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?

(A) 31 (B) 32 (C) 33 (D) 34 (E) 35

2006 AMC 10 A, Problem #1— 2006 AMC 12 A, Problem #1— “Start by multiplying the number of sandwiches and the cost of sandwiches.”

Solution (A) Five sandwiches cost 5 · 3 = 15 dollars and eight sodas cost 8 · 2 = 16 dollars. Together they cost 15 + 16 = 31 dollars.

Difficulty: Easy NCTM Standard: Number and Operations Standard: understand meanings of operations and how they relate to one another. Mathworld.com Classification: Algebra > Products > Product Define x ⊗ y = x3 − y. What is h ⊗ (h ⊗ h) ?

(A) −h (B) 0 (C) h (D) 2h (E) h3

2006 AMC 10 A, Problem #2— 2006 AMC 12 A, Problem #2— “If we set x = h, then what should y equal?”

Solution (C) By the definition we have

h ⊗ (h ⊗ h) = h ⊗ (h3 − h) = h3 − (h3 − h) = h.

Difficulty: Medium NCTM Standard: Algebra Standard: understand patterns, relations, and functions Mathworld.com Classification: Foundations of Mathematics > Set Theory > Set Properties > Operation The ratio of Mary’s age to Alice’s age is 3 : 5. Alice is 30 years old. How many years old is Mary?

(A) 15 (B) 18 (C) 20 (D) 24 (E) 50

2006 AMC 10 A, Problem #3— 2006 AMC 12 A, Problem #3— “Put the ratio of their ages into a fraction.”

Solution (B) Mary is (3/5)(30) = 18 years old.

Difficulty: Easy NCTM Standard: Number and Operations Standard: understand meanings of operations and how they relate to one another Mathworld.com Classification: Number Theory > Arithmetic > Fractions > Ratio A digital watch displays hours and minutes with am and pm. What is the largest possible sum of the digits in the display?

(A) 17 (B) 19 (C) 21 (D) 22 (E) 23

2006 AMC 10 A, Problem #4— 2006 AMC 12 A, Problem #4— “What is the largest possible sum of the minutes digits?”

Solution (E) The largest possible sum of the two digits representing the minutes is 5 + 9 = 14, occurring at 59 minutes past each hour. The largest possible single digit that can represent the hour is 9. This exceeds the largest possible sum of two digits that can represent the hour, which is 1 + 2 = 3. Therefore, the largest possible sum of all the digits is 14 + 9 = 23, occurring at 9:59.

Difficulty: Medium-easy NCTM Standard: Problem Solving Standard: apply and adapt a variety of appropriate strategies to solve problems Mathworld.com Classification: Calculus and Analysis > Calculus > Maxima and Minima > Maximum 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?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

2006 AMC 10 A, Problem #5— 2006 AMC 12 A, Problem #5— “How much did each slice of plain pizza cost?”

Solution (D) Each slice of plain pizza cost $1. Dave paid $2 for the anchovies in addition to $5 for the five slices of pizza that he ate, for a total of $7. Doug paid only $3 for the three slices of pizza that he ate. Hence Dave paid 7 − 3 = 4 dollars more than Doug.

Difficulty: Medium-easy NCTM Standard: Problem Solving Standard: solve problems that arise in mathematics and in other contexts Mathworld.com Classification: Number Theory > Arithmetic > Fractions > Fraction The 8×18 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 ? D C y

A B

(A) 6 (B) 7 (C) 8 (D) 9 (E) 10 2006 AMC 10 A, Problem #7— 2006 AMC 12 A, Problem #6— “What must the side length of the square be?”

Solution (A) Let E represent the end of the cut on DC, and let F represent the end of the cut on AB. For a square to be formed, we must have

DE = y = FB and DE + y + FB = 18, so y = 6.

The rectangle that is formed by this cut is indeed a square, since the original rectangle has area 8 · 18 = 144, and√ the rectangle that is formed by this cut has a side of length 2 · 6 = 12 = 144.

E C D y E C E D y

B y F y

A F y B A F

Difficulty: Medium-easy NCTM Standard: Standard: apply transformations and use symmetry to analyze mathematical situations Mathworld.com Classification: Geometry > Plane Geometry > Geometric Similarity > Congruent 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?

