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• Math books
  ...combinatorics, and number theory, along with sets of accompanying practice problems at the end of every section. ...88606&sprefix=after+school+maths+%2Caps%2C268&sr=8-2 100 Challenging Maths Problems]
  24 KB (3,177 words) - 12:53, 20 February 2024
• Logic books
  These '''Logic books''' are recommended by [[Art of Problem Solving]] administrators and m === Digital Logic===
  1 KB (151 words) - 14:14, 19 December 2008
• AMC 8
  ...ltiple Choice|difficulty=1 - 1.5|breakdown=<u>Problems 1 - 12</u>: 1<br><u>Problems 13 - 25</u>: 1.5}} The AMC 8 exam is a 25 problem exam. There are 40 minutes given in the exam. Problems increase in difficulty as the problem number increases.
  4 KB (558 words) - 22:25, 28 April 2024
• Principle of Inclusion-Exclusion
  ...math>|A\cap B|</math>, that's just putting four guys in order. By the same logic as above, this is <math>2!\binom{6}{4}=30</math>. Again, <math>|A\cap C|</m [[2011 AMC 8 Problems/Problem 6]]
  9 KB (1,703 words) - 07:25, 24 March 2024
• Constructive counting
  The exact same logic applies to the third digit; it can be any of the <math>8</math> digits exce ''[[2001 AMC 12 Problems/Problem 16 | 2001 AMC 12 Problem 16]]: A spider has one sock and one shoe f
  12 KB (1,896 words) - 23:55, 27 December 2023
• Casework
  ...emonstrate casework in action. Unlike the selections in this article, most problems cannot be completely solved through casework. However, it is crucial as an While there are problems where casework produces the most elegant solution, in those where a shorter
  5 KB (709 words) - 10:28, 19 February 2024
• Complementary counting
  ...don't want, then subtracts that from the total number of possibilities. In problems that involve complex or tedious [[casework]], complementary counting is oft ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive intege
  8 KB (1,192 words) - 17:20, 16 June 2023
• Mathematical Kangaroo
  ...e with 5 possible answer choices. The remaining levels have tests with 30 problems, each multiple choice with 5 possible answer choices. ...thin the second one-third of the test is worth 4 points, and the remaining problems are worth 5 points. No penalty or partial credit is given to unanswered or
  6 KB (936 words) - 15:38, 22 February 2024
• 2006 AIME I Problems/Problem 13
  We now see a pattern! Using the exact same logic, we can condense this whole messy thing into a closed form: [[Category:Intermediate Number Theory Problems]]
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AMC 12A Problems/Problem 17
  By logic <math>s=3</math> and <math>sr=5 \implies r=5/3.</math> [[Category:Introductory Geometry Problems]]
  6 KB (958 words) - 23:29, 28 September 2023
• Set
  ==Problems== ([[2005 AMC 12A Problems/Problem 13|Source]])
  11 KB (2,021 words) - 00:00, 17 July 2011
• 1985 AIME Problems/Problem 14
  ...{4} = 10(p-10)</math>. This equation is the same as above, and by the same logic, the answer is <math>n=\boxed{25}</math>. * [[AIME Problems and Solutions]]
  5 KB (772 words) - 22:14, 18 June 2020
• 1985 AIME Problems/Problem 12
  ~Azjps (Fundamental Logic) Here is a similar problem from another AIME test: [[2003 AIME II Problems/Problem 13|2003 AIME II Problem 13]], in which we have an equilateral trian
  17 KB (2,837 words) - 13:34, 4 April 2024
• 1986 AIME Problems/Problem 13
  ...additional T's among the first <math>4</math> spaces. We can use identical logic to show why we can divide the two H's among the last four spaces to get exa * [[AIME Problems and Solutions]]
  4 KB (772 words) - 21:09, 7 May 2024
• 1986 AIME Problems/Problem 9
  ...AD + BD' = 425 - DD'</math>. Thus <math>DD' = 425 - d</math>. By the same logic, <math>EE' = 450 - d</math>. [[Category:Intermediate Geometry Problems]]
  11 KB (1,850 words) - 18:07, 11 October 2023
• 1986 AIME Problems/Problem 7
  ...h term is <math>729</math> plus the <math>36</math>th term. Using the same logic, the <math>36</math>th term is <math>243</math> plus the <math>4</math>th t * [[AIME Problems and Solutions]]
  5 KB (866 words) - 00:00, 22 December 2022
• 1988 AIME Problems/Problem 13
  ...x^2-x-1)</math> into <math>1</math>. We can similarly continue to use this logic, by repeatedly cancelling out the middle term, and obtain the process: [[Category:Intermediate Algebra Problems]]
  10 KB (1,585 words) - 03:58, 1 May 2023
• 1989 AIME Problems/Problem 10
  By identical logic, we can find similar expressions for the sums of the other two cotangents: Identical logic works for the other two angles in the triangle. So, the cotangent of any a
  8 KB (1,401 words) - 21:41, 20 January 2024
• 1989 AIME Problems/Problem 1
  ~Novus677 (Fundamental Logic) ~Vrjmath (Fundamental Logic)
  4 KB (523 words) - 00:12, 8 October 2021
• 1991 AIME Problems/Problem 12
