangent circles, as shown. What is the sum of the areas of these circles?

$12\pi$

$\dfrac{25\pi}{2}$

$13\pi$

$\dfrac{27\pi}{2}$

$14\pi$

Answer: E

Concepts:tangent circles system of equations circle area

Difficulty rating: 1330

Small Hint:

The radii at two adjacent vertices sum to the side length between them

Big Hint:

Solve $r + s = 3,\ r + t = 4,\ s + t = 5$

Solution:

If $r, s, t$ are the radii at the vertices, then $r + s = 3,\ r + t = 4,\ s + t = 5.$ Adding all three gives $r + s + t = 6,$ so $r = 1,\ s = 2,\ t = 3.$

The sum of the areas is $\pi(1^2 + 2^2 + 3^2) = 14\pi.$

Thus, the correct answer is E.

14.

Two farmers agree that pigs are worth $\$300$ and that goats are worth $\$210.$ When one farmer owes the other money, he pays the debt in pigs or goats, with “change” received in the form of goats or pigs as necessary. (For example, a $\$390$ debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?

$\$5$

$\$10$

$\$30$

$\$90$

$\$210$

Answer: C

Concepts:greatest common divisor Diophantine Equation

Difficulty rating: 1580

Small Hint:

A resolvable debt has the form $300p + 210g$ for integers $p, g$

Big Hint:

Every such value is a multiple of $\gcd(300, 210)$

Solution:

A debt $D$ is resolvable if and only if $D = 300p + 210g$ $= 30(10p + 7g)$ for integers $p, g.$ Thus $D$ is a multiple of $\gcd(300, 210) = 30,$ so no smaller positive debt works.

A debt of $\$30$ is achievable since $30 = 300(-2) + 210(3),$ i.e. give $3$ goats and receive $2$ pigs in change.

Thus, the correct answer is C.

15.

Suppose $\cos x = 0$ and $\cos(x + z) = \tfrac{1}{2}.$ What is the smallest possible positive value of $z?$

$\dfrac{\pi}{6}$

$\dfrac{\pi}{3}$

$\dfrac{\pi}{2}$

$\dfrac{5\pi}{6}$

$\dfrac{7\pi}{6}$

Answer: A

Concepts:trigonometry

Difficulty rating: 1590

Small Hint:

$\cos x = 0$ means $x$ is an odd multiple of $\tfrac{\pi}{2}$

Big Hint:

$\cos(x + z) = \tfrac{1}{2}$ means $x + z = 2n\pi \pm \tfrac{\pi}{3};$ minimize $z = (x+z) - x$

Solution:

Since $\cos x = 0,$ we have $x = \tfrac{\pi}{2} + k\pi.$ Since $\cos(x + z) = \tfrac{1}{2},$ we have $x + z = 2n\pi \pm \tfrac{\pi}{3}.$

On the unit circle, the smallest positive angular separation between an odd multiple of $\tfrac{\pi}{2}$ and an angle congruent to $\pm\tfrac{\pi}{3}$ is $\tfrac{\pi}{6}.$ It is attained by taking $x = -\tfrac{\pi}{2}$ and $x + z = -\tfrac{\pi}{3},$ which gives $z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6}.$

Thus, the correct answer is A.

16.

Circles with centers $A$ and $B$ have radii $3$ and $8,$ respectively. A common internal tangent intersects the circles at $C$ and $D,$ respectively. Lines $AB$ and $CD$ intersect at $E,$ and $AE = 5.$ What is $CD?$

$13$

$\dfrac{44}{3}$

$\sqrt{221}$

$\sqrt{255}$

$\dfrac{55}{3}$

Answer: B

Difficulty rating: 1760

Small Hint:

Radii $AC$ and $BD$ are perpendicular to the tangent, so $\triangle ACE \sim \triangle BDE$

Big Hint:

$CE = \sqrt{AE^2 - AC^2} = 4,$ and $\tfrac{DE}{CE} = \tfrac{BD}{AC}$

Solution:

The radii satisfy $AC \perp CD$ and $BD \perp CD.$ By the Pythagorean theorem, $CE = \sqrt{5^2 - 3^2} = 4.$

Since $\triangle ACE \sim \triangle BDE,$ we get $\tfrac{DE}{CE} = \tfrac{BD}{AC} = \tfrac{8}{3},$ so $DE = 4 \cdot \tfrac{8}{3} = \tfrac{32}{3}.$ Then $\begin{gathered} CD = CE + DE \\ = 4 + \frac{32}{3} \\ = \frac{44}{3}. \end{gathered}$

Thus, the correct answer is B.

