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- == Problem == ...{4} \le 250</math>, so we have <math>n+1 = 4, 4 \cdot 3^2, \ldots, 4 \cdot 13^2, 4\cdot 3^2 \cdot 5^2</math>.10 KB (1,702 words) - 22:23, 25 July 2024
- == Problem == ...olutions so <math>P(x)</math> is at least quadratic. Let us first try this problem out as if <math>P(x)</math> is a quadratic polynomial. Thus <math>P(n)-(n+34 KB (642 words) - 02:14, 1 June 2024
- == Problem == ...assign to <math>(0,1,0)</math>. (We will see how this correlates with the problem.) Then define for each lattice point <math>(i,j)</math> its triplet thus:5 KB (897 words) - 00:21, 29 July 2022
- == Problem == {{AIME box|year=2004|n=I|num-b=12|num-a=14}}2 KB (298 words) - 20:02, 4 July 2013
- == Problem == {{AIME box|year=2004|n=II|num-b=12|num-a=14}}3 KB (486 words) - 22:15, 7 April 2023
- == Problem == This solution is not a proof and since the AIME does not require proof, it is okay to finsh the answer like this. If you wo6 KB (1,083 words) - 21:10, 10 January 2025
- == Problem == Find the value of <math>10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).</math>3 KB (473 words) - 12:06, 18 December 2018
- == Problem == {{AIME box|year=1985|num-b=12|num-a=14}}5 KB (795 words) - 22:27, 19 December 2024
- == Problem == {{AIME box|year=1986|num-b=12|num-a=14}}4 KB (772 words) - 21:09, 7 May 2024
- == Problem == Thus, our problem can be restated: What is the probability that in a sequence of 31 distinct8 KB (1,269 words) - 10:55, 26 June 2024
- == Problem == ...erefore, <math>c_{14} = - 2</math>. Undergoing a similar process, <math>c_{13} = 3</math>, <math>c_{12} = - 5</math>, <math>c_{11} = 8</math>, and we see10 KB (1,595 words) - 16:30, 24 August 2024
- == Problem == {{AIME box|year=1989|num-b=12|num-a=14}}2 KB (274 words) - 04:07, 17 December 2023
- == Problem == {{AIME box|year=1990|num-b=12|num-a=14}}5 KB (762 words) - 01:18, 10 February 2023
- == Problem == ...inds that <math>n=44</math> is the largest possible integer satisfying the problem conditions.7 KB (1,328 words) - 20:24, 5 February 2024
- == Problem == {{AIME box|year=1992|num-b=12|num-a=14}}4 KB (703 words) - 23:13, 30 August 2024
- == Problem == From the problem statement, <math>AB=3t</math>, and <math>CD=t</math>. Since <math>\Delta AB11 KB (1,811 words) - 20:23, 29 December 2024
- == Problem == ...e equation by <math>t^{10}</math>, <math>1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1</math>.3 KB (383 words) - 20:30, 16 June 2024
- == Problem == This is a pretty easy problem just to bash. Since the max number we can get is <math>7</math>, we just ne2 KB (287 words) - 01:25, 12 December 2019
- == Problem == Because the problem asks for a ratio, we can divide each side length by <math>\sqrt{3}</math> t3 KB (521 words) - 01:18, 25 February 2016
- ==Problem== {{AIME box|year=1997|num-b=12|num-a=14}}7 KB (1,225 words) - 19:56, 4 August 2021
Page text matches
- ...e many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theo <cmath>5-12-13</cmath>6 KB (978 words) - 12:02, 6 March 2025
- \\frac{x+5-13}{x+5}+4&\ge 3 \1-\frac{13}{x+5}+4&\ge 312 KB (1,806 words) - 06:07, 19 June 2024
- ...ed States at the [[International Mathematics Olympiad]] (IMO). While most AIME participants are high school students, some bright middle school students a High scoring AIME students are invited to take the prestigious [[United States of America Mat5 KB (669 words) - 17:19, 11 March 2025
- == Problems == ...largest area that this triangle can have? ([[1992 AIME Problems/Problem 13|AIME 1992]])3 KB (583 words) - 21:20, 2 August 2024
- ...nd make the equation factorable. It can be used to solve more than algebra problems, sometimes going into other topics such as number theory. == Fun Practice Problems ==4 KB (682 words) - 10:25, 18 February 2025
- ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====9 KB (1,703 words) - 01:20, 7 December 2024
- * <math>13! = 6227020800</math> ==Problems==10 KB (809 words) - 16:40, 17 March 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME II Problems]]1 KB (133 words) - 12:32, 22 March 2011
- draw(graph(f,(7+2*sqrt(13))/3,(7-2*sqrt(13))/3),red+linewidth(1)); ==Problems==3 KB (551 words) - 16:22, 13 September 2023
- == Problems == * [[1991 AIME Problems/Problem 12]]1 KB (179 words) - 19:41, 3 January 2025
- This is a problem where constructive counting is not the simplest way to proceed. This next e ...proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution.13 KB (2,018 words) - 15:31, 10 January 2025
- * [[2020 AMC 10A Problems/Problem 24]] * [[1985 AIME Problems/Problem 13]]6 KB (923 words) - 17:39, 30 September 2024
- ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is8 KB (1,192 words) - 17:20, 16 June 2023
- ...t a factor of <math>151</math>, so we can move on. In fact, because <math>13 > \sqrt{151}</math>, we can declare <math>151</math> as prime and stop. ==Problems==3 KB (496 words) - 22:14, 5 January 2024
- == Example Problems == * [[2003_AIME_I_Problems/Problem_13 | 2003 AIME I Problem 13]]4 KB (547 words) - 17:23, 30 December 2020
- ...of the two preceding it. The first few terms are <math>1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...</math>. ==Problems==7 KB (1,111 words) - 14:57, 24 June 2024
- == Problems == *[[2007 AMC 12A Problems/Problem 18]]5 KB (860 words) - 15:36, 10 December 2023
- ==Problems== ([[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Source]])4 KB (658 words) - 21:52, 4 February 2025
- == Practice Problem == ...ath>\omega_{C}</math>. If the sides of triangle <math>ABC</math> are <math>13,</math> <math>14,</math> and <math>15,</math> the radius of <math>\omega</m3 KB (533 words) - 13:51, 2 September 2024
- ...s for any given Diophantine equations. This is known as [[Hilbert's tenth problem]]. The answer, however, is no. == Problems ==9 KB (1,434 words) - 01:15, 4 July 2024
