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• 1987 AIME Problems/Problem 8
  ...is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>? <cmath>\begin{align*}104(n+k) &< 195n< 105(n+k)\\
  2 KB (393 words) - 16:59, 16 December 2020
• 1987 AIME Problems/Problem 7
  ...3, 4), (2, 3, 4), (3, 3, 4), (3, 2, 4), (3, 1, 4)</math> and <math>(3, 0, 4)</math>. ...ach of the forms <math>(3, 3, n)</math>, <math>(3, n, 3)</math> and <math>(n, 3, 3)</math>.
  3 KB (547 words) - 22:54, 4 April 2016
• 1987 AIME Problems/Problem 3
  ...product of the distinct proper divisors of <math>n</math>. A number <math>n</math> is ''nice'' in one of two instances: ...visors are <math>p</math> and <math>q</math>, and <math>p(n) = p \cdot q = n</math>.
  3 KB (511 words) - 09:29, 9 January 2023
• 1987 AIME Problems/Problem 1
  ...non-negative]] [[integer]]s is called "simple" if the [[addition]] <math>m+n</math> in base <math>10</math> requires no carrying. Find the number of sim ...e then fixed). Thus, the number of [[ordered pair]]s will be <math>(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}</math>.
  1 KB (191 words) - 14:42, 17 September 2016
• 1988 AIME Problems/Problem 13
  Let <math>F_n</math> represent the <math>n</math>th number in the Fibonacci sequence. Therefore, x^2 - x - 1 = 0&\Longrightarrow x^n = F_n(x), \ n\in N \\
  10 KB (1,585 words) - 03:58, 1 May 2023
• 1988 AIME Problems/Problem 10
  ...ly one vertex of a square/hexagon/octagon, we have that <math>V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48</math>. ...ron must be a diagonal of that face. Each square contributes <math>\frac{n(n-3)}{2} = 2</math> diagonals, each hexagon <math>9</math>, and each octagon
  5 KB (811 words) - 19:10, 25 January 2021
• 1988 AIME Problems/Problem 9
  ...equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework: ...0}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution.
  6 KB (893 words) - 08:15, 2 February 2023
• 1988 AIME Problems/Problem 5
  ...99}</math> is an integer multiple of <math>10^{88}</math>. Find <math>m + n</math>. ...math>\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}</math>, and <math>m + n = \boxed{634}</math>.
  822 bytes (108 words) - 22:21, 6 November 2016
• 1988 AIME Problems/Problem 4
  Suppose that <math>|x_i| < 1</math> for <math>i = 1, 2, \dots, n</math>. Suppose further that What is the smallest possible value of <math>n</math>?
  2 KB (394 words) - 10:21, 27 January 2024
• 1988 AIME Problems/Problem 3
  1) <math>\log_a b^n=n\log_a b</math>. 2) <math>\log_{a^n} b=\frac{1}{n}\log_a b</math>.
  3 KB (481 words) - 21:52, 18 November 2020
• 1988 AIME Problems/Problem 2
  ...ts of <math>k</math>. For <math>n \ge 2</math>, let <math>f_n(k) = f_1(f_{n - 1}(k))</math>. Find <math>f_{1988}(11)</math>. We see that <math>f_{1}(11)=4</math>
  696 bytes (103 words) - 19:16, 27 February 2018
• 1989 AIME Problems/Problem 15
  real x = 0.4, y = 0.2, z = 1-x-y; label("$X$", X, N);
  13 KB (2,091 words) - 00:20, 26 October 2023
• 1989 AIME Problems/Problem 14
  ...ressed in the base <math>-n+i</math> using the integers <math>0,1,2,\ldots,n^2</math> as digits. That is, the equation <center><math>r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0</math></center>
  2 KB (408 words) - 17:28, 16 September 2023
• 1989 AIME Problems/Problem 12
  C=origin; B=(8,0); D=IP(CR(C,6.5),CR(B,8)); A=(4,-3); P=midpoint(A--B); Q=midpoint(C--D); ...p); dot("$C$",C,left,p); dot("$D$",D,up,p); dot("$M$",P,dir(-45),p); dot("$N$",Q,0.2*(Q-P),p);
  2 KB (376 words) - 13:49, 1 August 2022
• 1989 AIME Problems/Problem 11
