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• 2016 AIME II Problems/Problem 15
  2 KB (277 words) - 20:02, 19 April 2023
• 1975 AHSME Problems/Problem 15
  887 bytes (148 words) - 23:42, 24 June 2019
• 1964 AHSME Problems/Problem 15
  == Problem 15==
  2 KB (260 words) - 21:01, 23 July 2019
• 1954 AHSME Problems/Problem 15
  == Problem 15==
  441 bytes (54 words) - 01:26, 28 February 2020
• 1965 AHSME Problems/Problem 15
  298 bytes (49 words) - 11:12, 10 June 2016
• 1977 AHSME Problems/Problem 15
  1 KB (200 words) - 21:30, 1 May 2024
• 2016 AMC 8 Problems/Problem 15
  1 KB (156 words) - 22:58, 17 May 2024
• 1979 AHSME Problems/Problem 15
  == Problem 15 ==
  1 KB (240 words) - 13:21, 6 January 2017
• 2017 AMC 10A Problems/Problem 15
  ...al length of <math>2-0=2</math>. Therefore, the probability is <math>1.5/2=15/20=\boxed{\frac{3}{4} \space \text{(C)}}.</math>
  4 KB (533 words) - 19:01, 15 March 2024
• 2017 AMC 12A Problems/Problem 15
  Notice that <math>\frac{5\pi}{4}<\frac{3.15*5}{4}<4<\frac{3\pi}{2}, f(\frac{5\pi}{4})=3-\frac{3\sqrt{2}}{2}>0 .</math>
  3 KB (564 words) - 14:12, 23 October 2021
• 2017 AMC 10B Problems/Problem 15
  ...xtbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}</math>
  4 KB (656 words) - 01:20, 4 December 2023
• 2017 AMC 12B Problems/Problem 15
  9 KB (1,416 words) - 14:30, 23 November 2023
• 2017 AIME I Problems/Problem 15
  ==Problem 15== [[File:2017 AIME I 15.png|530px|right]]
  22 KB (3,622 words) - 17:11, 6 January 2024
• 2017 AIME II Problems/Problem 15
  [[File:2017 AIME II 15.png|300px|right]]
  6 KB (971 words) - 02:08, 22 January 2024
• 1953 AHSME Problems/Problem 15
  ==Problem 15==
  1 KB (211 words) - 14:41, 27 January 2020
• 2017 AMC 8 Problems/Problem 15
  4 KB (613 words) - 11:36, 24 December 2023
• 2008 iTest Problems/Problem 15
  1,002 bytes (137 words) - 13:25, 22 June 2018
• 2018 AMC 10A Problems/Problem 15
  label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S);
  2 KB (342 words) - 17:10, 15 October 2023
• 2018 AMC 12A Problems/Problem 15
  46 bytes (5 words) - 16:05, 8 February 2018
• 2018 AMC 12B Problems/Problem 15
  ...e a multiple of <math>3</math>, but does not contain 3. We have <math>(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87 ...3) = 15</math> cases. For <math>\underline{3AB}</math>, we also have <math>15</math> cases, but when <math>B=3, 9</math>, <math>A</math> can equal <math>
  8 KB (1,244 words) - 08:19, 30 May 2023