17.

Square $ABCD$ has side length $s,$ a circle centered at $E$ has radius $r,$ and $r$ and $s$ are both rational. The circle passes through $D,$ and $D$ lies on $\overline{BE}.$ Point $F$ lies on the circle, on the same side of $\overline{BE}$ as $A.$ Segment $AF$ is tangent to the circle, and $AF = \sqrt{9 + 5\sqrt{2}}.$ What is $\frac{r}{s}?$

$\dfrac{1}{2}$

$\dfrac{5}{9}$

$\dfrac{3}{5}$

$\dfrac{5}{3}$

$\dfrac{9}{5}$

Answer: B

Concepts:power of a point tangent line coordinate geometry

Difficulty rating: 1910

Small Hint:

The tangent length satisfies $AF^2 = AE^2 - r^2$

Big Hint:

Place $B$ at the origin; then $9 + 5\sqrt{2} = s^2 + rs\sqrt{2},$ and match rational and irrational parts

Solution:

Set $B = (0, 0),$ $C = (s, 0),$ $A = (0, s),$ $D = (s, s),$ so that $E = \left(s + \tfrac{r}{\sqrt{2}},\ s + \tfrac{r}{\sqrt{2}}\right)$ lies on ray $BD.$

Since $AF$ is tangent to the circle, $AF^2 = AE^2 - r^2.$ Computing $AE^2$ and simplifying gives $9 + 5\sqrt{2} = s^2 + rs\sqrt{2}.$

Because $r$ and $s$ are rational, the rational and irrational parts match: $s^2 = 9$ and $rs = 5.$ Thus $s = 3,\ r = \tfrac{5}{3},$ and $\frac{r}{s} = \tfrac{5}{9}.$

Thus, the correct answer is B.

18.

The function $f$ has the property that for each real number $x$ in its domain, $\frac{1}{x}$ is also in its domain and $f(x) + f\!\left(\frac{1}{x}\right) = x.$ What is the largest set of real numbers that can be in the domain of $f?$

$\{x \mid x \neq 0\}$

$\{x \mid x \lt 0\}$

$\{x \mid x \gt 0\}$

$\{x \mid x \neq -1,\ x \neq 0,\ x \neq 1\}$

$\{-1, 1\}$

Answer: E

Concepts:functional equation substitution

Difficulty rating: 1890

Small Hint:

Substitute $\frac{1}{x}$ in place of $x$ to get a second equation

Big Hint:

Comparing the two equations forces $x = \frac{1}{x}$

Solution:

Replacing $x$ by $\frac{1}{x}$ gives $f\!\left(\tfrac{1}{x}\right) + f(x) = \tfrac{1}{x}.$ Together with $f(x) + f\!\left(\tfrac{1}{x}\right) = x,$ this requires $x = \tfrac{1}{x},$ so $x = \pm 1.$

Both values are consistent, with $f(1) = \tfrac{1}{2}$ and $f(-1) = -\tfrac{1}{2}.$ So the largest possible domain is $\{-1, 1\}.$

Thus, the correct answer is E.

19.

Circles with centers $(2, 4)$ and $(14, 9)$ have radii $4$ and $9,$ respectively. The equation of a common external tangent to the circles can be written in the form $y = mx + b$ with $m \gt 0.$ What is $b?$

$\dfrac{908}{119}$

$\dfrac{909}{119}$

$\dfrac{130}{17}$

$\dfrac{911}{119}$

$\dfrac{912}{119}$

Answer: E

Concepts:tangent line trigonometric identity coordinate geometry

Difficulty rating: 1960

Small Hint:

Each circle is tangent to the $x$ -axis, so the $x$ -axis is one common external tangent

Big Hint:

The two external tangents meet on the line through the centers, whose slope is $\tfrac{5}{12} = \tan\theta;$ the other tangent has slope $\tan 2\theta$

Solution:

Each circle’s radius equals its center’s $y$ -coordinate, so both are tangent to the $x$ -axis, which is a common external tangent. The two external tangents meet at the $x$ -intercept of the line through the centers.

That line has slope $\tfrac{9 - 4}{14 - 2} = \tfrac{5}{12} = \tan\theta$ and passes through $(2, 4),$ meeting the $x$ -axis at $\left(-\tfrac{38}{5}, 0\right).$

The other tangent makes angle $2\theta$ with the $x$ -axis, so its slope is $\tan 2\theta = \frac{2 \cdot \tfrac{5}{12}}{1 - \left(\tfrac{5}{12}\right)^2} = \frac{120}{119}.$ Then $b = \tfrac{120}{119} \cdot \tfrac{38}{5} = \tfrac{912}{119}.$

Thus, the correct answer is E.