  By similar logic, we have <math>APOS</math> is a cyclic quadrilateral. Let <math>AP = x</mat [[Category:Intermediate Geometry Problems]]
  8 KB (1,270 words) - 23:36, 27 August 2023
• 1992 AIME Problems/Problem 15
  With this logic, we realize that the desired quantity is simply <math>\left \lfloor \frac{7 [[Category:Intermediate Number Theory Problems]]
  2 KB (358 words) - 01:54, 2 October 2020
• 1994 AIME Problems/Problem 10
  ...dot AD}</math> and that <math>BC = \sqrt{AB \cdot BD}</math>. Applying the logic with the established values of k, we get <math>AC = 29k \cdot \sqrt{29^2 + [[Category:Intermediate Geometry Problems]]
  3 KB (534 words) - 16:23, 26 August 2018
• 1995 AIME Problems/Problem 15
  Applying similar logic, we get the equations: [[Category:Intermediate Combinatorics Problems]]
  6 KB (979 words) - 13:20, 11 April 2022
• 2001 AIME I Problems/Problem 15
  ..., we are once again left with <math>2</math> choices to move to by similar logic to the first case. However, after we move back to the red tetrahedron, our [[Category:Intermediate Combinatorics Problems]]
  11 KB (1,837 words) - 18:53, 22 January 2024
• 2001 AIME I Problems/Problem 14
  Now using the same logic as above we can find <math>M_n = M_{n-2} + M_{n-3}</math> ( the cases are 0 ...an replace the number nineteen with any number without a big effect on the logic that we use to solve the problem. This pits the problem as a likely candida
  13 KB (2,298 words) - 19:46, 9 July 2020
• 2001 AIME I Problems/Problem 6
  The logic is that, if <math>n</math> is rolled, then the number of valid cases is the [[Category:Intermediate Combinatorics Problems]]
  11 KB (1,729 words) - 20:50, 28 November 2023
• 2003 AIME II Problems/Problem 6
  ...that, both “half” triangles created also have the same area. The same logic can be applied to all other triangles). [[Category: Intermediate Geometry Problems]]
  5 KB (787 words) - 17:38, 30 July 2022
• 2000 AIME II Problems/Problem 15
  ...quence is equal to the original one, simply written backwards. By the same logic as before, we may rewrite this sequence as [[Category:Intermediate Trigonometry Problems]]
  3 KB (469 words) - 21:14, 7 July 2022
• 2006 Romanian NMO Problems/Grade 9/Problem 4
  ...s <math>n\leq 3</math>, which is clearly false. This is a contradiction in logic, so the three people in second place could not have won exactly two matches *[[2006 Romanian NMO Problems]]
  2 KB (325 words) - 15:53, 11 December 2011
• 2007 AIME I Problems/Problem 10
  ...te, and this implies that <math>k</math> must be even. We can use the same logic for <math>c</math> and <math>d</math>. Therefore, the coefficient of <math> [[Category:Intermediate Combinatorics Problems]]
  13 KB (2,328 words) - 00:12, 29 November 2023
• ITest
  ...est is varied, ranging from simple arithmetic problems to complex Olympiad problems. Winning teams earn recent technology prizes like a video game console of ...luding [[algebra]], [[geometry]], pre-[[calculus]], [[probability]], and [[logic]]. Graphing calculators are allowed, but computers are not allowed.
  2 KB (264 words) - 20:31, 13 December 2018
• 2005 AMC 12A Problems/Problem 13
  Not too bad with some logic and the awesome guess and check. Let <math>A=6</math>. Then let <math>B=7,E [[Category:Introductory Algebra Problems]]
  3 KB (430 words) - 18:52, 11 July 2020
• Primary Mathematics World Contest
  *Individual Round: The individual round consists of 15 problems, numbered I1-I15. *Team Round: The team round consists of 10 problems, numbered T1-T10.
  1 KB (183 words) - 13:57, 15 October 2018
• 2004 AMC 12A Problems/Problem 17
  {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #17]] and [[2004 AMC 10A Problems/Problem 24|2004 AMC 10A #24]]}} ~Azjps (Fundamental Logic)
  2 KB (233 words) - 08:14, 6 September 2021
• 2007 AMC 10A Problems/Problem 20
  ~Azjps (Fundamental Logic) ~Rbhale12 (Fundamental Logic)
  4 KB (663 words) - 07:32, 4 November 2022
• 2008 AMC 12B Problems/Problem 25
  ...ude of <math>\triangle{CBY}</math> so <math>BC = BY.</math> Using the same logic on <math>\triangle{BCX}</math> gives <math>BC = BX \iff BYXC</math> is a rh [[Category:Introductory Geometry Problems]]
  12 KB (2,015 words) - 20:54, 9 October 2022
• 2008 AIME I Problems/Problem 3
  == Solution 4 (Logic) == [[Category:Intermediate Number Theory Problems]]
  3 KB (564 words) - 22:15, 28 November 2023
• 2008 AIME I Problems/Problem 11
  Applying this logic to all of the other cases gives us <math>\binom{7}{3}</math>, <math>\binom{ [[Category:Intermediate Combinatorics Problems]]
  7 KB (1,173 words) - 22:39, 28 November 2023
• 2008 AIME II Problems/Problem 14
  ...<math>k=\frac{2}{\sqrt{3}}</math>. This inequality must hold by the above logic, and in fact, the inequality reaches equality when <math>x=0</math>. Thus, [[Category:Intermediate Algebra Problems]]
  8 KB (1,387 words) - 11:56, 29 January 2024
• How should I prepare?