Page text matches

• 1986 AIME Problems/Problem 12
  ...ments. Let <math>S</math> be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of <math>S</math> have the same sum. What ...> must have more than 4 elements, otherwise its sum would be at most <math>15+14+13+12=54</math>.
  2 KB (364 words) - 19:41, 1 September 2020
• 1986 AIME Problems/Problem 11
  ...ath>1</math>). By the [[Binomial Theorem]], this is <math>(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}</math>. ...y^2</math> term is <math>{2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}</math>. This is actually the sum of the first 16 triangular numbers, which
  6 KB (872 words) - 16:51, 9 June 2023
• 1986 AIME Problems/Problem 10
  ...ath>a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14</math> so <math>a+b+c\geq 15</math>. Checking each of the multiples of <math>222</math> from <math>15\cdot222</math> to <math>18\cdot222</math> by subtracting <math>N</math> fro
  3 KB (565 words) - 16:51, 1 October 2023
• 1986 AIME Problems/Problem 9
  By similar triangles, <math>BE'=\frac{d}{510}\cdot450=\frac{15}{17}d</math> and <math>EC=\frac{d}{425}\cdot450=\frac{18}{17}d</math>. Sinc ...510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}</math>.
  11 KB (1,850 words) - 18:07, 11 October 2023
• 1986 AIME Problems/Problem 8
  ...\log 1 + \log 2 + \log 4 + \log 5 +\ldots + \log 1000000 = \log (2^05^0)(2^15^0)(2^25^0)\cdots (2^65^6).</cmath> Each power of <math>2</math> appears <ma
  3 KB (487 words) - 20:52, 16 September 2020
• 1986 AIME Problems/Problem 2
  ...right)^2}{2\cdot 2\sqrt5\cdot2\sqrt6}=\frac{16}{8\sqrt{30}}=\sqrt{\frac{2}{15}},</cmath> so <math>\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{\frac{13}{15}}.</math>
  3 KB (460 words) - 00:44, 5 February 2022
• 1987 AIME Problems/Problem 14
  {{AIME box|year=1987|num-b=13|num-a=15}}
  7 KB (965 words) - 10:42, 12 April 2024
• 1987 AIME Problems/Problem 10
  ...ance at rate <math>r</math> from the escalator, while Bob is getting <math>15</math> seconds of help at rate <math>r</math>. Solving for <math>r</math>,
  7 KB (1,187 words) - 16:21, 27 January 2024
• 1987 AIME Problems/Problem 8
  ...or which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>? <cmath>\begin{array}{ccccc}\frac{15}{8} &>& \frac{k + n}{n} &>& \frac{13}{7}\\
  2 KB (393 words) - 16:59, 16 December 2020
• 1987 AIME Problems/Problem 4
  ...h>|x - 60| = 0</math>, which is when <math>x = 60</math> and <math>y = \pm 15</math>. Since the graph is [[symmetry|symmetric]] about the y-axis, we just ...using the [[Shoelace Theorem]], we get <math>2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}</math>.
  2 KB (371 words) - 17:25, 13 February 2024
• 1987 AIME Problems/Problem 3
  ...6,\ 2^3 = 8,\ 2 \cdot 5 = 10,</math> <math>\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,</math> <math>\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,</math>
  3 KB (511 words) - 09:29, 9 January 2023
• 1988 AIME Problems/Problem 14
  {{AIME box|year=1988|num-b=13|num-a=15}}
  4 KB (700 words) - 17:21, 3 May 2021
• 1988 AIME Problems/Problem 13
  ...+ F_{15}) + 1 &\Longrightarrow (aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q \\ &\Longrightarrow aF_{17} + bF_{16} = 0 \text{ and } aF_{16} + bF_{15} + 1 = 0 \\
  10 KB (1,585 words) - 03:58, 1 May 2023
• 1988 AIME Problems/Problem 11
  <math>504 = 3m + 15</math>
  2 KB (422 words) - 00:22, 6 September 2020
• 1988 AIME Problems/Problem 6
  ...use the arithmetic progression from left to right has difference <math>x - 15</math>. Therefore, we have <math>x = 50</math>, and because the desired ast
  5 KB (878 words) - 23:06, 20 November 2023
• 1989 AIME Problems/Problem 15
  ...ac{[APC] + [BPC]}{[APB]} = 3</math>, so <math>CP = \dfrac{3}{4} \cdot CF = 15</math>. ...ed out, so <math>w_C = 1</math> and <math>w_F = 3</math>. Thus, <math>CP = 15</math> and <math>PF = 5</math>.
  13 KB (2,091 words) - 00:20, 26 October 2023
• 1989 AIME Problems/Problem 14
  {{AIME box|year=1989|num-b=13|num-a=15}}
  2 KB (408 words) - 17:28, 16 September 2023
• 1989 AIME Problems/Problem 11
  <math>n = 4: 5000+15*116 = 6740</math>
  5 KB (851 words) - 18:01, 28 December 2022
• 1989 AIME Problems/Problem 6
  label("$P$",(6,15),N); label("$X$",(12.5,15),N);
  6 KB (980 words) - 15:08, 14 May 2024
• 1989 AIME Problems/Problem 4
  <math>m = 15</math> gives a solution for k. <math>10 + 5a = 15^3</math>
  3 KB (552 words) - 12:41, 3 March 2024

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