20.

A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?

$\dfrac{1}{2187}$

$\dfrac{1}{729}$

$\dfrac{2}{243}$

$\dfrac{1}{81}$

$\dfrac{5}{243}$

Answer: C

Concepts:basic probability graph theory casework

Difficulty rating: 2070

Small Hint:

There are $3^7$ equally likely $7$ -move walks; count those visiting all $8$ vertices

Big Hint:

There are $3$ choices for the first move and $2$ for the second, after which the path is nearly forced

Solution:

From the start there are $3^7$ equally likely $7$ -move walks. For a walk visiting all $8$ vertices, there are $3$ choices for the first move and $2$ for the second, since it cannot return to the starting vertex.

Label cube vertices by three-bit strings. By symmetry, after fixing those first two moves we may take the first three vertices to be $000,001,011.$ A branch check gives exactly these three completions: $\begin{aligned} &010,110,111,101,100,\\ &010,110,100,101,111,\\ &111,101,100,110,010. \end{aligned}$ Thus there are $3 \cdot 2 \cdot 3 = 18$ such walks.

The probability is $\dfrac{18}{3^7} = \dfrac{18}{2187} = \dfrac{2}{243}.$

Thus, the correct answer is C.

21.

Let

$\tiny S_1 = \{(x, y) \mid \log_{10}(1 + x^2 + y^2) \le 1 + \log_{10}(x + y)\}$

and

$\tiny S_2 = \{(x, y) \mid \log_{10}(2 + x^2 + y^2) \le 2 + \log_{10}(x + y)\}.$

What is the ratio of the area of $S_2$ to the area of $S_1?$

$98$

$99$

$100$

$101$

$102$

Answer: E

Concepts:logarithm completing the square circle area

Difficulty rating: 2180

Small Hint:

For $j = 1, 2,$ rewrite the condition as $j + x^2 + y^2 \le 10^j(x + y)$

Big Hint:

Complete the square to get disks, then compare their squared radii

Solution:

For $j = 1, 2,$ the condition becomes $j + x^2 + y^2 \le 10^j(x + y),$ i.e. $\begin{gathered} \left(x - \frac{10^j}{2}\right)^2 \\ {}+ \left(y - \frac{10^j}{2}\right)^2 \\ \le \frac{10^{2j}}{2} - j. \end{gathered}$

These are disks with squared radii $\tfrac{100}{2} - 1 = 49$ for $S_1$ and $\tfrac{10000}{2} - 2 = 4998$ for $S_2.$

The logarithms require $x+y>0.$ For each disk, the squared distance from its center to the line $x+y=0$ is $\tfrac{10^{2j}}{2},$ which is $j$ more than the squared radius. Hence both disks lie entirely in $x+y>0,$ so no points were added when the inequalities were rewritten. The area ratio is $\dfrac{4998}{49} = 102.$

Thus, the correct answer is E.

22.

A circle of radius $r$ is concentric with and outside a regular hexagon of side length $2.$ The probability that three entire sides of the hexagon are visible from a randomly chosen point on the circle is $\frac{1}{2}.$ What is $r?$

$2\sqrt{2} + 2\sqrt{3}$

$3\sqrt{3} + \sqrt{2}$

$2\sqrt{6} + \sqrt{3}$

$3\sqrt{2} + \sqrt{6}$

$6\sqrt{2} - \sqrt{3}$

Answer: D

Concepts:geometric probability regular polygon trigonometry

Difficulty rating: 2340

Small Hint:

The complementary arcs, from which only two whole sides are visible, total half the circle; by symmetry each measures $30^\circ$

Big Hint:

The distance from the center to a side (the apothem) is $\sqrt{3},$ and $\sqrt{3} = r\sin 15^\circ$

Solution:

Place the hexagon at the center of the circle. Corresponding to each vertex is an arc from which only the two sides meeting there are entirely visible. These six congruent arcs make up the complementary probability $\tfrac{1}{2},$ so each arc measures $30^\circ.$

Take the arc centered on the ray from the center $O$ through a vertex $A,$ and let $P$ be its upper endpoint. Then $\angle POA = 15^\circ,$ and at $P$ a third side is just becoming visible, so $P$ lies on that side’s supporting line. Its distance from $O$ is the apothem $\sqrt{3}.$

Hence $\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4},$ giving $r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}.$

Thus, the correct answer is D.

