MathDB

2017 AIME Problems

Part of AIME Problems

Subcontests

(15)
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Median-ception

Consider arrangements of the 99 numbers 1,2,3,…,91, 2, 3, \dots, 9 in a 3×33 \times 3 array. For each such arrangement, let a1a_1, a2a_2, and a3a_3 be the medians of the numbers in rows 11, 22, and 33 respectively, and let mm be the median of {a1,a2,a3}\{a_1, a_2, a_3\}. Let QQ be the number of arrangements for which m=5m = 5. Find the remainder when QQ is divided by 10001000.

The Importance of Being Earnestrongly connected graph

Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way).
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Wot n' Subset Formation

Call a set SS product-free if there do not exist a,b,c∈Sa, b, c \in S (not necessarily distinct) such that ab=ca b = c. For example, the empty set and the set {16,20}\{16, 20\} are product-free, whereas the sets {4,16}\{4, 16\} and {2,8,16}\{2, 8, 16\} are not product-free. Find the number of product-free subsets of the set {1,2,3,4,5,6,7,8,9,10}\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}.

Anna Kareninested internally tangent circles

Circle C0C_0 has radius 11, and the point A0A_0 is a point on the circle. Circle C1C_1 has radius r<1r<1 and is internally tangent to C0C_0 at point A0A_0. Point A1A_1 lies on circle C1C_1 so that A1A_1 is located 90∘90^{\circ} counterclockwise from A0A_0 on C1C_1. Circle C2C_2 has radius r2r^2 and is internally tangent to C1C_1 at point A1A_1. In this way a sequence of circles C1,C2,C3,...C_1,C_2,C_3,... and a sequence of points on the circles A1,A2,A3,...A_1,A_2,A_3,... are constructed, where circle CnC_n has radius rnr^n and is internally tangent to circle Cn−1C_{n-1} at point An−1A_{n-1}, and point AnA_n lies on CnC_n 90∘90^{\circ} counterclockwise from point An−1A_{n-1}, as shown in the figure below. There is one point BB inside all of these circles. When r=1160r=\frac{11}{60}, the distance from the center of C0C_0 to BB is mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm+n.
[asy] size(6cm); real r = 0.8;
pair nthCircCent(int n){ pair ans = (0, 0); for(int i = 1; i <= n; ++i) ans += rotate(90 * i - 90) * (r^(i - 1) - r^i, 0); return ans; }
void dNthCirc(int n){ draw(circle(nthCircCent(n), r^n)); }
dNthCirc(0); dNthCirc(1); dNthCirc(2); dNthCirc(3);
dot("A0A_0", (1, 0), dir(0)); dot("A1A_1", nthCircCent(1) + (0, r), dir(135)); dot("A2A_2", nthCircCent(2) + (-r^2, 0), dir(0)); [/asy]
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Wot n' Imagination

Let z1=18+83iz_1 = 18 + 83i, z2=18+39i,z_2 = 18 + 39i, and z3=78+99i,z_3 = 78 + 99i, where i=−1i = \sqrt{-1}. Let zz be the unique complex number with the properties that z3−z1z2−z1⋅z−z2z−z3\frac{z_3 - z_1}{z_2 - z_1} \cdot \frac{z - z_2}{z - z_3} is a real number and the imaginary part of zz is the greatest possible. Find the real part of zz.

Jane Eyrectangle

Rectangle ABCDABCD has side lengths AB=84AB=84 and AD=42AD=42. Point MM is the midpoint of AD‾\overline{AD}, point NN is the trisection point of AB‾\overline{AB} closer to AA, and point OO is the intersection of CM‾\overline{CM} and DN‾\overline{DN}. Point PP lies on the quadrilateral BCONBCON, and BP‾\overline{BP} bisects the area of BCONBCON. Find the area of △CDP\triangle{CDP}.
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I got 99 problems and this is one.

Let a10=10a_{10} = 10, and for each integer n>10n >10 let an=100an−1+na_n = 100a_{n - 1} + n. Find the least n>10n > 10 such that ana_n is a multiple of 9999.

Moby Deck of Cards

A special deck of cards contains 4949 cards, each labeled with a number from 11 to 77 and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and still have at least one card of each color and at least one card with each number is pq\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+qp+q.
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American Remainder Theorem

When each of 702, 787, and 855 is divided by the positive integer mm, the remainder is always the positive integer rr. When each of 412, 722, and 815 is divided by the positive integer nn, the remainder is always the positive integer s≠rs \neq r. Fine m+n+r+sm+n+r+s.

Pride and Prejudisemifinals

Teams T1T_1, T2T_2, T3T_3, and T4T_4 are in the playoffs. In the semifinal matches, T1T_1 plays T4T_4 and T2T_2 plays T3T_3. The winners of those two matches will play each other in the final match to determine the champion. When TiT_i plays TjT_j, the probability that TiT_i wins is ii+j\frac{i}{i+j}, and the outcomes of all the matches are independent. The probability that T4T_4 will be the champion is pq\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+qp+q.
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Wot n' Geometric Combination

Two real numbers aa and bb are chosen independently and uniformly at random from the interval (0,75)(0, 75). Let OO and PP be two points on the plane with OP=200OP = 200. Let QQ and RR be on the same side of line OPOP such that the degree measures of ∠POQ\angle POQ and ∠POR\angle POR are aa and bb respectively, and ∠OQP\angle OQP and ∠ORP\angle ORP are both right angles. The probability that QR≤100QR \leq 100 is equal to mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm + n.

Eugene Oneginteger outputs of polynomial

Find the number of positive integers nn less than 20172017 such that 1+n+n22!+n33!+n44!+n55!+n66! 1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!} is an integer.
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Wot n' Triangle Minimization

The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths 232\sqrt3, 55, and 37\sqrt{37}, as shown, is mpn\tfrac{m\sqrt{p}}{n}, where mm, nn, and pp are positive integers, mm and nn are relatively prime, and pp is not divisible by the square of any prime. Find m+n+pm+n+p. [asy] size(5cm); pair C=(0,0),B=(0,2*sqrt(3)),A=(5,0); real t = .385, s = 3.5*t-1; pair R = A*t+B*(1-t), P=B*s; pair Q = dir(-60) * (R-P) + P; fill(P--Q--R--cycle,gray); draw(A--B--C--A^^P--Q--R--P); dot(A--B--C--P--Q--R); [/asy]

Don Quixotetrahedron

Tetrahedron ABCDABCD has AD=BC=28AD=BC=28, AC=BD=44AC=BD=44, and AB=CD=52AB=CD=52. For any point XX in space, define f(X)=AX+BX+CX+DXf(X)=AX+BX+CX+DX. The least possible value of f(X)f(X) can be expressed as mnm\sqrt{n}, where mm and nn are positive integers, and nn is not divisible by the square of any prime. Find m+nm+n.
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