  ...your current abilities, you will want to start out with different practice problems, different books, and in different areas of the forums. You should also try .../102-Combinatorial-Problems-Titu-Andreescu/dp/0817643176 102 Combinatorial Problems] by Titu Andreescu & Zuming Feng
  13 KB (1,926 words) - 11:22, 30 November 2023
• 1997 PMWC Problems/Problem T5
  [[Category:Logic Problems]]
  1 KB (263 words) - 22:58, 13 April 2013
• 2002 AMC 10A Problems/Problem 20
  ...{AD}{CD}=3=\frac{GA}{HC}</math> which means <math>HC=5</math>. By the same logic, <math>JE=3</math>, so <math>\frac{HC}{JE}=\frac{5}{3}</math>. [[Category:Introductory Geometry Problems]]
  3 KB (439 words) - 22:15, 9 June 2023
• 1997 PMWC Problems/Problem I2
  [[Category:Logic Problems]]
  1 KB (183 words) - 19:32, 10 March 2015
• 1985 AJHSME Problems/Problem 25
  [[Category:Logic Problems]]
  2 KB (300 words) - 09:07, 22 January 2023
• 1986 AJHSME Problems/Problem 17
  We can solve this problem using logic. [[Category:Introductory Number Theory Problems]]
  3 KB (457 words) - 15:02, 4 April 2021
• 1986 AJHSME Problems/Problem 22
  [[Category:Logic Problems]]
  1 KB (257 words) - 21:30, 3 July 2013
• 1987 AJHSME Problems/Problem 20
  [[Category:Introductory Number Theory Problems]] [[Category:Logic Problems]]
  1 KB (170 words) - 16:44, 24 May 2023
• 2009 AMC 10B Problems/Problem 22
  [[Category: Introductory Geometry Problems]] ...gorean theorem, <cmath>MR = \sqrt{1^2+2^2} = \sqrt{5}.</cmath> By the same logic, <math>MQ=\sqrt{5}</math>. The area of <math>\triangle QMR</math> is the ar
  12 KB (1,868 words) - 03:36, 30 September 2023
• 1987 AJHSME Problems/Problem 17
  [[Category:Logic Problems]]
  934 bytes (156 words) - 23:53, 4 July 2013
• 2009 AIME I Problems/Problem 1
  ...>, we get <math>999</math>, but the integers must be distinct. By the same logic, the smallest geometric number is <math>124</math>. The largest geometric n [[Category:Intermediate Number Theory Problems]]
  2 KB (360 words) - 03:32, 16 January 2023
• 2009 AIME II Problems/Problem 7
  ...>3</math> on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator o [[Category: Intermediate Number Theory Problems]]
  8 KB (1,312 words) - 16:23, 30 March 2024
• Championnat international des jeux mathématiques et logiques
  ...s jeux mathématiques et logiques'') is a contest that consists of several problems that tests the players' logical capabilities, run jointly by national feder Puzzle] -like problems. Finally, the championship in several countries provides a way to select, v
  3 KB (498 words) - 08:52, 1 September 2022
• 1991 AJHSME Problems/Problem 20
  Using logic, <math>a+b+c= 10</math>, therefore <math>b+a+1</math>(from the carry over)< [[Category:Introductory Algebra Problems]]
  1 KB (203 words) - 13:06, 16 August 2020
• 2010 AMC 12A Problems/Problem 3
  ==Video Solution 1 (Logic and Word Analysis)== [[Category:Introductory Geometry Problems]]
  2 KB (310 words) - 17:12, 1 April 2024
• 2010 AMC 12A Problems/Problem 13
  ...there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the gra [[Category:Introductory Geometry Problems]]
  9 KB (1,622 words) - 20:53, 11 September 2023
• 2010 AIME I Problems/Problem 10
  ...that there are <math>8</math> possible pairs in this case. Using the same logic, there are <math>10</math> ways for <math>a_2 = 11, 12 \ldots 19</math>. Fo [[Category:Intermediate Number Theory Problems]]
  7 KB (1,147 words) - 21:58, 23 January 2024
• 2010 USAJMO Problems/Problem 1
  ...ways of sorting them. The same goes for <math>m=2, 3, etc.</math> by this logic. Note that the <math>P(n)</math> as stated by the problem requires a <math> [[Category:Olympiad Number Theory Problems]]
  12 KB (2,338 words) - 20:30, 13 February 2024
• 2011 AMC 10A Problems/Problem 22
  ...t of ways to arrange colors among the vertices of a square. Using the same logic as above, there are <math>5^3=125</math> ways, except we must subtract the [[Category:Intermediate Combinatorics Problems]]
  14 KB (2,425 words) - 09:13, 5 November 2023
• 1999 AMC 8 Problems/Problem 18
  ...th>108</math> students who take the AMC8 meet in the evening to talk about problems and eat an average of two cookies apiece. Walter and Gretel are baking Bonn ...ts. These students eat <math>81 \cdot 2 = 162</math> cookies. Follow the logic of the second paragraph above to find that there needs to be <math>11</math
  2 KB (296 words) - 02:00, 28 February 2022
• 1992 AHSME Problems/Problem 17
  ...is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that <math>N\equiv 19+20+21+22+...+91+92\pmod{3}</math>, and adding [[Category: Introductory Number Theory Problems]]
  2 KB (342 words) - 13:19, 4 January 2021
• 2000 USAMO Problems/Problem 3
  By similar logic, we get <math>R+1</math> optimal strategies if <math>2WR = 3RB</math>, and [[Category:Olympiad Combinatorics Problems]]
  4 KB (800 words) - 21:24, 28 August 2019
• 2012 USAMO Problems/Problem 6
  ...) \le 8</math>.) For all <math>2 \le n < 8</math> however, we may use some logic to first layout a plan. Since for <math>n=6,n=4,</math> and <math>n=2</math *[[USAMO Problems and Solutions]]
  6 KB (1,081 words) - 13:37, 21 June 2023
• 1988 USAMO Problems/Problem 5
  On the other hand, we can use the same logic to show that [[Category:Olympiad Algebra Problems]]