(A) 7 (B) 8 (C) 9 (D) 10 (E) 11

2006 AMC 12 A, Problem #7— “Let Danielle be x years old. Find Sally and Mary’s age.”

Solution (B) Let Danielle be x years old. Sally is 40% younger, so she is 0.6x years old. Mary is 20% older than Sally, so Mary is 1.2(0.6x) = 0.72x years old. The sum of their ages is 23.2 = x+0.6x+0.72x = 2.32x years, so x = 10. Therefore Mary’s age is 0.72x = 7.2 years, and she will be 8 on her next birthday.

Difficulty: Medium NCTM Standard: Number and Operations Standard: understand meanings of operations and how they relate to one another Mathworld.com Classification: Number Theory > Arithmetic > Fractions > Percent How many sets of two or more consecutive positive integers have a sum of 15?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

2006 AMC 10 A, Problem #9— 2006 AMC 12 A, Problem #8— “Look at the possibilities for each value of n, where n is the number of consecutive positive integers.”

Solution (C) First note that, in general, the sum of n consecutive integers is n times their median. If the sum is 15, we have the following cases: if n = 2, then the median is 7.5 and the two integers are 7 and 8; if n = 3, then the median is 5 and the three integers are 4, 5, and 6; if n = 5, then the median is 3 and the five integers are 1, 2, 3, 4, and 5. Because the sum of four consecutive integers is even, 15 cannot be written in such a manner. Also, the sum of more than five consecutive integers must be more than 1+2+3+4+5 = 15. Hence there are 3 sets satisfying the condition. Note: It can be shown that the number of sets of two or more consecutive positive integers having a sum of k is equal to the number of odd positive divisors of k, excluding 1.

Difficulty: Medium NCTM Standard: Number and Operations Standard: understand meanings of operations and how they relate to one another Mathworld.com Classification: Number Theory > Special Numbers > Digit-Related Numbers > Consecutive Number Sequences 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?

(A) 10 (B) 12 (C) 15 (D) 18 (E) 20

2006 AMC 12 A, Problem #9— “Let p be the cost (in cents) of a pencil, and let s be the cost (in cents) of a set of one pencil and one eraser.”

Solution (A) Let p be the cost (in cents) of a pencil, and let s be the cost (in cents) of a set of one pencil and one eraser. Because Oscar buys 3 sets and 10 extra pencils for $1.00, we have

3s + 10p = 100.

Thus 3s is a multiple of 10 that is less than 100, so s is 10, 20, or 30. The corresponding values of p are 7, 4, and 1. Since the cost of a pencil is more than half the cost of the set, the only possibility is s = 10.

Difficulty: Easy NCTM Standard: Algebra Standard: represent and analyze mathematical situations and structures using algebraic symbols Mathworld.com Classification: Algebra > Algebraic Equations > Linear Equation p √ For how many real values of x is 120 − x an integer?

(A) 3 (B) 6 (C) 9 (D) 10 (E) 11

2006 AMC 10 A, Problem #10— 2006 AMC 12 A, Problem #10— p √ “For what values of x does 120 − x exist?”

p √ √Solution (E) Suppose that k = 120 − x is an integer. Then 0 ≤ k ≤ 120, and because k is an integer, we have 0 ≤ k ≤ 10. Thus there are 11 possible integer values of k. For each such k, the corresponding value of x is (120 − k2)2. Because (120 − k2)2 is positive and decreasing for 0 ≤ k ≤ 10, the 11 values of x are distinct.

AMC 10 Difficulty: Hard AMC 10 Difficulty: Medium-hard NCTM Standard: Problem Solving Standard: apply and adapt a variety of appropriate strategies to solve problems Mathworld.com Classification: Calculus and Analysis > Functions > Domain Which of the following describes the graph of the equation (x + y)2 = x2 + y2 ?