  Let the mode be <math>x</math>, which we let appear <math>n > 1</math> times. We let the arithmetic mean be <math>M</math>, and the sum ...t| = \left|\frac{S+xn}{121}-x\right| = \left|\frac{S}{121}-\left(\frac{121-n}{121}\right)x\right|
  5 KB (851 words) - 18:01, 28 December 2022
• 1989 AIME Problems/Problem 10
  pair A = (0,0), B = (3, 0), C = (1, 4); draw(rightanglemark(C,P, B, 4));
  8 KB (1,401 words) - 21:41, 20 January 2024
• 1989 AIME Problems/Problem 9
  ...h that <cmath>133^5+110^5+84^5+27^5=n^{5}.</cmath> Find the value of <math>n</math>. n^5&\equiv0\pmod{2}, \\
  6 KB (874 words) - 15:50, 20 January 2024
• 1989 AIME Problems/Problem 8
  ...tion is of the form <cmath>f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7</cmath> for some <math>k\in\{1,2,3\}.</math> and we wish to find <math>f(4).</math>
  8 KB (1,146 words) - 04:15, 20 November 2023
• 1989 AIME Problems/Problem 6
  D(B--A); D(A--C); D(B--C,dashed); MP("A",A,SW);MP("B",B,SE);MP("C",C,N);MP("60^{\circ}",A+(0.3,0),NE);MP("100",(A+B)/2);MP("8t",(A+C)/2,NW);MP("7t t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}.\end{align*}</cmath>
  6 KB (980 words) - 15:08, 14 May 2024
• 1989 AIME Problems/Problem 4
  ...h> equal <math>a+1</math>, <math>a+2</math>, <math>a+3</math>, and <math>a+4</math>, respectively. Call the square and cube <math>k^2</math> and <math>m Let the numbers be <math>a,a+1,a+2,a+3,a+4.</math> When then know <math>3a+6</math> is a perfect cube and <math>5a+10<
  3 KB (552 words) - 12:41, 3 March 2024
• 1989 AIME Problems/Problem 3
  ...eger]] and <math>d</math> is a single [[digit]] in [[base 10]]. Find <math>n</math> if <center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center>
  3 KB (499 words) - 22:17, 29 March 2024
• 1989 AIME Problems/Problem 1
  == Solution 4 (Symmetry with Generalization) == ...ht) - 1</math>, which is easier to compute. Either way, plugging in <math>n=29.5</math> gives <math>\boxed{869}</math>.
  4 KB (523 words) - 00:12, 8 October 2021
• 1990 AIME Problems/Problem 15
  ax^4 + by^4 &= 42. ...h>(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})</cmath>
  4 KB (644 words) - 16:24, 28 May 2023
• 1990 AIME Problems/Problem 14
  label("$P$", P, N); label("P", P, N);
  7 KB (1,086 words) - 08:16, 29 July 2023
• 1990 AIME Problems/Problem 11
  ...r which <math>n^{}_{}!</math> can be expressed as the [[product]] of <math>n - 3_{}^{}</math> [[consecutive]] positive integers. ...\sqrt{a!}</math>, which decreases as <math>a</math> increases. Thus, <math>n = 23</math> is the greatest possible value to satisfy the given conditions.
  3 KB (519 words) - 09:28, 28 June 2022
• 1990 AIME Problems/Problem 9
  Call the number of ways of flipping <math>n</math> coins and not receiving any consecutive heads <math>S_n</math>. Noti ...ath>n-1</math> flips must fall under one of the configurations of <math>S_{n-1}</math>.
  3 KB (425 words) - 12:36, 12 May 2024
• 1990 AIME Problems/Problem 7
  pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);
  8 KB (1,319 words) - 11:34, 22 November 2023
• 1990 AIME Problems/Problem 6
  ...otal number of fish in September is <math>125 \%</math>, or <math>\frac{5}{4}</math> times the total number of fish in May. ...s the number of fish in May. Solving for <math>n</math>, we see that <math>n = \boxed{840}</math>
  2 KB (325 words) - 13:16, 26 June 2022
• 1990 AIME Problems/Problem 5
  ...gral divisors, including <math>1_{}^{}</math> and itself. Find <math>\frac{n}{75}</math>. ...to the least power. Therefore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}</math>.