  8 KB (1,348 words) - 09:44, 25 June 2022
• 1974 AHSME Problems/Problem 13
  [[Category:Logic Problems]]
  1 KB (234 words) - 12:43, 5 July 2013
• 1978 USAMO Problems/Problem 5
  ...s who do not share a language with A. Let one of them be B. Using the same logic, one the remaining 4, let it be C, does not share a language with B. Now A, [[Category:Olympiad Combinatorics Problems]]
  902 bytes (147 words) - 12:18, 4 April 2015
• 1980 USAMO Problems/Problem 4
  ...AC| = |BC|</math> and <math>|AD| = |BD|</math>. Finally, applying the same logic to faces <math>ACB</math> and <math>ACD</math> we get <math>|AB| = |AD|</ma [[Category:Olympiad Geometry Problems]]
  2 KB (426 words) - 19:12, 20 March 2023
• 2013 AMC 10B Problems/Problem 25
  {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #23]] and [[2013 AMC 10B Problems|2013 AMC 10B #25]]}} Using the same logic for <math>b</math>, if <math>b < 18</math>, <math>2N \equiv 2b \pmod {36}</
  10 KB (1,623 words) - 15:44, 31 August 2022
• 2013 AIME I Problems/Problem 6
  2. In the 4-size: same logic gets you <math>\frac{1}{55}</math>, since we have <math>4</math> places for [[Category:Intermediate Combinatorics Problems]]
  5 KB (831 words) - 17:20, 9 January 2024
• 1966 AHSME Problems/Problem 20
  [[Category:Introductory Logic Problems]]
  589 bytes (97 words) - 02:30, 15 September 2014
• 1967 AHSME Problems/Problem 28
  [[Category:Introductory Logic Problems]]
  723 bytes (112 words) - 01:40, 16 August 2023
• 2014 AMC 10A Problems/Problem 10
  {{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #9]] and [[2014 AMC 10A Problems|2014 AMC 10A #10]]}} ...<math>3^\text{rd}</math> one or the <math>1^\text{st} + 2</math>. With the logic, we find that <math>b=a+2</math>. <math>b+2=(a+2)+2=\boxed{a+4}</math>.
  2 KB (320 words) - 00:07, 27 June 2023
• 1993 AHSME Problems/Problem 30
  Using the same logic, we notice that this sequence cycles and that since <math>x_0 = x_5</math> [[Category: Intermediate Algebra Problems]]
  1 KB (214 words) - 21:22, 1 June 2020
• 1993 AHSME Problems/Problem 27
  ...ath> are both tangent to the circle P at some point. We can apply the same logic to the other side as well to get <math>CI = 5 - 3x</math>. Finally, since w [[Category: Intermediate Geometry Problems]]
  3 KB (420 words) - 11:10, 18 July 2020
• 1968 AHSME Problems/Problem 10
  [[Category: Introductory Logic Problems]]
  617 bytes (91 words) - 01:52, 16 August 2023
• 1968 AHSME Problems/Problem 15
  ...would never become one of our prime factors. C is ruled out with the same logic as A. Lastly E is ruled out because we have already proved that 3 is possib [[Category: Introductory Number Theory Problems]]
  1 KB (195 words) - 09:24, 1 January 2024
• 1990 AHSME Problems/Problem 20
  ...e BAF = 90^\circ</math>, then <math>\beta = \angle BAF</math>. By the same logic, <math>\angle ABF = \alpha</math>. As a result, <math>\triangle AED \sim \triangle BFA</math>. By the same logic, <math>\triangle CFB \sim \triangle DEC</math>.
  2 KB (378 words) - 17:05, 23 March 2020
• 2008 UNC Math Contest II Problems/Problem 1
  Using the same logic we used for the case involving just one <math>2</math>, we can see that if [[Category:Introductory Algebra Problems]]
  4 KB (715 words) - 12:53, 23 December 2019
• 1988 AHSME Problems/Problem 7
  ...d just want to write down an answer. So, this is quite unreliable. You can logic it out. It doesn't make sense for the first three options to be the answer [[Category: Introductory Algebra Problems]]
  1 KB (196 words) - 21:15, 16 September 2021
• 1987 AHSME Problems/Problem 29
  We can use this logic to go backwards until we reach <math>t_1 = 1</math>, like so: [[Category: Intermediate Algebra Problems]]
  2 KB (326 words) - 12:46, 31 March 2018
• 1986 AHSME Problems/Problem 19
  ...t part of her route. For her <math>y</math>-coordinate, we can use similar logic to find that the coordinate is <math>\sqrt{3} + 0 - \frac{\sqrt{3}}{2} = \ [[Category: Intermediate Geometry Problems]]
  2 KB (240 words) - 20:45, 9 October 2017
• Searching the community
  ...und:yellow"> highlighted in yellow</span>. For example, when searching for problems involving a [[circle]], one would simply type "[[circle]]" into the indicat ...ces for different terms simultaneously; for example, to search for [[AMC]] problems using [[Simon's Favorite Factoring Trick]], search for posts containing "[[
  21 KB (3,334 words) - 23:53, 23 March 2022
• 2016 AMC 12B Problems/Problem 16
  ==Solution 3 (Logic)== [[Category:Introductory Number Theory Problems]]
  8 KB (1,396 words) - 07:17, 31 March 2023
• Gmaas
  ...roblem. Therefore, I proved that you cannot use the Games theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Games can turn things ...oblems, which has around 73,000 bytes. EDIT: Gmaas has beaten 2008 most iT Problems! It is the longest AoPS Wiki article ever. Gmaas's article has 89,119 bytes
  69 KB (11,805 words) - 20:49, 18 December 2019
• 1964 AHSME Problems/Problem 12
  [[Category:Introductory Logic Problems]]
  1 KB (170 words) - 18:13, 23 July 2019
• 2017 AMC 10B Problems/Problem 3
  <math>\textbf{(D)}</math> The same logic as above, but when <math>-\frac{1}{2}<y<0</math> this time. [[Category:Introductory Algebra Problems]]