(A) the empty set (B) one point (C) two lines (D) a (E) the entire plane

2006 AMC 10 A, Problem #11— 2006 AMC 12 A, Problem #11— “Expand (x + y)2.”

Solution (C) The equation (x+y)2 = x2 +y2 is equivalent to x2 +2xy + y2 = x2 + y2, which reduces to xy = 0. Thus the graph of the equation consists of the two lines that are the coordinate axes.

AMC 10 Difficulty: Hard AMC 12 Difficulty: Medium-Hard NCTM Standard: Algebra Standard: use mathematical models to represent and understand quantitative relationships Mathworld.com Classification: Applied Mathematics > Data Visualization > Function Graph A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside 18 20 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?

(A) 171 (B) 173 (C) 182 (D) 188 (E) 210

2006 AMC 10 A, Problem #14— 2006 AMC 12 A, Problem #12— “The bottom of the largest ring is 2 cm below the top of the next ring.”

Solution (B) The top of the largest ring is 20 cm above its bottom. That point is 2 cm below the top of the next ring, so it is 17 cm above the bottom of the next ring. The additional distances to the bottoms of the remaining rings are 16 cm, 15 cm,..., 1 cm. Thus the total distance is 17 · 18 20 + (17 + 16 + ··· + 2 + 1) = 20 + = 20 + 17 · 9 = 173 cm. 2

OR The required distance is the sum of the outside diameters of the 18 rings minus a 2-cm overlap for each of the 17 pairs of consecutive rings. This equals 20 · 21 (3+4+5+···+20)−2·17 = (1+2+3+4+5+···+20)−3−34 = −37 = 173 cm. 2

AMC 10 Difficulty: Medium-hard AMC 12 Difficulty: Medium NCTM Standard: Measurement Standard: understand measurable attributes of objects and the units, systems, and processes of measurement Mathworld.com Classification: Geometry > Distance > Distance The vertices of a 3 – 4 – 5 right triangle are the centers of three mutually externally , as shown. What is the sum of the areas of these circles? A 3 4 B C 5

25π 27π (A) 12π (B) 2 (C) 13π (D) 2 (E) 14π

2006 AMC 12 A, Problem #13— “Label the radii. What do we know about the right triangle?”

Solution (E) Let r, s, and t be the radii of the circles centered at A, B, and C, respectively. Then r + s = 3, r + t = 4, and s + t = 5, from which r = 1, s = 2, and t = 3. Thus the sum of the areas of the circles is

π(12 + 22 + 32) = 14π.

Difficulty: Medium NCTM Standard: Geometry Standard: analyze characteristics and properties of two- and three- dimensional geometric shapes and develop mathematical arguments about geometric relationships Mathworld.com Classification: Geometry > Plane Geometry > Triangles > Special Triangles > Other Triangles > Right Triangle 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?

(A) $5 (B) $10 (C) $30 (D) $90 (E) $210

2006 AMC 10 A, Problem #22— 2006 AMC 12 A, Problem #14— “If a debt of D dollars can be resolved in this way, then integers p and g must exist with D = 300p + 210g = 30(10p + 7g). ”

Solution (C) If a debt of D dollars can be resolved in this way, then integers p and g must exist with

D = 300p + 210g = 30(10p + 7g).

As a consequence, D must be a multiple of 30, and no positive debt less than $30 can be resolved. A debt of $30 can be resolved since

30 = 300(−2) + 210(3).

This is done by giving 3 goats and receiving 2 pigs.

AMC 10 Difficulty: Medium-hard AMC 12 Difficulty: Medium NCTM Standard: Algebra Standard: use mathematical models to represent and understand quantitative relationships Mathworld.com Classification: Number Theory > Diophantine Equations > Diophantine Equation Suppose cos x = 0 and cos(x + z) = 1/2. What is the smallest possible positive value of z ?