  1 KB (175 words) - 03:45, 21 January 2023
• 1990 AIME Problems/Problem 3
  ...the interior angle of a regular sided [[polygon]] is <math>\frac{(n-2)180}{n}</math>. {{AIME box|year=1990|num-b=2|num-a=4}}
  3 KB (516 words) - 19:18, 16 April 2024
• 1991 AIME Problems/Problem 15
  <math>\sum_{k=1}^n \sqrt{(2k-1)^2+a_k^2},</math> ...}</math> for which <math>S_n^{}</math> is also an integer. Find this <math>n^{}_{}</math>.
  4 KB (658 words) - 16:58, 10 November 2023
• 1991 AIME Problems/Problem 13
  Solving the resulting quadratic equation <math>r^{2}-rt+t(t-1)/4=0</math>, for <math>r</math> in terms of <math>t</math>, one obtains that ...e present case <math>t\leq 1991</math>, and so one easily finds that <math>n=44</math> is the largest possible integer satisfying the problem conditions
  7 KB (1,328 words) - 20:24, 5 February 2024
• 1991 AIME Problems/Problem 12
  ...t terms, denote the [[perimeter]] of <math>ABCD^{}_{}</math>. Find <math>m+n^{}_{}</math>. ...\(Q\)",Q,E);label("\(R\)",R,SW);label("\(S\)",S,W); label("\(15\)",B/2+P/2,N);label("\(20\)",B/2+Q/2,E);label("\(O\)",O,SW); </asy></center>
  8 KB (1,270 words) - 23:36, 27 August 2023
• 1991 AIME Problems/Problem 11
  ...es. The area of one circle is thus <math>\pi(2 - \sqrt {3})^{2} = \pi (7 - 4 \sqrt {3})</math>, so the area of all <math>12</math> circles is <math>\pi ...nce, the radius <math>r_{}^{}=R\sin(\pi/n)</math>. The total area <math>A_{n}^{}</math> of the <math>n_{}^{}</math> circles is thus given by
  4 KB (740 words) - 19:33, 28 December 2022
• 1991 AIME Problems/Problem 10
  ...es the partial sums of <math>P_b</math> (in other words, <math>S_b = \sum_{n=1}^{b} P_b</math>): aab & 4 & 2 & 3 \\
  5 KB (813 words) - 06:10, 25 February 2024
• 1991 AIME Problems/Problem 9
  ...c mn,</math> where <math>\frac mn</math> is in lowest terms. Find <math>m+n^{}_{}.</math> ...ot will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = \boxed{044}</math>.
  10 KB (1,590 words) - 14:04, 20 January 2023
• 1991 AIME Problems/Problem 3
  ...may write <math>A_{k}^{}={N\choose k}x^{k}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>. Taking logarithms in both sides of this last equati ...ft[\prod_{j=1}^{k}\frac{(N-j+1)x}{j}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .
  5 KB (865 words) - 12:13, 21 May 2020
• 1991 AIME Problems/Problem 2
  ...}^{}</math> has [[edge | sides]] <math>\overline {AB}</math> of [[length]] 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</m pair A=(0,0),B=(4,0),C=(4,3),D=(0,3);
  4 KB (595 words) - 12:51, 17 June 2021
• 1992 AIME Problems/Problem 15
  ...that the decimal representation of <math>m!</math> ends with exactly <math>n</math> zeroes. How many positive integers less than <math>1992</math> are n ...a multiple of <math>5</math>, <math>f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)</math>.
  2 KB (358 words) - 01:54, 2 October 2020
• 1992 AIME Problems/Problem 13
  ...h> then plugging in <math>a+n</math> for <math>x</math> gives us <math>{(a+n)^2 \ge 0}</math>). ===Solution 4 (Involves Basic Calculus)===
  4 KB (703 words) - 02:40, 29 December 2023
• 1992 AIME Problems/Problem 9
  ...<math>n^{}_{}</math> are relatively prime positive integers, find <math>m+n^{}_{}</math>. ...to find that <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>.