  1 KB (227 words) - 04:25, 4 December 2020
• 2017 AMC 12B Problems/Problem 21
  Thus the sum of the six numbers equals to <math>30</math>. We apply the logic above in a similar way for the sum of the scores from the first test to the [[Category:Intermediate Number Theory Problems]]
  8 KB (1,381 words) - 20:33, 9 November 2023
• 2018 AMC 10A Problems/Problem 14
  ...o the explanation given in Solution 6. This is the algebraic reason to the logic of if you get a test score with a lower percentage than your average (no ma [[Category:Intermediate Number Theory Problems]]
  10 KB (1,414 words) - 11:59, 27 October 2023
• 2018 AMC 10A Problems/Problem 20
  ...e|[[2018 AMC 10A Problems/Problem 20|2018 AMC 10A #20]] and [[2018 AMC 12A Problems/Problem 15|2018 AMC 12A #15]]}} ...ow, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are <math>4</math> different letters, so there are <
  5 KB (827 words) - 21:31, 24 October 2023
• 2018 AMC 10A Problems/Problem 19
  ...1 \pmod {10}</math> because <math>19^{2a}=361^{a}</math>, and by the same logic as why <math>11^{\text{any positive integer}}\equiv 1 \pmod {10}</math>, we [[Category:Introductory Number Theory Problems]]
  7 KB (1,137 words) - 07:19, 31 March 2023
• 2018 AMC 10A Problems/Problem 22
  Finally, <math>\gcd (c,d) = 54</math>, so using the same logic, <math>m</math> is a multiple of <math>3</math> and is relatively prime to [[Category:Intermediate Number Theory Problems]]
  10 KB (1,599 words) - 04:51, 6 August 2023
• 2018 AMC 10A Problems/Problem 5
  ...ate|[[2018 AMC 10A Problems/Problem 5|2018 AMC 10A #5]] and [[2018 AMC 12A Problems/Problem 4|2018 AMC 12A #4]]}} [[Category:Logic Problems]]
  2 KB (357 words) - 14:38, 3 July 2023
• 2018 AMC 10B Problems/Problem 19
  {{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #14]] and [[2018 AMC 10B Problems|2018 AMC 10B #19]]}} .... Therefore, <math>j=38</math>. Now, since <math>j-1=37</math>, by similar logic, <math>37=(1+k)(a-1)</math>, so <math>k=36</math> and Joey will be <math>38
  5 KB (888 words) - 05:20, 12 April 2024
• 2018 AMC 10B Problems/Problem 25
  ...[[2018 AMC 10B Problems#Problem 25 | 2018 AMC 10B #25]] and [[2018 AMC 12B Problems#Problem 24|2018 AMC 12B #24]]}} By following similar logic, we can find that there is one solution between <math>1</math> ad <math>2</
  10 KB (1,543 words) - 12:51, 29 January 2024
• 2018 AIME I Problems/Problem 13
  ...^2}{2bc}}2} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}}.</cmath> Applying similar logic to <math>\sin \tfrac B2</math> and <math>\sin\tfrac C2</math> and simplifyi [[Category:Intermediate Geometry Problems]]
  13 KB (2,200 words) - 21:36, 6 January 2024
• Yandex Q.E.D. Math Challenge 2018
  ...year we welcome people around the world to test and develop their math and logic skills in the Yandex 60-minute Q.E.D on March 24th, 2018. *Ability to solve logic problems
  988 bytes (146 words) - 10:38, 22 March 2018
• 2018 AIME II Problems/Problem 10
  ...are <math>2 \cdot 2 \cdot 2 = 8</math> solutions for this case, by similar logic to '''Subcase 3.1'''. [[Category: Intermediate Combinatorics Problems]]
  11 KB (1,796 words) - 14:39, 5 November 2022
• 1963 AHSME Problems/Problem 37
  ...h>P</math> is between <math>P_1</math> and <math>P_7</math>. Using similar logic, <math>P</math> must be between <math>P_3</math> and <math>P_5</math> in or [[Category:Intermediate Geometry Problems]]
  1 KB (225 words) - 18:53, 15 July 2018
• 2006 iTest Problems/Problem 4
  ...8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}} [[Category:Introductory Logic Problems]]
  2 KB (243 words) - 22:59, 27 November 2018
• 2019 AIME II Problems/Problem 11
  ...math>\implies O_{1}B = O_{1}A=\frac{147}{16\sqrt{5}}</cmath> Using similar logic we obtain <math>O_{2}A =\frac{189}{16\sqrt{5}}.</math> [[Category: Intermediate Geometry Problems]]
  12 KB (1,985 words) - 19:52, 28 January 2024
• 2019 AMC 10B Problems/Problem 25
  ...e|[[2019 AMC 10B Problems#Problem 25|2019 AMC 10B #25]] and [[2019 AMC 12B Problems#Problem 23|2019 AMC 12B #23]]}} '''Case 3''': Four <math>11</math> blocks arranged. Using the same logic as Case 2, we have <math>{7\choose4} = 35</math> ways to arrange four <math
  11 KB (1,763 words) - 11:51, 27 February 2024
• 2019 AMC 10B Problems/Problem 23
  ...e|[[2019 AMC 10B Problems#Problem 23|2019 AMC 10B #23]] and [[2019 AMC 12B Problems#Problem 20|2019 AMC 12B #20]]}} ==Video Solution by The Power of Logic==
  6 KB (967 words) - 10:25, 20 December 2023
• 2019 AIME I Problems/Problem 4
  <math>\textbf{Case 4}</math>: Three substitutions. Using similar logic as <math>\textbf{Case 3}</math>, we get <math>(11\cdot 11)\cdot (11\cdot 10 [[Category:Intermediate Combinatorics Problems]]
  4 KB (551 words) - 10:32, 5 February 2022
• 2019 AIME I Problems/Problem 8
  ==Video Solution By The Power Of Logic== [[Category:Intermediate Algebra Problems]]
  10 KB (1,878 words) - 13:19, 1 February 2024
• 2019 AIME I Problems/Problem 10
  ==Video Solution by The Power of Logic== ~The Power of Logic
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• 2019 AIME I Problems/Problem 14
  ==Video Solution by The Power Of Logic== [[Category:Intermediate Number Theory Problems]]
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• What is the definition of Pure Mathematics?