π π π 5π 7π (A) 6 (B) 3 (C) 2 (D) 6 (E) 6

2006 AMC 12 A, Problem #15— “What are the possible values of x and x + z?”

Solution (A) Because cos x = 0 and cos(x + z) = 1/2, it follows that x = mπ/2 for some odd integer m and x + z = 2nπ ± π/3 for some integer n. Therefore mπ π π π z = 2nπ − ± = kπ + ± 2 3 2 3 for some integer k. The smallest value of k that yields a positive value for z is 0, and the smallest positive value of z is π/2 − π/3 = π/6. OR Let O denote the center of the unit circle. Because cos x = 0, the terminal side of an angle of measure x, measured counterclockwise from the positive x-axis, intersects the circle at A = (0, 1) or B = (0, −1). A C

O

D B Because cos(x + z) = 1/2, the terminal side of an angle of measure x + z ³ √ ´ ³ √ ´ 1 3 1 3 intersects the circle at C = 2 , 2 or D = 2 , − 2 . Thus all angles of positive measure z = (x + z) − x can be measured counterclockwise from either OA or OB to either OC or OD. The smallest such angle is ∠BOD, which has measure π/6 and is attained, for example, when x = −π/2 and x + z = −π/3.

Difficulty: Medium-hard NCTM Standard: Problem Solving Standard: apply and adapt a variety of appropriate strategies to solve problems Mathworld.com Classification: Calculus and Analysis > Functions > Periodic Function Circles with centers A and B have radii 3 and 8, respectively. A common internal tangent touches the circles at C and D, as shown. Lines AB and CD intersect at E, and AE = 5. What is CD ? D

A E B C

44 √ √ 55 (A) 13 (B) (C) 221 (D) 255 (E) 3 3 2006 AMC 10 A, Problem #23— 2006 AMC 12 A, Problem #16— “Use the to determine CE.”

Solution (B) Radii AC and BD are each perpendicular to CD. By the Pythagorean Theorem, √ CE = 52 − 32 = 4.

Because 4ACE and 4BDE are similar, DE BD BD 8 32 = , so DE = CE · = 4 · = . CE AC AC 3 3 Therefore 32 44 CD = CE + DE = 4 + = . 3 3

AMC 10 Difficulty: Hard AMC 12 Difficulty: Medium NCTM Standard: Geometry Standard: analyze characteristics and properties of two- and three- dimensional geometric shapes and develop mathematical arguments about geometric relationships Mathworld.com Classification: Geometry > Plane Geometry > Circles > Tangent Circles 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 BE. Point F lies on the circle, on the same side of BE as A. Segment AF is tangent to the p √ circle, and AF = 9 + 5 2. What is r/s ?

F

E A D

B C

1 5 3 5 9 (A) 2 (B) 9 (C) 5 (D) 3 (E) 5 2006 AMC 12 A, Problem #17— “Look at right triangle AF E.”

Solution³ (B) Let B ´= (0, 0), C = (s, 0), A = (0, s), D = (s, s), and E = s + √r , s + √r . Apply the Pythagorean Theorem to 4AF E to 2 2 obtain µ ¶ µ ¶ ³ √ ´ r 2 r 2 r2 + 9 + 5 2 = s + √ + √ , 2 2 √ √ from which 9 + 5 2 = s2 + rs 2. Because r and s are rational, it follows that s2 = 9 and rs = 5, so r/s = 5/9. OR Extend AD past D to meet the circle at G 6= D. Because E is collinear√ with B and D, 4EDG is an isosceles right triangle. Thus DG = r 2. By the Power of a Point Theorem,

√ ³ √ ´ √ 9+5 2 = AF 2 = AD·AG = AD·(AD + DG) = s s + r 2 = s2+rs 2.

As in the first solution, conclude that r/s = 5/9.

Difficulty: Hard NCTM Standard: Geometry Standard: analyze characteristics and properties of two- and three- dimensional geometric shapes and develop mathematical arguments about geometric relationships Mathworld.com Classification: Geometry > Plane Geometry > Circles > Circle The function f has the property that for each real number x in its domain, 1/x is also in its domain and µ ¶ 1 f(x) + f = x. x What is the largest set of real numbers that can be in the domain of f ?

(A) {x | x 6= 0} (B) {x | x < 0} (C) {x | x > 0} (D) {x | x 6= −1 and x 6= 0 and x 6= 1} (E) {−1, 1}

2006 AMC 12 A, Problem #18— 1 “If x is in the domain, what do we know about x?”

Solution (E) The conditions on f imply that both µ ¶ µ ¶ µ ¶ µ ¶ 1 1 1 1 1 x = f(x) + f and = f + f = f + f(x). x x x 1/x x

Thus if x is in the domain of f, then x = 1/x, so x = ±1. The conditions are satisfied if and only if f(1) = 1/2 and f(−1) = −1/2.