  5 KB (874 words) - 10:27, 22 August 2021
• 1992 AIME Problems/Problem 8
  ...4-a_3,\ldots)</math>, whose <math>n^{\mbox{th}}_{}</math> term is <math>a_{n+1}-a_n^{}</math>. Suppose that all of the terms of the sequence <math>\Delt <cmath> a_{n} = \frac{1}{2}(n-19)(n-92) </cmath>
  5 KB (778 words) - 21:36, 3 December 2022
• 1992 AIME Problems/Problem 4
  \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} ...iangle]] do three consecutive entries occur that are in the ratio <math>3 :4 :5</math>?
  3 KB (476 words) - 14:13, 20 April 2024
• 1992 AIME Problems/Problem 3
  ...on, so that <math>\frac{n}{2n}=\frac{1}{2}</math>, and <math>\frac{n+3}{2n+4}>\frac{503}{1000}</math>. ..., <math>1000n+3000>1006n+2012</math>, so <math>n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}</math>. Thus, the answer is <math>\boxed{164}</math>.
  2 KB (251 words) - 08:05, 2 January 2024
• 1992 AIME Problems/Problem 2
  ...> potential ascending numbers, one for each [[subset]] of <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9\}</math>. ...count each case individually: <math>\binom{n}{0}+\binom{n}{1}+...\binom{n}{n}</math> so the 2 statements are equivalent. Therfore we have <math>2^9-\bin
  2 KB (336 words) - 05:18, 4 November 2022
• 2006 AMC 10B Problems/Problem 22
  ...math>, <math>J</math>, and <math>N</math> are positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? ...h>, so the peanut butter and jam for <math>N</math> sandwiches costs <math>N(4B+5J)\cent</math>.
  2 KB (394 words) - 00:51, 25 November 2023
• 1993 AIME Problems/Problem 15
  ...</math> and <math>n\,</math> are relatively prime integers. Find <math>m + n\,</math>. A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H);
  3 KB (449 words) - 21:39, 21 September 2023
• 1993 AIME Problems/Problem 14
  ...ath>\sqrt{N}\,</math>, for a positive integer <math>N\,</math>. Find <math>N\,</math>. ...s of this rectangle be <math>A(4,y)</math>, <math>B(-x,3)</math>, <math>C(-4,-y)</math> and <math>D(x,-3)</math> for nonnegative <math>x,y</math>. Then
  3 KB (601 words) - 09:25, 19 November 2023
• 1993 AIME Problems/Problem 13
  ...0</math>. So, <math>\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}</math>. ...n^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}</math>.
  8 KB (1,231 words) - 20:06, 26 November 2023
• 1993 AIME Problems/Problem 12
  ...and <math>P_n\,</math> is the most recently obtained point, then <math>P_{n + 1}^{}</math> is the midpoint of <math>\overline{P_n L}</math>. Given tha ...e coordinates stay within the triangle. We have <cmath>P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\bmod{560},\ 2y_n\bmod{420})</cmath>
  4 KB (611 words) - 13:59, 15 July 2023
• 1993 AIME Problems/Problem 11
  ...atively prime positive integers. What are the last three digits of <math>m+n\,</math>? ...frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\cdots = \frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}</math>, and the probability of the second person winning is <
  7 KB (1,058 words) - 20:57, 22 December 2020
• 1993 AIME Problems/Problem 8
  ...ion is double counted, except the case where both <math>m</math> and <math>n</math> contain all <math>6</math> elements of <math>S.</math> So our final ...possibilities; this is because it must contain all of the "missing" <math>n - k</math> elements and thus has a choice over the remaining <math>k.</math
  9 KB (1,400 words) - 14:09, 12 January 2024
• 1993 AIME Problems/Problem 7
  ...ot6^2}{C(1000,6)\cdot6!}=\frac14,</cmath> from which the answer is <math>1+4=\boxed{005}.</math> ...ath>p = \frac{3 + 2}{20} = \frac{1}{4}</math>, and the answer is <math>1 + 4 = \boxed{005}</math>.
