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Factor 180 = 2² · 3² · 5. You need 5 digits (each 1–9) whose product is 180. First find all unordered multisets of 5 digits from {1,...,9} with product 180. Then count the arrangements of each multiset (divide 5! by the factorial of each repeated digit's count).

First understand f: it is 0 for x < 1 and 2x−2 for x ≥ 1 (so f(x) = 2(x−1) for x ≥ 1). To solve f(f(f(f(x)))) = x, work backwards. Let g = f∘f∘f∘f. For each case, track carefully which branch of f applies at each step. Start with f(f(x)) = x and build up.

Interior angle of a regular pentagon is 108°. Triangle PRS is equilateral with P inside the pentagon, so all angles are 60°. Find angle PRT, then use the fact that T and V are vertices of the regular pentagon to find angle PTV. Work through the angles at each vertex systematically.

Any triangulation of a 2003-gon with non-crossing diagonals uses exactly 2001 triangles. For a triangle inscribed in a circle to be acute, the center must be inside the triangle. Count how many triangles in the triangulation can be acute — equivalently, can contain the center of the circle. At most one triangle can contain the center, so at least 2000 triangles are non-acute... but can we do better?

Let the side length be s. Place the square with O at origin: O=(0,0), P=(s,0), E=(s,s), N=(0,s). T is on NO, say T=(t,s). Area of triangle TOP = (1/2)|det([P−T, O−T])| = 62 and area of triangle TEN = (1/2) · t · s... Wait, re-read: OPEN is the square with vertices O, P, E, N in order. Set up coordinates carefully and use the area conditions to find s.

Each student can be represented by a binary string of length 10 (membership in each club). The first condition says all strings are distinct. The second condition about any three students is more subtle — it rules out certain configurations. Think about what the conditions imply for the set of characteristic vectors. What structure do they form?

Let x, y, z be the number of committees of each type (5 senators; 4 senators + 4 aides; 2 senators + 12 aides). Set up two equations from the constraints: every senator serves on 5 committees, every aide serves on 3 committees. Count total senator-slots and aide-slots in two ways.

Let h be the water height. The cone's radius-to-height ratio is 3/12 = 1/4, so at height h the water radius is h/4. Volume: V = πh³/48. Differentiate: dV/dt = πh²/16 · dh/dt. Since dV/dt = −h (rate out), set up the ODE and separate variables to find h(t). Solve for when h = 0.

The vertical distance between the graphs is f(x) = (2 + sin x) − cos x = 2 + sin x − cos x. Write sin x − cos x = √2 sin(x − π/4). The minimum value of 2 + √2 sin(x − π/4) occurs when sin(x − π/4) = −1.

Count palindromes by number of digits. 1-digit: all 9 (1–9). 2-digit: first digit determines the second, so 9 choices. 3-digit: first and last digits are equal, middle is free — how many choices are there? Add them all up.

Since AB = 24 and OA = OB = 15 (radii), use the power of a point. C lies on ray AB beyond B, with BC = 28. So AC = AB + BC = 52. Power of point C with respect to circle O is CA · CB = OC² − r². Compute OC using this relationship.

To guarantee at least one of each color, think about the worst case: you could pull out many socks without getting one of each color. The worst case is pulling out all socks of three colors before getting any of the fourth. Which color has the fewest socks? You need to exhaust all socks of the three most common colors and then pick one more.

A 3-engine plane crashes if 2 or more engines fail: P₃ = C(3,2)p²(1−p) + p³. A 5-engine plane crashes if 3 or more fail: P₅ = C(5,3)p³(1−p)² + C(5,4)p⁴(1−p) + p⁵. Set P₃ = P₅ and solve. Factor out p² (p=0 is trivial). The resulting equation in p should factor nicely.

Place the right angle at the origin, with legs along the axes. For a rectangle with one side along a leg of length a (say the leg of length 3), if the rectangle has width x along that leg and height h, similar triangles give h = b(1 − x/a) where b is the other leg. Maximize the area xh. Try all cases (side along the leg of length 3 or 4).

Since triangle ABC is equilateral and angles EAC = EBC, point E lies on the circumcircle of triangle ABC (angles subtended by AC are equal). Use Ptolemy's theorem on the cyclic quadrilateral formed, along with the given lengths BE = 5 and CE = 12.

A circle of radius 3 has area 9π. A square of side 2 has area 4. The circle crosses the center of the square, so the center of the circle is at the center of the square. Find the area of intersection using integration or geometric decomposition, then compute (area of circle − overlap) + (area of square − overlap).

First factor 2001. Then use Euler's totient function: the count of integers from 1 to n coprime to n is φ(n) = n · ∏(1 − 1/p) over prime factors p of n. Scale appropriately for the range 1 to 2000.

A match ends when one player reaches 21 wins. Let p = 0.3 (Frank) and q = 0.7 (Joe). The total number of games ranges from 21 to 41. Think of the match as a negative binomial problem. The expected number of games equals E[games until Frank gets 21] weighted by P(Frank wins) plus E[games until Joe gets 21] weighted by P(Joe wins). Use the negative binomial distribution.

Use the scalar triple product formula: V = |det([b−a, c−a, d−a])| / 6, where a, b, c, d are the four vertices. Form the 3×3 matrix with rows (b−a), (c−a), (d−a) and compute its determinant.

Let the two circles have radii r and R (r < R) with the same center. The longest chord entirely within the ring is tangent to the inner circle. If the chord is tangent to the inner circle, its length is 2√(R² − r²). The area of the ring is π(R² − r²) = A. Express the chord length in terms of A.