Difficulty: Hard NCTM Standard: Problem Solving Standard: apply and adapt a variety of appropriate strategies to solve problems Mathworld.com Classification: Calculus and Analysis > Functions > Domain 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 > 0. What is b ?

9

(14,9) 4 (2,4)

908 909 130 911 912 (A) 119 (B) 119 (C) 17 (D) 119 (E) 119

2006 AMC 12 A, Problem #19— “What do we know about the containing the centers of the two circles?”

Solution (E) The slope of the line l containing the centers of the circles is 5/12 = tan θ, where θ is the acute angle between the x-axis and line l. The equation of line l is y − 4 = (5/12)(x − 2). This line and the two common external are concurrent. Because one of these tangents is the x-axis, the point of intersection is the x-intercept of line l, which is (−38/5, 0). The acute angle between the x-axis and the other tangent is 2θ, so the slope of that tangent is

5/12 120 tan 2θ = 2 · = . 1 − (5/12)2 119

Thus the equation of that tangent is y = (120/119) (x + (38/5)), and 120 38 912 b = · = . 119 5 119

Difficulty: Hard NCTM Standard: Geometry Standard: specify locations and describe spatial relationships using coordinate geometry and other representational systems Mathworld.com Classification: Geometry > Plane Geometry > Circles > Tangent Circles 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?

1 1 2 1 5 (A) 2187 (B) 729 (C) 243 (D) 81 (E) 243

2006 AMC 10 A, Problem #25— 2006 AMC 12 A, Problem #19— “What is the probability after 1 move (2 moves, 3 moves, . . . ) that the bug has visited 2 (3, 4, . . . ) different vertices?”

Solution (C) At each vertex there are three possible locations that the bug can travel to in the next move, so the probability that the bug will visit three different vertices after two moves is 2/3. Label the first three vertices that the bug visits as A to B to C, as shown in the diagram. In order to visit every vertex, the bug must travel from C to either G or D. The bug travels to G with probability 1/3, and from there the bug must visit the vertices F , E, H, D in that order. Each of these choices has probability 1/3 of occurring. So the probability that the path continues in the form C → G → F → E → H → D  1 5 is 3 . Alternatively, the bug could travel from C to D with probability 1/3, and then travel to H, which also occurs with probability 1/3. From H the bug could go either to G or to E, with probability 2/3, and from there to the two remaining vertices, each with probability 1/3. So the probability that the path continues in one of the forms

% E → F → G H G C → D → H & G → F → E E F  4 is 2 1 . D 3 3 C Hence the bug will visit every vertex in seven moves with probability   "      #       A B 2 1 5 2 1 4 2 1 2 1 4 2 + = + = . 3 3 3 3 3 3 3 3 243

Difficulty: Hard NCTM Standard: Data Analysis and Probability Standard: understand and apply basic concepts of probability Mathworld.com Classification: Probability and Statistics > Probability > Probability Let

2 2 S1 = {(x, y) | log10(1 + x + y ) ≤ 1 + log10(x + y)} and

2 2 S2 = {(x, y) | log10(2 + x + y ) ≤ 2 + log10(x + y)}.

What is the ratio of the area of S2 to the area of S1?

(A) 98 (B) 99 (C) 100 (D) 101 (E) 102

2006 AMC 12 A, Problem #21— “These inequalities define regions bounded by circles.”

Solution (E) For j = 1 and 2, the given inequality is equivalent to

j + x2 + y2 ≤ 10j(x + y),

or to µ ¶ µ ¶ 10j 2 10j 2 102j x − + y − ≤ − j, 2 2 2 provided that x + y > 0. These inequalities define regions bounded by circles. For j = 1 the circle has center√ (5, 5) and radius 7. For j = 2 the circle has center (50, 50) and radius 4998. In each case the center is on the line y = x in the first quadrant, and the radius is less than the distance from the center to the origin. Thus x + y > 0 at each interior point of each circle, as was required to ensure the equivalence of the inequalities. The squares of the radii of the circles are 49 and 4998 for j = 1 and 2, respectively. Therefore the ratio of the area of S2 to that of S1 is (4998π)/(49π) = 102.