  5 KB (772 words) - 09:04, 7 January 2022
• 1993 AIME Problems/Problem 6
  ...series of consecutive integers as <math>a,\ b,\ c</math>. Therefore, <math>n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55</math>. Simpli Let the desired integer be <math>n</math>. From the information given, it can be determined that, for positive
  3 KB (524 words) - 18:06, 9 December 2023
• 1993 AIME Problems/Problem 5
  ...r [[integer]]s <math>n \ge 1\,</math>, define <math>P_n(x) = P_{n - 1}(x - n)\,</math>. What is the [[coefficient]] of <math>x\,</math> in <math>P_{20} Using the formula for the sum of the first <math>n</math> numbers, <math>1 + 2 + \cdots + 20 = \frac{20(20+1)}{2} = 210</math>
  2 KB (355 words) - 13:25, 31 December 2018
• 1993 AIME Problems/Problem 4
  ...n</math>. From <math>a + d = b + c</math>, <math>d = b + c - a = a + 2m + n</math>. ...+ m</math>, <math>c = a + m + n</math>, and <math>d = b + c - a = a + 2m + n</math> into <math>bc - ad = 93</math>,
  8 KB (1,343 words) - 16:27, 19 December 2023
• 1993 AIME Problems/Problem 3
  ...many contestants caught <math>n\,</math> fish for various values of <math>n\,</math>. <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
• 1993 AIME Problems/Problem 2
  ...ast, etc. If the candidate went <math>n^{2}_{}/2</math> miles on the <math>n^{\mbox{th}}_{}</math> day of this tour, how many miles was he from his star ...,\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}</math>, and so on. The E/W displacement is thus <math>1^
  2 KB (241 words) - 11:56, 13 March 2015
• 1994 AIME Problems/Problem 15
  ...0.7), dashes = linetype("8 6")+linewidth(0.7)+blue, bluedots = linetype("1 4") + linewidth(0.7) + blue; D(D(MP("A",A)) -- D(MP("B",B)) -- D(MP("C",C,N)) -- cycle);
  4 KB (717 words) - 22:20, 3 June 2021
• 1994 AIME Problems/Problem 14
  ...), C = MP("C",D((1,0))), A = MP("A",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180); ...,C+.4*expi(beta*pi/180)));MP("\beta",B,(5,1.2),fontsize(9));MP("\alpha",C,(4,1.2),fontsize(9));
  2 KB (303 words) - 00:03, 28 December 2017
• 1994 AIME Problems/Problem 13
  ...i}{10}\right)</math> after expanding. Here <math>k</math> ranges from 0 to 4 because two angles which sum to <math>2\pi</math> are involved in the produ The expression to find is <math>\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}</math>.
  3 KB (375 words) - 23:46, 6 August 2021
• 1994 AIME Problems/Problem 12
  ...row. Then <math>6|n</math>, and our goal is to maximize the value of <math>n</math>. ...th>, which we can easily verify works, and the answer is <math>\frac{13}{6}n^2 = \boxed{702}</math>.
  3 KB (473 words) - 17:06, 1 January 2024
• 1994 AIME Problems/Problem 11
  ...tower 94 bricks tall. Each brick can be oriented so it contributes <math>4''\,</math> or <math>10''\,</math> or <math>19''\,</math> to the total heigh We have the smallest stack, which has a height of <math>94 \times 4</math> inches. Now when we change the height of one of the bricks, we eithe
  4 KB (645 words) - 15:12, 15 July 2019
• 1994 AIME Problems/Problem 10
  ...and <math>n\,</math> are relatively prime positive integers. Find <math>m+n.\,</math> ...\frac{BC}{AB} = \frac{29^2 x}{29x^2} = \frac{29}{421}</math>, and <math>m+n = \boxed{450}</math>.
  3 KB (534 words) - 16:23, 26 August 2018
• 1994 AIME Problems/Problem 5
  ...<math>n\,</math>. (If <math>n\,</math> has only one digits, then <math>p(n)\,</math> is equal to that digit.) Let ...a three-digit number (so <math>5 \equiv 005</math>), and since our <math>p(n)</math> ignores all of the zero-digits, replace all of the <math>0</math>s
  2 KB (275 words) - 19:27, 4 July 2013
• 1994 AIME Problems/Problem 3
  <cmath>T_{n-1} + T_n = n^2,</cmath> where <math>T_n = 1+2+...+n = \frac{n(n+1)}{2}</math> is the <math>n</math>th triangular number.
