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| ==Problem 1== | | ==Problem 1== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | In <math>\triangle ABC</math> with <math>AB=AC,</math> point <math>D</math> lies strictly between <math>A</math> and <math>C</math> on side <math>\overline{AC},</math> and point <math>E</math> lies strictly between <math>A</math> and <math>B</math> on side <math>\overline{AB}</math> such that <math>AE=ED=DB=BC.</math> The degree measure of <math>\angle ABC</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> |
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| [[2020 AIME I Problems/Problem 1 | Solution]] | | [[2020 AIME I Problems/Problem 1 | Solution]] |
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| ==Problem 2== | | ==Problem 2== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | There is a unique positive real number <math>x</math> such that the three numbers <math>\log_8(2x),\log_4x,</math> and <math>\log_2x,</math> in that order, form a geometric progression with positive common ratio. The number <math>x</math> can be written as <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> |
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| [[2020 AIME I Problems/Problem 2 | Solution]] | | [[2020 AIME I Problems/Problem 2 | Solution]] |
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| ==Problem 3== | | ==Problem 3== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | A positive integer <math>N</math> has base-eleven representation <math>\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}</math> and base-eight representation <math>\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},</math> where <math>a,b,</math> and <math>c</math> represent (not necessarily distinct) digits. Find the least such <math>N</math> expressed in base ten. |
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| [[2020 AIME I Problems/Problem 3 | Solution]] | | [[2020 AIME I Problems/Problem 3 | Solution]] |
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| ==Problem 4== | | ==Problem 4== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | Let <math>S</math> be the set of positive integers <math>N</math> with the property that the last four digits of <math>N</math> are <math>2020,</math> and when the last four digits are removed, the result is a divisor of <math>N.</math> For example, <math>42{,}020</math> is in <math>S</math> because <math>4</math> is a divisor of <math>42{,}020.</math> Find the sum of all the digits of all the numbers in <math>S.</math> For example, the number <math>42{,}020</math> contributes <math>4+2+0+2+0=8</math> to this total. |
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| [[2020 AIME I Problems/Problem 4 | Solution]] | | [[2020 AIME I Problems/Problem 4 | Solution]] |
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| ==Problem 5== | | ==Problem 5== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | Six cards numbered <math>1</math> through <math>6</math> are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order. |
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| [[2020 AIME I Problems/Problem 5 | Solution]] | | [[2020 AIME I Problems/Problem 5 | Solution]] |
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| ==Problem 6== | | ==Problem 6== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | A flat board has a circular hole with radius <math>1</math> and a circular hole with radius <math>2</math> such that the distance between the centers of the two holes is <math>7</math>. Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
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| [[2020 AIME I Problems/Problem 6 | Solution]] | | [[2020 AIME I Problems/Problem 6 | Solution]] |
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| ==Problem 7== | | ==Problem 7== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | A club consisting of <math>11</math> men and <math>12</math> women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as <math>1</math> member or as many as <math>23</math> members. Let <math>N</math> be the number of such committees that can be formed. Find the sum of the prime numbers that divide <math>N.</math> |
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| [[2020 AIME I Problems/Problem 7 | Solution]] | | [[2020 AIME I Problems/Problem 7 | Solution]] |
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| ==Problem 8== | | ==Problem 8== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | A bug walks all day and sleeps all night. On the first day, it starts at point <math>O,</math> faces east, and walks a distance of <math>5</math> units due east. Each night the bug rotates <math>60^\circ</math> counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the point <math>P.</math> Then <math>OP^2=\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math> |
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| [[2020 AIME I Problems/Problem 8 | Solution]] | | [[2020 AIME I Problems/Problem 8 | Solution]] |
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| ==Problem 9== | | ==Problem 9== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | Let <math>S</math> be the set of positive integer divisors of <math>20^9.</math> Three numbers are chosen independently and at random with replacement from the set <math>S</math> and labeled <math>a_1,a_2,</math> and <math>a_3</math> in the order they are chosen. The probability that both <math>a_1</math> divides <math>a_2</math> and <math>a_2</math> divides <math>a_3</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m.</math> |
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| [[2020 AIME I Problems/Problem 9 | Solution]] | | [[2020 AIME I Problems/Problem 9 | Solution]] |
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| ==Problem 10== | | ==Problem 10== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | Let <math>m</math> and <math>n</math> be positive integers satisfying the conditions |
| + | |
| + | <math>\quad\bullet\ \gcd(m+n,210)=1,</math> |
| + | |
| + | <math>\quad\bullet\ m^m</math> is a multiple of <math>n^n,</math> and |
| + | |
| + | <math>\quad\bullet\ m</math> is not a multiple of <math>n.</math> |
| + | |
| + | Find the least possible value of <math>m+n.</math> |
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| [[2020 AIME I Problems/Problem 10 | Solution]] | | [[2020 AIME I Problems/Problem 10 | Solution]] |
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| ==Problem 11== | | ==Problem 11== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | For integers <math>a,b,c</math> and <math>d,</math> let <math>f(x)=x^2+ax+b</math> and <math>g(x)=x^2+cx+d.</math> Find the number of ordered triples <math>(a,b,c)</math> of integers with absolute values not exceeding <math>10</math> for which there is an integer <math>d</math> such that <math>g(f(2))=g(f(4))=0.</math> |