  Some definitions hit almost all the areas of math, but some are too broad and logic, for example, often fits into the definition. === Problems ===
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• 1963 AHSME Problems/Problem 26
  [[Category:Introductory Logic Problems]]
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• GMAAS
  ...herefore, I proved that you cannot use the Almighty Gmaas theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Almighty Gmaas can tur ...ich has around 73,000 bytes. EDIT: Almighty Gmaas has beaten 2008 most iT Problems! It is the longest AoPS Wiki article ever. Almighty Gmaas's article has 89
  85 KB (13,954 words) - 17:25, 22 March 2024
• 2019 AMC 8 Problems/Problem 11
  ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)== [[Category: Introductory Combinatorics Problems]]
  3 KB (477 words) - 20:42, 13 February 2024
• 2019 AMC 8 Problems/Problem 22
  ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)== [[Category:Introductory Algebra Problems]]
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• 2019 AMC 8 Problems/Problem 24
  This question is extremely similar to [[1971 AHSME Problems/Problem 26]]. ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
  27 KB (4,256 words) - 19:30, 17 January 2024
• 2023 AIME I Problems/Problem 15
  ..., once again, <math>5\mid a^2+b^2</math>. If <math>b=6</math>, by the same logic, <math>12\sqrt3<a<24</math>, so <math>a=23</math>, where we run into the sa [[Category:Intermediate Algebra Problems]]
  8 KB (1,299 words) - 17:37, 3 June 2023
• 2006 iTest Problems/Problem 10
  For the fifth figure, we use the same logic to find that each large triangle has <math>15+14+...+1=120</math> numbers ...h> medium triangles, and <math>3</math> big triangles. Then, using similar logic to above, we find there are <math>3+2+1=6</math> dots per small triangle, <
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• 1976 AHSME Problems/Problem 27
  ~Someonenumber011 (Fundamental Logic) [[Category:AHSME Problems]]
  3 KB (452 words) - 13:41, 8 September 2021
• G285 2021 Fall Problem Set
  Welcome to the Fall Problem Set! There are <math>15</math> problems, <math>10</math> multiple-choice, and <math>5</math> free-response. Larry is playing a logic game. In this game, Larry counts <math>1,2,3,6, \cdots </math>, and removes
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• 2020 AIME II Problems/Problem 14
  Using this logic, we can find the number of solutions to <math>f(x) = y</math>. For every in [[Category: Intermediate Algebra Problems]]
  7 KB (1,186 words) - 15:31, 5 January 2024
• 2021 AIME I Problems/Problem 1
  ==Video Solution by Power of Logic== [[Category:Introductory Combinatorics Problems]]
  5 KB (780 words) - 00:58, 23 August 2022
• 2021 AIME I Problems/Problem 2
  ==Video Solution by Power of Logic== [[Category:Intermediate Geometry Problems]]
  8 KB (1,294 words) - 00:59, 23 August 2022
• 2021 AIME I Problems/Problem 3
  ==Video Solution by Power of Logic== [[Category:Introductory Number Theory Problems]]
  7 KB (1,174 words) - 08:21, 13 May 2023
• 2021 AIME I Problems/Problem 11
  ...'s why it works out: Let the length of diagonal BD be a. Then, by the same logic as given in the solution by fidgetboss_4000, <math>x=\frac{33+a^{2}}{2a}</m This problem is kinda similar to [[2021 AIME II Problems/Problem 12]]
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• 2022 AIME I Problems/Problem 3
  ...right triangles. Therefore, <math>AD = DW = 333</math>. We can apply this logic to triangles <math>BCQ</math> and <math>XCQ</math> as well, giving us <math [[Category:Intermediate Geometry Problems]]
  24 KB (3,832 words) - 20:59, 2 March 2024
• 2022 AIME I Problems/Problem 9
  ==Video Solution (Power of Logic)== [[Category:Intermediate Combinatorics Problems]]
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• 2022 AIME I Problems/Problem 13
  <cmath>\text{(Also, I'm not going to give you explanations for the other logic equations.)}</cmath> [[Category:Intermediate Number Theory Problems]]
  7 KB (1,207 words) - 21:53, 2 January 2024
• 2021 AMC 12A Problems/Problem 3
  ...ate|[[2021 AMC 10A Problems/Problem 3|2021 AMC 10A #3]] and [[2021 AMC 12A Problems/Problem 3|2021 AMC 12A #3]]}} ==Solution 3 (Vertical Addition and Logic)==
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• 2021 AMC 12A Problems/Problem 4
  ...ate|[[2021 AMC 10A Problems/Problem 7|2021 AMC 10A #7]] and [[2021 AMC 12A Problems/Problem 4|2021 AMC 12A #4]]}} ==Solution 1 (Comprehensive Explanation of Logic)==