  2 KB (252 words) - 11:12, 3 July 2023
• 1994 AIME Problems/Problem 2
  ...ath>, where <math>m</math> and <math>n</math> are integers. Find <math>m + n</math>. ...adratic formula]] shows that the answer is <math>\frac{16 \pm \sqrt{16^2 + 4 \cdot 240}}{2} = 8 \pm \sqrt{304}</math>. Discard the negative root, so our
  2 KB (272 words) - 03:53, 23 January 2023
• 1994 AIME Problems/Problem 1
  ...h term, so <math>n = 4 + (997-1) \cdot 3 = 2992</math>. The value of <math>n^2 - 1 = 2992^2 - 1 \pmod{1000}</math> is <math>\boxed{063}</math>.
  946 bytes (139 words) - 21:05, 1 September 2023
• 1995 AIME Problems/Problem 15
  ...and <math>n_{}</math> are relatively prime positive integers, find <math>m+n</math>. ...s can occur in a row, so the sequence is blocks of <math>1</math> to <math>4</math> <tt>H</tt>'s separated by <tt>T</tt>'s and ending in <math>5</math>
  6 KB (979 words) - 13:20, 11 April 2022
• 1995 AIME Problems/Problem 14
  ...}</math> is not divisible by the square of any prime number. Find <math>m+n+d.</math> ...; D(MP("A",A,NW)--MP("B",B,SE)); D(MP("C",C,NE)--MP("D",D,SW)); D(MP("E",E,N)); D(C--B--O--E,d);D(O--D(MP("F",F,NE)),d); MP("39",(B+F)/2,NE);MP("30",(C+
  3 KB (484 words) - 13:11, 14 January 2023
• 1995 AIME Problems/Problem 13
  Let <math>f(n)</math> be the integer closest to <math>\sqrt[4]{n}.</math> Find <math>\sum_{k=1}^{1995}\frac 1{f(k)}.</math> ...}\right)^4 \right\rfloor</math> values of <math>n</math> for which <math>f(n) = k</math>. Expanding using the [[binomial theorem]],
  2 KB (287 words) - 01:25, 12 December 2019
• 1995 AIME Problems/Problem 12
  ...> where <math>m_{}</math> and <math>n_{}</math> are integers, find <math>m+n.</math> // n = normal to plane
  8 KB (1,172 words) - 21:57, 22 September 2022
• 1995 AIME Problems/Problem 10
  Let our answer be <math>n</math>. Write <math> n = 42a + b </math>, where <math>a, b</math> are positive integers and <math> ...math>5</math> is the only prime divisible by <math>5</math>. We get <math> n = 215</math> as our largest possibility in this case.
  3 KB (436 words) - 19:26, 2 September 2023
• 1995 AIME Problems/Problem 9
  ...ntsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE
  7 KB (1,181 words) - 13:47, 3 February 2023
• 1995 AIME Problems/Problem 8
  ..._{y=1}^{99} \left\lfloor\frac{100-y}{y(y+1)} \right\rfloor = 49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.</cmath> ...+1)<100</math>. This yields the equations <math>x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9</math>.
  4 KB (646 words) - 17:37, 1 January 2024
• 1995 AIME Problems/Problem 7
  Given that <math>(1+\sin t)(1+\cos t)=5/4</math> and ...ath>m_{}</math> and <math>n_{}</math> [[relatively prime]], find <math>k+m+n.</math>
  3 KB (427 words) - 09:23, 13 December 2023
• 1995 AIME Problems/Problem 5
  ...lues of <math>a, b, c,</math> and <math>d_{},</math> the equation <math>x^4+ax^3+bx^2+cx+d=0</math> has four non-real roots. The product of two of the ...conjugate of <math>m</math>, and <math>n'</math> be the conjugate of <math>n</math>. Then,
  3 KB (451 words) - 15:02, 6 September 2021
• 1995 AIME Problems/Problem 4
  <cmath>PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}</cmath> D(A--MP("A_9",G,N)); D(B--MP("A_3",F,N)); D(C--MP("A_6",D,N)); D(A--P); D(rightanglemark(A,G,P,12));
  3 KB (605 words) - 11:30, 5 May 2024
• 1995 AIME Problems/Problem 3
  ...h> and <math>n</math> are relatively prime positive integers, find <math>m+n.</math> ...)</math>, so the number of steps the object may have taken is either <math>4</math> or <math>6</math>.