| + | |
| [[2020 AIME I Problems/Problem 11 | Solution]] | | [[2020 AIME I Problems/Problem 11 | Solution]] |
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| ==Problem 12== | | ==Problem 12== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | Let <math>n</math> be the least positive integer for which <math>149^n-2^n</math> is divisible by <math>3^3\cdot5^5\cdot7^7.</math> Find the number of positive integer divisors of <math>n.</math> |
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| [[2020 AIME I Problems/Problem 12 | Solution]] | | [[2020 AIME I Problems/Problem 12 | Solution]] |
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| ==Problem 13== | | ==Problem 13== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | Point <math>D</math> lies on side <math>\overline{BC}</math> of <math>\triangle ABC</math> so that <math>\overline{AD}</math> bisects <math>\angle BAC.</math> The perpendicular bisector of <math>\overline{AD}</math> intersects the bisectors of <math>\angle ABC</math> and <math>\angle ACB</math> in points <math>E</math> and <math>F,</math> respectively. Given that <math>AB=4,BC=5,</math> and <math>CA=6,</math> the area of <math>\triangle AEF</math> can be written as <math>\tfrac{m\sqrt{n}}p,</math> where <math>m</math> and <math>p</math> are relatively prime positive integers, and <math>n</math> is a positive integer not divisible by the square of any prime. Find <math>m+n+p.</math> |
| + | |
| [[2020 AIME I Problems/Problem 13 | Solution]] | | [[2020 AIME I Problems/Problem 13 | Solution]] |
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| ==Problem 14== | | ==Problem 14== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | Let <math>P(x)</math> be a quadratic polynomial with complex coefficients whose <math>x^2</math> coefficient is <math>1.</math> Suppose the equation <math>P(P(x))=0</math> has four distinct solutions, <math>x=3,4,a,b.</math> Find the sum of all possible values of <math>(a+b)^2.</math> |
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| [[2020 AIME I Problems/Problem 14 | Solution]] | | [[2020 AIME I Problems/Problem 14 | Solution]] |
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| ==Problem 15== | | ==Problem 15== |
− | Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule! Pakistowners rule!
| + | Let <math>\triangle ABC</math> be an acute triangle with circumcircle <math>\omega,</math> and let <math>H</math> be the intersection of the altitudes of <math>\triangle ABC.</math> Suppose the tangent to the circumcircle of <math>\triangle HBC</math> at <math>H</math> intersects <math>\omega</math> at points <math>X</math> and <math>Y</math> with <math>HA=3,HX=2,</math> and <math>HY=6.</math> The area of <math>\triangle ABC</math> can be written in the form <math>m\sqrt{n},</math> where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n.</math> |
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2020 AIME I (Answer Key) Printable version | AoPS Contest Collections • PDF
|
Instructions
- This is a 15-question, 3-hour examination. All answers are integers ranging from
to , inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
- No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
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1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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Problem 1
In
with
point
lies strictly between
and
on side
and point
lies strictly between
and
on side
such that
The degree measure of
is
where
and
are relatively prime positive integers. Find
Solution
Problem 2
There is a unique positive real number
such that the three numbers
and
in that order, form a geometric progression with positive common ratio. The number
can be written as
where
and
are relatively prime positive integers. Find
Solution
Problem 3
A positive integer
has base-eleven representation
and base-eight representation
where
and
represent (not necessarily distinct) digits. Find the least such
expressed in base ten.
Solution
Problem 4
Let
be the set of positive integers
with the property that the last four digits of
are
and when the last four digits are removed, the result is a divisor of
For example,
is in
because
is a divisor of
Find the sum of all the digits of all the numbers in
For example, the number
contributes
to this total.
Solution
Problem 5
Six cards numbered
through
are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.
Solution
Problem 6
A flat board has a circular hole with radius
and a circular hole with radius
such that the distance between the centers of the two holes is
. Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is
, where
and
are relatively prime positive integers. Find
.
Solution
Problem 7
A club consisting of
men and
women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as
member or as many as
members. Let
be the number of such committees that can be formed. Find the sum of the prime numbers that divide
Solution
Problem 8
A bug walks all day and sleeps all night. On the first day, it starts at point
faces east, and walks a distance of
units due east. Each night the bug rotates
counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the point
Then
where
and
are relatively prime positive integers. Find
Solution
Problem 9
Let
be the set of positive integer divisors of
Three numbers are chosen independently and at random with replacement from the set
and labeled
and
in the order they are chosen. The probability that both
divides
and
divides
is
where
and
are relatively prime positive integers. Find
Solution
Problem 10
Let
and
be positive integers satisfying the conditions
is a multiple of
and
is not a multiple of
Find the least possible value of
Solution
Problem 11
For integers
and
let
and
Find the number of ordered triples
of integers with absolute values not exceeding
for which there is an integer
such that
Solution
Problem 12
Let
be the least positive integer for which
is divisible by
Find the number of positive integer divisors of
Solution
Problem 13
Point
lies on side
of
so that
bisects
The perpendicular bisector of
intersects the bisectors of
and
in points
and
respectively. Given that
and
the area of
can be written as
where
and
are relatively prime positive integers, and
is a positive integer not divisible by the square of any prime. Find
Solution
Problem 14
Let
be a quadratic polynomial with complex coefficients whose
coefficient is
Suppose the equation
has four distinct solutions,
Find the sum of all possible values of
Solution
Problem 15
Let
be an acute triangle with circumcircle
and let
be the intersection of the altitudes of
Suppose the tangent to the circumcircle of
at
intersects
at points
and
with
and
The area of
can be written in the form
where
and
are positive integers, and
is not divisible by the square of any prime. Find
Solution
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.