  7 KB (1,027 words) - 22:45, 28 October 2022
• 2021 AMC 12A Problems/Problem 12
  ...e|[[2021 AMC 12A Problems/Problem 12|2021 AMC 12A #12]] and [[2021 AMC 10A Problems/Problem 14|2021 AMC 10A #14]]}} == Video Solution by Power Of Logic (Using Vieta's Formulas) ==
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• 2021 AMC 12B Problems/Problem 15
  ...e|[[2021 AMC 10B Problems#Problem 20|2021 AMC 10B #20]] and [[2021 AMC 12B Problems#Problem 15|2021 AMC 12B #15]]}} ==Video Solution by The Power of Logic==
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• 2021 AMC 12B Problems/Problem 25
  ...e|[[2021 AMC 10B Problems#Problem 25|2021 AMC 10B #25]] and [[2021 AMC 12B Problems#Problem 25|2021 AMC 12B #25]]}} Here I give a more logic way to show how to find the upper bound of <math>m. </math>
  17 KB (2,687 words) - 23:43, 7 November 2022
• 1957 AHSME Problems/Problem 19
  Using this same logic, <math>10011_2</math> would be <math>1*2^4 + 1*2^1 + 1 * 2^0 = \boxed{\text [[Category:AHSME]][[Category:AHSME Problems]]
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• 2021 USAJMO Problems/Problem 1
  ==Video Solution by the Power of Logic== ~The Power of Logic
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• 2022 AIME II Problems/Problem 1
  ==Video Solution by Power of Logic== [[Category:Introductory Number Theory Problems]]
  4 KB (626 words) - 14:34, 19 November 2023
• 2022 AIME II Problems/Problem 2
  ==Video Solution by Power of Logic== [[Category:Intermediate Combinatorics Problems]]
  4 KB (673 words) - 16:15, 24 January 2023
• 2022 AIME II Problems/Problem 4
  ==Video Solution by Power of Logic== [[Category:Intermediate Algebra Problems]]
  5 KB (778 words) - 18:14, 30 January 2024
• 2022 AIME II Problems/Problem 6
  ==Video Solution by The Power of Logic== [[Category:Intermediate Algebra Problems]]
  4 KB (689 words) - 12:57, 10 September 2023
• 2022 AIME II Problems/Problem 8
  ==Video Solution by The Power of Logic== [[Category:Intermediate Number Theory Problems]]
  13 KB (2,136 words) - 22:39, 31 December 2023
• 2022 AIME II Problems/Problem 13
  ==Video Solution by The Power of Logic== [[Category:Intermediate Algebra Problems]]
  7 KB (1,065 words) - 01:50, 10 May 2024
• Rational root theorem
  ...th>, which implies <math>p | a_0</math> by [[Euclid's lemma]]. Via similar logic in modulo <math>q</math>, <math>q|a_n</math>, as required. <math>\square</m Here are some problems with solutions that utilize the rational root theorem.
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• 2021 Fall AMC 12B Problems/Problem 24
  ==Video Solution by Power of Logic(Trig and Power of a point)== [[Category:Intermediate Geometry Problems]]
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• 2021 Fall AMC 12B Problems/Problem 23
  ==Video Solution by The Power Of Logic== [[Category:Intermediate Combinatorics Problems]]
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• 2023 AIME I Problems/Problem 5
  ...ta</math>. Therefore, <math>2r^2 \sin \theta = 56</math>. Applying similar logic to <math>\triangle BPD</math>, we get <math>2r^2 \sin (90^\circ - \theta) = [[Category:Intermediate Geometry Problems]]
  19 KB (3,107 words) - 23:31, 17 January 2024
• 2022 AMC 10A Problems/Problem 14
  ...e|[[2022 AMC 10A Problems/Problem 14|2022 AMC 10A #14]] and [[2022 AMC 12A Problems/Problem 10|2022 AMC 12A #10]]}} In general, for <math>1,\ldots,2n</math>, the same logic yields answer: <math>\left\lfloor\dfrac{n}{2}\right\rfloor! \cdot \left\lce
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• 2022 AMC 10A Problems/Problem 24
  ...e|[[2022 AMC 10A Problems/Problem 24|2022 AMC 10A #24]] and [[2022 AMC 12A Problems/Problem 24|2022 AMC 12A #24]]}} ==Solution 4 (Fake solve, incorrect logic, correct answer by coincidence)==
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• 2022 AMC 10B Problems/Problem 22
  ...e|[[2022 AMC 10B Problems/Problem 22|2022 AMC 10B #22]] and [[2022 AMC 12B Problems/Problem 21|2022 AMC 12B #21]]}} ==Video Solution by The Power of Logic(#20-#21)==
  7 KB (1,202 words) - 01:15, 10 June 2023
• 2022 AMC 10B Problems/Problem 21
  ...e|[[2022 AMC 10B Problems/Problem 21|2022 AMC 10B #21]] and [[2022 AMC 12B Problems/Problem 20|2022 AMC 12B #20]]}} ==Video Solution by The Power of Logic(#20-#21)==
  7 KB (1,219 words) - 23:07, 19 September 2023
• 2022 AMC 10B Problems/Problem 23
  ...e|[[2022 AMC 10B Problems/Problem 23|2022 AMC 10B #23]] and [[2022 AMC 12B Problems/Problem 22|2022 AMC 12B #22]]}} ==Video Solution by The Power of Logic (#22 and #23)==
