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A souvenir shop sells solid metal sculptures of Achilles in two sizes. One is 6 cm tall and one is 8 cm tall. Proportions between the two sculptures are preserved. The smaller sculpture has a mass of 486 grams. What is the mass, in grams, of the larger sculpture? (A) 648 (B) 864 (C) 972 (D) 1152 (E) 1458
D. 1152. Mass scales as the cube of the linear scale factor, so 486(8/6)³ = 1152.
C. 6 − π/6. Eight octants with disjoint interiors have the volume of one sphere of radius 1/2. Subtract its volume, π/6, from 6.
B. 4053. Each of 2026 chords crosses two others, for 2026 distinct crossings. Each chord adds one region plus one per crossing with an earlier chord. The total is 1 + 2026 + 2026.
C. 4. Division gives n − 2 + 3/(n + 2). Thus n + 2 can be any of −3, −1, 1, or 3.
C. 30. The product ab is 6 log₂(3) log₃(5) = 6 log₂(5), and c = log₂(5)/5.
B. 3. Write 1/[k(k + 5)] as (1/5)[1/k − 1/(k + 5)]. The sum telescopes to 19/500 = 0.038.
D. 5/16. The second toss must be heads. HHH has seven valid endings. HHT requires a fourth head and then has three valid endings. There are 10 valid outcomes out of 32.
B. 1/2. The equatorial square has area 2 and sits 1 unit from the apex. The halfway slice has half the side length and one-quarter the area.
B. 80. From 10a + 13b = 1013, we have b ≡ 1 (mod 10). Increasing b lowers a + b. The largest feasible b is 71, giving a = 9.
B. 152√3. Put the circle center at (0,0) and D at (−5,0). Subtract x² + y² = 361 from (x + 5)² + y² = 256 to find x = −13. The height is 8√3 and the base is 38.
B. 513. If t statements are true, the true statements are exactly those with k ≤ 1026 − t. Solve t = 1026 − t.
D. √3. Polly’s displacement vectors turn 240° between her moves, so each group of three sums to zero. The tenth is a unit-circle chord of length 2 sin(60°) = √3.
D. 6/19. Conditioning on the first draw gives A = (2/5)B and B = 2/5 + (3/5)A. Thus A = 4/19 and B = 10/19.
B. (0.21, 0.22). The event x₁² + x₂² > 4 is the part of [0,2]² outside a radius-2 quarter circle. Its probability is 1 − π/4 ≈ 0.2146.
A. −3. Pair the supplementary angles after combining the logs. The product is [sin(π/8) cos(π/8)]² = 1/8.
E. 145. Count by how many values are from 8–10. With zero, one, or two, one of the 1–7 slots is determined modulo 7, giving 49, 63, and 27. With three, the residue patterns (1,3,3) and (2,2,3) give six more.
C. 448. For k right steps, there are 5 − k starting columns and 4^k choices of step heights. The heights force the vertical segments. Sum (5 − k)4^k for k = 1 through 4.
E. 40/3. Unroll the pole. A turn of rise h has ribbon area 4√(100 + h²), half the area 10h of the pole per turn. Solve 4√(100 + h²) = 5h.
C. (3,4,8). C starts above A and B. C, D, and E tie at term 4, and C leads at terms 5 and 6. Averaging preserves the lead from term 7 on.
D. Infinitely many concentric circles and hyperbolas. Rewrite the cosine as sin(π/2 − y²). The two sine cases give x² + y² = π/2 + 2kπ or x² − y² = π/2 + 2kπ.
D. 17. The generating function for triangular numbers is x/(1 − x)³. At x = 1/3 it equals 9/8.
A. 20√3 − 24. Each card rotates by 30°, giving 12 congruent sectors. With A = (0,0) and B = (1,0), one sector has outer corners (√3,1) and (1,√3), and their outer edges meet at (1,2 − 1/√3). The sector area is 5/√3 − 2.
C. 375. Each condition forces one binary bit to be 1, so N = 2^200 − 1. Euler’s theorem gives N ≡ 0 (mod 125), and N ≡ 7 (mod 8). The unique residue modulo 1000 satisfying both is 375.
A. 10π. For t = r², the offsets from the side midpoints to the tangency points are half the differences of vertex tangent lengths. The largest offset is the sum of the other two, so √(19 − t) = √(11 − t) + √(14 − t). The only feasible root is t = 10.
D. 43. The count f(2^m) is 1 plus the sum of f(d) over divisors d > 1 of m, since the bottom base is 2^(m/d). As 8192 = 2^13, add 1 to the sum of f(2^j) for j = 1 through 13, which is 42.