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- == Problem == ...>3</math> digits. Checking the other cases, we see that it must have <math>3</math> digits.4 KB (622 words) - 22:47, 13 October 2024
- #REDIRECT [[2006 AIME I Problems/Problem 3]]44 bytes (5 words) - 12:21, 14 June 2009
- == Problem == {{AIME box|year=2005|n=II|num-b=2|num-a=4}}3 KB (581 words) - 21:19, 22 September 2024
- == Problem == ...so <math>n</math> must be in the form <math>n=p\cdot q</math> or <math>n=p^3</math> for distinct [[prime number]]s <math>p</math> and <math>q</math>.2 KB (249 words) - 09:37, 23 January 2024
- == Problem == {{AIME box|year=2004|n=I|num-b=2|num-a=4}}1 KB (156 words) - 17:56, 1 January 2016
- == Problem == ...s close together as possible, which occurs when the smaller block is <math>3 \times 7 \times 11</math>. Then the extra layer makes the entire block <ma2 KB (377 words) - 11:53, 10 March 2014
- == Problem == ...rt{x^2+18x+45}</math>, we have <math>n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0</math>. Because the square root of a real number can't be negative, th3 KB (532 words) - 13:05, 29 November 2024
- == Problem == ...smaller [[triangle]]s <math>t_{1}</math>, <math>t_{2}</math>, and <math>t_{3}</math> in the figure, have [[area]]s <math>4</math>, <math>9</math>, and <4 KB (726 words) - 13:39, 13 August 2023
- == Problem == ...nd <math>c</math> are [[positive integer]]s which satisfy <math>c=(a + bi)^3 - 107i</math>, where <math>i^2 = -1</math>.1 KB (205 words) - 18:58, 10 March 2024
- == Problem == == Solution 3 (less trig required, use of quadratic formula) ==3 KB (545 words) - 23:44, 12 October 2023
- == Problem == #:If we let <math>n=p^3</math> with <math>p</math> prime, then its proper divisors are <math>p</mat3 KB (511 words) - 09:29, 9 January 2023
- == Problem == 2^{3 \log_2(\log_8x)} &= \log_2x\3 KB (481 words) - 21:52, 18 November 2020
- == Problem == == Solution 3 ==4 KB (584 words) - 14:38, 11 August 2024
- == Problem == ...th>P_2^{}</math> be a regular <math>s~\mbox{gon}</math> <math>(r\geq s\geq 3)</math> such that each [[interior angle]] of <math>P_1^{}</math> is <math>\3 KB (516 words) - 19:18, 16 April 2024
- == Problem == ===Solution 3===5 KB (865 words) - 12:13, 21 May 2020
- == Problem == ...ches won, so that <math>\frac{n}{2n}=\frac{1}{2}</math>, and <math>\frac{n+3}{2n+4}>\frac{503}{1000}</math>.2 KB (251 words) - 08:05, 2 January 2024
- == Problem == <center><math>\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \2 KB (364 words) - 00:05, 9 July 2022
- == Problem == ...some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triang2 KB (252 words) - 11:12, 3 July 2023
- == Problem == ...tt>. The sequences <tt>N,N,N,E,E,S</tt> can be permuted in <math>\frac{6!}{3!2!1!} = 60</math> ways. However, if the first four steps of the sequence ar3 KB (602 words) - 23:15, 16 June 2019
- == Problem == ...Favorite Factoring Trick]], we rewrite as <math>[(x+7)(y-3)]^n = (x+7)^n(y-3)^n</math>. Both [[binomial expansion]]s will contain <math>n+1</math> non-l3 KB (515 words) - 04:29, 27 November 2023
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- ...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 B = (4, 3);6 KB (978 words) - 12:02, 6 March 2025
- ...idual articles often have sample problems and solutions for many levels of problem solvers. Many also have links to books, websites, and other resources rele * CompetifyHub's Problem Sets [//competifyhub.com/resources/ Free Competition Resources for Grades 114 KB (1,913 words) - 23:52, 6 March 2025
- ...n particular, notice that although <math>3 > 2</math>, we must have <math>-3 < -2</math>. In particular, when multiplying or dividing by negative quant ..., in the example <math>x \ge \frac{3}{2}</math>, the value <math>x = \frac{3}{2}</math> satisfies the inequality because the inequality is nonstrict.12 KB (1,806 words) - 06:07, 19 June 2024
- ...are invited to take the [[American Invitational Mathematics Examination]] (AIME). ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC.4 KB (596 words) - 04:53, 3 February 2025
- ...ke the more challenging [[American Invitational Mathematics Examination]] (AIME). ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC.5 KB (646 words) - 04:52, 3 February 2025
- ...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
- ...o participate in the [[Math Olympiad Summer Program]] (MOP), a challenging 3 to 4 week math program for the brightest students in the country. ...dministered by the [[Mathematical Association of America]] (MAA). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the MAA.6 KB (836 words) - 04:57, 3 February 2025
- ...e Spring Semester to determine the team each year. The 6 practices include 3 individual tests to help determine the team and some lectures on certain ma * Performance on problems at practice sessions.22 KB (3,532 words) - 11:25, 27 September 2024
- == Problems == ...largest area that this triangle can have? ([[1992 AIME Problems/Problem 13|AIME 1992]])3 KB (583 words) - 21:20, 2 August 2024
- ...math> was added to this, then we would have a [[perfect square]], <math>(x-3)^2=x^2-6x+9</math>. To do this, add <math>7</math> to each side of the equ <math>x^2-6x+9=7\Rightarrow(x-3)^2=7\Rightarrow x-3=\pm{\sqrt7}\Rightarrow x=3\pm\sqrt{7}</math>2 KB (422 words) - 16:20, 5 March 2023
- ...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
- ...as a great introductory video to combinations, permutations, and counting problems in general! [https://bit.ly/CombinationsAndPermutations Permutations & Comb Consider the set of letters A, B, and C. There are <math>3!</math> different [[permutations]] of those letters. Since order doesn't ma4 KB (638 words) - 21:55, 5 January 2025
- * <math>3! = 6</math> ...2</math> give another power of <math>p</math>. Those divisible by <math>p^3</math> give yet another power of <math>p</math>. Continuing in this manner10 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
- In contest problems, Fermat's Little Theorem is often used in conjunction with the [[Chinese Re Let <math>S = \{1,2,3,\cdots, p-1\}</math>. Then, we claim that the set <math>a \cdot S</math>, c15 KB (2,618 words) - 12:03, 19 February 2025
- 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
- ...<math>\phi{(n)}</math> is the number of integers in the range <math>\{1,2,3\cdots{,n}\}</math> which are relatively prime to <math>n</math>. If <math>{ ==Problems==4 KB (569 words) - 22:34, 30 December 2024
- == Problems == ...e integers. Determine <math>p + q</math>. ([[Mock AIME 3 Pre 2005 Problems/Problem 7|Source]])3 KB (543 words) - 19:35, 29 October 2024
- ...uality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures. == Problems ==6 KB (922 words) - 17:34, 13 January 2025