Model each professor's arrival time as a uniform random variable on [0, 120] minutes. They overlap if |X − Y| < 15. Draw the 120×120 square and find the area of the region where |X − Y| < 15 as a fraction of the total area.

List all pairs (a, b) with a + b divisible by 5. The possible sums divisible by 5 are 5 and 10. Count the pairs giving sum 5: (1,4),(2,3),(3,2),(4,1). Count the pairs giving sum 10: (4,6),(5,5),(6,4). Total favorable outcomes out of 36.

The 8 odd integers are distinct and positive, so they are at least 1, 3, 5, 7, 9, 11, 13, 15 — but their sum is 64, far more than 20. Wait — reread: 8 positive odd integers (not necessarily distinct) that are sets? Actually re-examine: distinct sets means we want unordered collections. Since they're odd and sum to 20 (even), and odd+odd=even, 8 odd numbers sum to an even number ✓. Start by finding the minimum sum and work from there.

Label the six seats 1 through 6. Students in seats 2–5 must pass by someone to reach an aisle unless all students between them and an aisle have already left. It's easier to count the complementary event: all students leave without passing anyone. This requires a specific ordering — think about which students can leave freely.

Matt's count leaves remainder 1 when divided by 2, 3, 4, 5, 6, 7, and 8. So his count is 1 more than a multiple of lcm(2,3,4,5,6,7,8). Compute that LCM, find multiples in [999, 1999], then determine the number of equal piles.

Let the three expressions be A = xy, B = 1 − x − y + xy = (1−x)(1−y), C = x + y − 2xy. Note A + B + C = 1. To minimize the largest of three quantities summing to 1, consider when two of them are equal (and at a maximum). Try setting A = C or B = C and solving.

Place the center of C at the origin. The two diameters form four sectors. The small circle S is tangent to both diameters and internally tangent to C. If the angle between diameters is 30°, the circle S sits in a 30° wedge. Use the formula for a circle inscribed in an angle: its center is at distance r/sin(θ/2) from the vertex, where θ is the angle.

Use the Conway leading number method or set up a Markov chain. Let states track the current longest suffix of each target sequence. Consider: if you have seen '10101', could '010101' have already appeared? Think carefully about overlaps between the two sequences.

The 45° north parallel lies at height h = R sin(45°) = R/√2 above the equator. Use the spherical cap volume formula: V = πh²(3R − h)/3. Then divide by the total sphere volume (4πR³/3).

Each word uses exactly one letter from each die. Analyze letter co-occurrences: D appears with A, O, G, I, R, E, V. A appears with D, L, T, R, E, N. Find letters that never appear together (must be on different dice). Build the dice by finding a 3-coloring of the letter graph.

Set up a system of equations. Notice that subtracting the first equation from the second eliminates one variable. What does 2 hamburgers + 2 milkshakes + 2 fries equal in terms of the two given equations?

Consider small cases first. If N = p², the only proper divisor is p, so the condition is vacuously satisfied... wait, N needs at least two proper divisors. Try N = p³ (proper divisors: p, p²; difference p²-p = p(p-1), must divide p³, so p-1 divides p², giving p=2, N=8). Try N = pq (proper divisors: p, q; difference |p-q| must divide pq). Try N = p²q, etc.

Count digit occurrences. For pages 1-9: each digit 1-9 appears once. For 10-99: count tens and units separately. For 100-549: count hundreds, tens, and units. Alternatively, use the formula: sum of digits from 1 to n.

Let AB = 10a + b and CD = 10c + d, or more simply let the first two digits be m and the last two be n. Then (m+n)² = 100m + n. So (m+n)² - (m+n) = 99m, giving (m+n)(m+n-1) = 99m. Let s = m+n, then s(s-1) = 99m and n = s - m = s - s(s-1)/99.

A number is divisible by 11 if the alternating sum of its digits is divisible by 11. For 100...001, the digits are 1, then k zeros, then 1. Use the divisibility rule for 11: 10 ≡ -1 (mod 11), so 10^(k+1) + 1 ≡ (-1)^(k+1) + 1 (mod 11). This is 0 when k+1 is odd, i.e., k is even.

Write n in base 3. The function M acts on the base-3 representation by shifting digits. If n = (a_k a_{k-1} ... a_0)_3, then M(n) corresponds to a specific digit manipulation. Find the pattern by computing small values.

By Vieta's formulas: a+b+c = -a, ab+bc+ca = b, abc = -c. Analyze cases based on whether c = 0 or c ≠ 0.

The number of compositions of n into k parts is C(n-1, k-1). Sum over k from 2 to n, or use the recurrence: each composition of n can be formed by adding 1 to the first part of a composition of n-1, or by prepending 1 to a composition of n-1.

Let v be the Concorde's speed without wind. London to New York: distance = 6000 km, time = 11:00 - 12:00 + 5h = 4h (accounting for time zone). So v = 6000/4 = 1500 km/h. New York to London: speed = v + 250 = 1750 km/h (with jet stream). Time = 6000/1750 hours. Add 5h time difference.

Rewrite as y² - x² = p, so (y-x)(y+x) = p. Since p is prime, the only factorizations are 1·p.

Use the identity 2cos(θ)=e^(iθ)+e^(-iθ) and the fact that the 11th roots of unity satisfy a cyclotomic polynomial.

Write the number as 7·10ⁿ + (10ⁿ - 1)/9. Consider the number modulo 3 and use properties of repunits.

To minimize the sum of all digits, use numbers with as many zeros as possible. Consider numbers like 100, 200, ..., 900 and fill the rest with single-digit numbers.

Use Vieta's formulas. If f(x)=(x-r)(x-s)(x-t), then f'(x) has roots at the critical points. The derivative roots being integers imposes strong constraints on r, s, t.