Difficulty: Hard NCTM Standard: Problem Solving Standard: apply and adapt a variety of appropriate strategies to solve problems Mathworld.com Classification: Calculus and Analysis > Special Functions > Logarithms > Logarithm 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 1/2. What is r ? √ √ √ √ √ √ √ √ (A) 2 2 + 2 3 (B) 3 3 + 2 (C) 2 6 + 3 (D) 3 2 + 6 √ √ (E) 6 2 − 3

2006 AMC 12 A, Problem #22— “Corresponding to each vertex of the hexagon, there is an arc on the circle from which only the two sides meeting at that vertex are visible.”

Solution (D) Place the hexagon in a coordinate plane with center at the origin O and vertex A at (2, 0). Let B, C, D, E, and F be the other vertices in counterclockwise order.

C B P (r, 0) O A

Corresponding to each vertex of the hexagon, there is an arc on the circle from which only the two sides meeting at that vertex are visible. The given probability condition implies that those arcs have a combined degree measure of 180◦, so by symmetry each is 30◦. One such arc is centered at (r, 0). Let P be the endpoint of this arc in the upper half-plane. Then ∠POA = 15◦. Side BC is visible from points immediately above P , so P is√ collinear with B and C. Because the perpendicular distance from O to BC is 3, we have √ 3 = r sin 15◦ = r sin(45◦ − 30◦) = r (sin 45◦ cos 30◦ − sin 30◦ cos 45◦) .

So √ Ã√ ! √ √ √ 2 3 1 6 − 2 3 = r · − = r · . 2 2 2 4 Therefore √ √ √ √ 4 3 4 3 6 + 2 √ √ √ √ r = √ √ = √ √ · √ √ = 18 + 6 = 3 2 + 6. 6 − 2 6 − 2 6 + 2

Difficulty: Hard NCTM Standard: Geometry Standard: analyze characteristics and properties of two- and three- dimensional geometric shapes and develop mathematical arguments about geometric relationships Mathworld.com Classification: Geometry > General Geometry > Concentric Given a finite sequence S = (a1, a2, . . . , an) of n real numbers, let A(S) be the sequence µ ¶ a + a a + a a + a 1 2 , 2 3 ,..., n−1 n 2 2 2 of n − 1 real numbers. Define A1(S) = A(S) and, for each m m−1 integer m, 2 ≤ m ≤ n − 1¡, define A (S) =¢ A(A (S)). 2 100 100 ¡Suppose¢ x > 0, and let S = 1, x, x , . . . , x . If A (S) = 1/250 , then what is x ?

√ √ √ √ 2 1 2 (A) 1 − 2 (B) 2 − 1 (C) 2 (D) 2 − 2 (E) 2 2006 AMC 12 A, Problem #23— “First investigate A2(S).”

Solution (B) For every sequence S = (a1, a2, . . . , an) of at least three terms, µ ¶ a + 2a + a a + 2a + a a + 2a + a A2(S) = 1 2 3 , 2 3 4 ,..., n−2 n−1 n . 4 4 4

Thus for m = 1 and 2, the coefficients of¡ the¢ terms¡ ¢ in¡ the¢ numerator of Am(S) are the binomial coefficients m , m ,..., m , and the ¡ ¢ ¡ ¢ 0 ¡1 ¢ m m m m m+1 denominator is 2 . Because r + r+1 = r+1 for all integers r ≥ 0, the coefficients of the terms in the numerators of Am+1(S) are ¡m+1¢ ¡m+1¢ ¡m+1¢ 0 , 1 ,..., m+1 for 2 ≤ m ≤ n − 2. The definition implies that the denominator of each term in Am+1(S) is 2m+1. For the given sequence, the sole term in A100(S) is µ ¶ µ ¶ 1 X100 100 1 X100 100 1 a = xm = (x + 1)100. 2100 m m+1 2100 m 2100 m=0 m=0

Therefore µ ¶ µ ¶ 1 (1 + x)100 = A100(S) = , 250 2100 √ so (1 + x)100 = 250, and because x > 0, we have x = 2 − 1.