23.

Given a finite sequence $S = (a_1, a_2, \ldots, a_n)$ of $n$ real numbers, let $A(S)$ be the sequence $\small \left(\frac{a_1 + a_2}{2}, \frac{a_2 + a_3}{2}, \ldots, \frac{a_{n-1} + a_n}{2}\right)$ of $n - 1$ real numbers. Define $A^1(S) = A(S)$ and, for each integer $m,$ $2 \le m \le n - 1,$ define $A^m(S) = A(A^{m-1}(S)).$ Suppose $x \gt 0,$ and let $S = (1, x, x^2, \ldots, x^{100}).$ If $A^{100}(S) = (\frac{1}{2^{50}}),$ then what is $x?$

$1 - \dfrac{\sqrt{2}}{2}$

$\sqrt{2} - 1$

$\dfrac{1}{2}$

$2 - \sqrt{2}$

$\dfrac{\sqrt{2}}{2}$

Answer: B

Concepts:binomial theorem recursion

Difficulty rating: 2400

Small Hint:

Applying $A$ repeatedly introduces binomial coefficients; $A^{100}(S)$ has a single term

Big Hint:

That term is $\dfrac{(1 + x)^{100}}{2^{100}}$

Solution:

Each application of $A$ averages adjacent terms, so after $100$ steps the single remaining term is $\begin{gathered} \frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m} x^m \\ = \frac{(1 + x)^{100}}{2^{100}}. \end{gathered}$

Setting this equal to $\dfrac{1}{2^{50}}$ gives $(1 + x)^{100} = 2^{50},$ so $1 + x = 2^{\frac{1}{2}} = \sqrt{2}.$ Since $x \gt 0,$ we get $x = \sqrt{2} - 1.$

Thus, the correct answer is B.

24.

The expression

$(x + y + z)^{2006} + (x - y - z)^{2006}$

is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

$6018$

$671{,}676$

$1{,}007{,}514$

$1{,}008{,}016$

$2{,}015{,}028$

Answer: D

Concepts:binomial theorem parity basic counting

Difficulty rating: 2340

Small Hint:

A monomial $x^a y^b z^c$ survives only when $a$ is even; the odd- $a$ terms cancel

Big Hint:

For each even $a$ from $0$ to $2006,$ count the possible values of $b$

Solution:

A term $x^a y^b z^c$ survives only when $a$ is even, since terms with odd $a$ cancel between the two expansions.

For each even $a$ with $0 \le a \le 2006,$ the exponent $b$ ranges over $2007 - a$ values and $c = 2006 - a - b$ is then determined. Summing over even $a:$ $\begin{gathered} (2007 - 0) + (2007 - 2) \\ {}+ \cdots + (2007 - 2006) \\ = 2007 + 2005 + \cdots + 1, \end{gathered}$ the sum of the first $1004$ odd positive integers, which is $1004^2 = 1{,}008{,}016.$

Thus, the correct answer is D.

25.

How many non-empty subsets $S$ of $\{1, 2, 3, \ldots, 15\}$ have the following two properties?

$(1)$ No two consecutive integers belong to $S.$

$(2)$ If $S$ contains $k$ elements, then $S$ contains no number less than $k.$

$277$

$311$

$376$

$377$

$405$

Answer: E

Concepts:subsets bijection combinations

Difficulty rating: 2550

Small Hint:

A $k$ -element valid set is a $k$ -subset of $\{k, k+1, \ldots, 15\}$ with no two consecutive

Big Hint:

Such sets biject with $k$ -subsets of a $(17 - 2k)$ -element set, counted by $\binom{17 - 2k}{k}$

Solution:

By property $(2),$ a valid $k$ -element set is a $k$ -subset of $\{k, k+1, \ldots, 15\}$ with no two consecutive elements.

Collapsing the gaps between chosen elements, these correspond bijectively to $k$ -subsets of a $(17 - 2k)$ -element set, counted by $\binom{17 - 2k}{k}.$ This is nonzero only for $k \le 5,$ so the total is $\begin{gathered} \binom{15}{1} + \binom{13}{2} \\ {}+ \binom{11}{3} + \binom{9}{4} \\ {}+ \binom{7}{5} \\ = 15 + 78 + 165 + 126 + 21 \\ = 405. \end{gathered}$

Thus, the correct answer is E.