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• 2019 AIME II Problems/Problem 10
  ==Problem== ...nds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.
  8 KB (1,172 words) - 18:21, 8 August 2024
• 2019 AIME I Problems/Problem 10
  ==Problem== ...-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 </cmath>can be expressed as <math>x^{2019} + 20x^{2018} + 19x^{2017}+g(x)</math>, where <math>g(x)</math> is a polyno
  10 KB (1,691 words) - 22:25, 31 January 2025

Page text matches

• Mock AMC
  ...ake. Sometimes, the administrator may ask other people to sign up to write problems for the contest. * Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.
  51 KB (6,175 words) - 21:41, 27 November 2024
• 2019 AIME I Problems/Problem 2
  ==Problem== ...ath> ones that form <math>P</math>, so <math>P = \frac{38}{380} = \frac{1}{10}</math>. Therefore the answer is <math>\frac{9}{20} \rightarrow \boxed{029}
  5 KB (831 words) - 18:47, 29 January 2025
• Mock MathCounts
  ...istered to the AoPS community, while others may be sourced from a group of problem writers. Different users may have a different way of participating; some ma ...ms from one subject. Having a group is also good so they can discuss which problems are good or need improvement, and fix errors. More than one person working
  26 KB (3,260 words) - 19:28, 15 August 2024
• Chicken McNugget Theorem
  ...n McNugget Theorem''' (or '''Postage Stamp Problem''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s <mat ...t Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest
  17 KB (2,823 words) - 23:06, 15 November 2024
• Ball-and-urn
  It is used to solve problems of the form: how many ways can one distribute <math>k</math> indistinguisha ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then
  5 KB (795 words) - 17:39, 31 December 2024
• Gmaas
  ...into anything. Using that fact, you can use the Games theorem to solve any problem. 10. Gmaas is superior to you. If you see Gmaas or a picture of the Great Gmaas
  69 KB (11,805 words) - 20:49, 18 December 2019
• 2018 AIME II
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2018 AIME II Problems|Entire Test]]
  1 KB (133 words) - 18:13, 18 March 2020
• 2019 AIME I
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2019 AIME I Problems|Entire Test]]
  1 KB (133 words) - 17:41, 29 March 2019
• 2019 AIME II
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2019 AIME II Problems|Entire Test]]
  1 KB (133 words) - 15:43, 22 March 2019
• 2020 AIME I
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2020 AIME I Problems|Entire Test]]
  1 KB (133 words) - 18:11, 18 March 2020
• 2018 AIME II Problems
  {{AIME Problems|year=2018|n=II}} ==Problem 1==
  9 KB (1,385 words) - 00:26, 21 January 2024
• 2019 AMC 10C Problems
  Here are the problems from the 2019 AMC 10C, a mock contest created by the AoPS user fidgetboss_4000. ==Problem 1==
  12 KB (1,917 words) - 12:14, 29 November 2021
• 2019 AIME II Problems/Problem 15
  ==Problem== ...n two distinct points, <math>X</math> and <math>Y</math>. Suppose <math>XP=10</math>, <math>PQ=25</math>, and <math>QY=15</math>. The value of <math>AB\c
  7 KB (1,129 words) - 16:27, 6 January 2025
• 2019 AIME II Problems/Problem 9
  ==Problem== ...pretty. Let <math>S</math> be the sum of positive integers less than <math>2019</math> that are <math>20</math>-pretty. Find <math>\tfrac{S}{20}</math>.
  3 KB (474 words) - 01:38, 22 December 2024
• 2019 AIME II Problems/Problem 10
  ==Problem== ...nds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.
  8 KB (1,172 words) - 18:21, 8 August 2024
• 2019 AIME II Problems/Problem 11
  ==Problem== ...\left(\frac{147}{16\sqrt{5}}\right)\cdot \frac{189}{16\sqrt{5}}\cdot \frac{10}{21}} =\sqrt{\left(\frac{168}{8\sqrt{5}}\right)^2 - \left(\frac{7 \cdot 189
  14 KB (2,229 words) - 14:57, 27 December 2024
• 2019 AIME I Problems
  {{AIME Problems|year=2019|n=I}} ==Problem 1==
  8 KB (1,331 words) - 06:57, 4 January 2021
• 2019 AIME I Problems/Problem 1
  ==Problem== ...3+\cdots+10^{321})-321</math>. We know the former will yield <math>1111....10</math>, so we only have to figure out what the last few digits are. There a
  3 KB (433 words) - 07:57, 9 February 2023
• 2019 AIME II Problems
  {{AIME Problems|year=2019|n=II}} ==Problem 1==
  7 KB (1,254 words) - 14:45, 21 August 2023
• 2019 AIME I Problems/Problem 3
  ==Problem== dot((10,0));
  6 KB (933 words) - 19:30, 29 January 2025

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