Difficulty: Hard NCTM Standard: Problem Solving Standard: apply and adapt a variety of appropriate strategies to solve problems Mathworld.com Classification: Number Theory > Sequences > Sequence 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?

(A) 6018 (B) 671,676 (C) 1,007,514 (D) 1,008,016 (E) 2,015,028 2006 AMC 12 A, Problem #24— “There is exactly one term in the simplified expression for every monomial of the form xaybzc, where a, b, and c are non-negative integers.”

Solution (D) There is exactly one term in the simplified expression for every monomial of the form xaybzc, where a, b, and c are non-negative integers, a is even, and a+b+c = 2006. There are 1004 even values of a with 0 ≤ a ≤ 2006. For each such value, b can assume any of the 2007 − a integer values between 0 and 2006 − a, inclusive, and the value of c is then uniquely determined as 2006 − a − b. Thus the number of terms in the simplified expression is

(2007 − 0) + (2007 − 2) + ··· + (2007 − 2006) = 2007 + 2005 + ··· + 1.

This is the sum of the first 1004 odd positive integers, which is 10042 = 1,008,016. OR The given expression is equal to X 2006! ¡ ¢ xaybzc + xa(−y)b(−z)c , a!b!c! where the sum is taken over all non-negative integers a, b, and c with a + b + c = 2006. ¡k+2¢ Because the number of non-negative integer solutions of a+b+c = k is 2 , the sum ¡2008¢ is taken over 2 terms, but those for which b and c have opposite parity have a sum of zero. If b is odd and c is even, then a is odd, so a = 2A+1, b = 2B +1, and c = 2C for some non-negative integers A, B, and C. Therefore 2A + 1 + 2B + 1 + 2C = 2006, ¡1004¢ so A + B + C = 1002. Because the last equation has 2 non-negative integer ¡1004¢ solutions, there are 2 terms for which b is odd and c is even. The number of terms for which b is even and c is odd is the same. Thus the number of terms in the simplified expression is µ ¶ µ ¶ 2008 1004 − 2 = 1004 · 2007 − 1004 · 1003 = 10042 = 1,008,016. 2 2

Difficulty: Hard NCTM Standard: Algebra Standard: understand patterns, relations, and functions Mathworld.com Classification: Discrete Mathematics > Combinatorics > Binomial Coefficients > Binomial Series How many non-empty subsets S of {1, 2, 3,..., 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.

(A) 277 (B) 311 (C) 376 (D) 377 (E) 405

2006 AMC 12 A, Problem #25— “First examine sets S with 1 element, then 2 elements, etc.”

Solution (E) For 1 ≤ k ≤ 15, the k-element sets with properties (1) and (2) are the k-element subsets of Uk = {k, k + 1,..., 15} that contain no two consecutive integers. If {a1, a2, . . . , ak} is such a set, with its elements listed in increasing order, then {a1 + k − 1, a2 + k − 2, . . . , ak−1 + 1, ak} is a k-element subset of U2k−1. Conversely, if {b1, b2, . . . , bk} is a k-element subset of U2k−1, with its elements listed in increasing order, then {b1 − k + 1, b2 − k + 2, . . . , bk−1 − 1, bk} is a set with properties (1) and (2). Thus for each k, the number of k-element sets with properties (1) and (2) is equal to the number of k-element subsets of the (17 − 2k)-element set U2k−1. Because k ≤ 17 − 2k only if k ≤ 5, the total number of such sets is µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ X5 17 − 2k 15 13 11 9 7 = + + + + k 1 2 3 4 5 k=1 = 15 + 78 + 165 + 126 + 21 = 405.

Difficulty: Hard NCTM Standard: Problem Solving Standard: build new mathematical knowledge through problem solving Mathworld.com Classification: Foundations of Mathematics > Set Theory > Sets > Subset