  10 KB (1,607 words) - 14:45, 12 June 2023
• 2022 AMC 10B Problems/Problem 25
  ...e|[[2022 AMC 10B Problems#Problem 25|2022 AMC 10B #25]] and [[2022 AMC 12B Problems#Problem 23|2022 AMC 12B #23]]}} ==Video Solution by The Power of Logic==
  9 KB (1,252 words) - 02:13, 30 December 2023
• 2022 AMC 10B Problems/Problem 19
  ...e|[[2022 AMC 10B Problems/Problem 19|2022 AMC 10B #19]] and [[2022 AMC 12B Problems/Problem 18|2022 AMC 12B #18]]}} == Video Solution by OmegaLearn (Using Logic and Casework) ==
  5 KB (621 words) - 03:22, 7 July 2023
• 2023 AIME I Problems/Problem 1
  ...a woman, so the probability is <math>\frac{8}{11}.</math> We use the same logic for the <math>3</math>rd, <math>4</math>th and <math>5</math>th men to get ...ould happen with a probability of <math>\frac{10}{12}</math> using similar logic. Doing this for the <math>4</math>th and <math>5</math>th men, we get proba
  11 KB (1,803 words) - 17:17, 12 January 2024
• 2023 AIME I Problems/Problem 6
  We can apply similar logic to compute <math>E_2</math> and get [[Category:Intermediate Combinatorics Problems]]
  11 KB (1,654 words) - 17:28, 31 January 2024
• 2023 AIME II Problems/Problem 1
  ==Video Solution by the Power of Logic(both #1 and #2)== [[Category:Introductory Algebra Problems]]
  2 KB (296 words) - 17:40, 29 February 2024
• 2023 AIME II Problems/Problem 2
  ==Video Solution by the Power of Logic(#1 and #2)== [[Category:Introductory Number Theory Problems]]
  1 KB (174 words) - 20:32, 8 June 2023
• 2023 AIME II Problems/Problem 5
  ==Video Solution by The Power of Logic== [[Category:Intermediate Number Theory Problems]]
  3 KB (474 words) - 20:13, 18 June 2023
• 2023 AIME II Problems/Problem 4
  ==Video Solution by The Power of Logic(#3 and #4)== [[Category:Intermediate Algebra Problems]]
  10 KB (1,593 words) - 07:23, 9 May 2024
• 2023 AIME II Problems/Problem 3
  ...gle PCA = 45^\circ</math>, so <math>\angle BPC = 135^\circ</math>. Similar logic shows <math>\angle APC = 135^\circ</math>. ==Video Solution by The Power of Logic(#3 and #4)==
  9 KB (1,354 words) - 21:16, 26 January 2024
• 2023 AIME II Problems/Problem 8
  ==Video Solution by The Power of Logic== [[Category:Intermediate Algebra Problems]]
  9 KB (1,284 words) - 23:37, 31 January 2024
• 2023 AIME II Problems/Problem 6
  ==Video Solution by The Power of Logic== [[Category:Intermediate Combinatorics Problems]]
  7 KB (1,245 words) - 12:48, 1 February 2024
• 2023 AIME II Problems/Problem 7
  ==Video Solution by The Power of Logic== [[Category:Intermediate Combinatorics Problems]]
  8 KB (1,232 words) - 18:43, 26 November 2023
• 2023 AIME II Problems/Problem 10
  ...ing 3s which aren't supposed to be connected, so it is impossible. Similar logic works for why all 1s can't be connected with all 2s. [[Category:Intermediate Combinatorics Problems]]
  5 KB (715 words) - 00:03, 31 December 2023
• Know the Facts: Gmaas
  ...roblem. Therefore, I proved that you cannot use the Gmaas theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Gmaas can turn things ...oblems, which has around 73,000 bytes. EDIT: Gmaas has beaten 2008 most iT Problems! It is the longest AoPS Wiki article ever. Gmaas's article has 89,119 bytes
  88 KB (14,928 words) - 13:54, 29 April 2024
• 2023 AMC 12B Problems/Problem 5
  ...e|[[2023 AMC 10B Problems/Problem 10|2023 AMC 10B #10]] and [[2023 AMC 12B Problems/Problem 5|2023 AMC 12B #5]]}} ==Solution 5 (Logic)==
  6 KB (870 words) - 23:20, 26 December 2023
• 2023 AMC 12B Problems/Problem 9
  ...e|[[2023 AMC 10B Problems/Problem 13|2023 AMC 10B #13]] and [[2023 AMC 12B Problems/Problem 9|2023 AMC 12B #9]]}} ==Solution 3 (Logic)==
  4 KB (643 words) - 21:27, 27 April 2024
• Problems Collection
  This is a page where you can share the problems you made (try not to use past exams). ...th>a,b,c</math> (WLOG, <math>a</math>) is <math>0</math>. Again, using the logic of solution 1 we see <math>p=\frac{1}{4}</math> so <math>p+q=\frac{7}{12}</
  43 KB (7,006 words) - 14:24, 19 February 2024
• 2024 AIME II Problems/Problem 9
  ...on't appear on the board, they will not occupy any of the columns. Similar logic can be applied to show that if white pieces occupy all of the rows, they wi [[Category:Intermediate Combinatorics Problems]]
  6 KB (995 words) - 12:52, 25 March 2024
• 2005 iTest Problems
  [[2005 iTest Problems/Problem 1|Solution]] [[2005 iTest Problems/Problem 2|Solution]]
  25 KB (3,738 words) - 10:53, 23 April 2024