  3 KB (602 words) - 23:15, 16 June 2019
• 1995 AIME Problems/Problem 1
  ...> and <math>n</math> are relatively prime positive integers. Find <math>m-n.</math> :<math>1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2</math>
  2 KB (302 words) - 19:29, 4 July 2013
• 1996 AIME Problems/Problem 15
  pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); D(MP("A",A,N)--MP("B",B)--MP("C",C)--MP("D",D,N)--cycle); D(B--D); D(A--C); D(MP("O",O,SE));
  5 KB (710 words) - 21:04, 14 September 2020
• 1996 AIME Problems/Problem 13
  ...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. BD^2 + DE^2 &= \frac{15}{4} \\
  3 KB (521 words) - 01:18, 25 February 2016
• 1996 AIME Problems/Problem 12
  ...imes because there are <math>5</math> places <math>a_n</math> and <math>a_{n + 1}</math> can be. To find all possible values for <math>|a_n - a_{n - 1}|</math> we have to compute
  5 KB (879 words) - 11:23, 5 September 2021
• 1996 AIME Problems/Problem 11
  Let <math>\mathrm {P}</math> be the product of the [[root]]s of <math>z^6+z^4+z^3+z^2+1=0</math> that have a positive [[imaginary]] part, and suppose tha 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\
  6 KB (1,022 words) - 20:23, 17 April 2021
• 1996 AIME Problems/Problem 9
  ...even lockers closed. Then he opens the lockers that are multiples of <math>4</math>, leaving only lockers <math>2 \pmod{8}</math> and <math>6 \pmod{8}</ .../math> and <math>6 \pmod{8}</math> are just lockers that are <math>2 \pmod{4}</math>. Edit by [[User: Yiyj1|Yiyj1]]
  3 KB (525 words) - 23:51, 6 September 2023
• 1996 AIME Problems/Problem 7
  fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5));
  4 KB (551 words) - 11:44, 26 June 2020
• 1996 AIME Problems/Problem 6
  ...th>m</math> and <math>n</math> are relatively prime integers. Find <math>m+n</math>. ...lity that one team wins all games is <math>5\cdot \left(\frac{1}{2}\right)^4=\frac{5}{16}</math>.
  3 KB (461 words) - 00:33, 16 May 2024
• 1996 AIME Problems/Problem 3
  ...[[integer]] <math>n</math> for which the expansion of <math>(xy-3x+7y-21)^n</math>, after like terms have been collected, has at least 1996 terms. ...>y</math> and so none of the terms will need to be collected. Hence <math>(n+1)^2 \ge 1996</math>, the smallest square after <math>1996</math> is <math>
  3 KB (515 words) - 04:29, 27 November 2023
• 1996 AIME Problems/Problem 2
  .../math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even integer? ...<math>n</math> must satisfy these [[inequality|inequalities]] (since <math>n < 1000</math>):
  1 KB (163 words) - 19:31, 4 July 2013
• 1997 AIME Problems/Problem 14
  ...n</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m+n</math>. ...\pi/1997</math>, and let <math>w</math> be the root corresponding to <math>n\theta=2n\pi/ 1997</math>. Then
  5 KB (874 words) - 22:30, 1 April 2022
• 1997 AIME Problems/Problem 11
  ...h>x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}</math>. What is the greatest integer that does not exceed <math>100x ...m_{n=1}^{44} \sin n} = \frac{\sum_{n=46}^{89} \sin n}{\sum_{n=1}^{44} \sin n} = \frac {\sin 89 + \sin 88 + \dots + \sin 46}{\sin 1 + \sin 2 + \dots + \s
  10 KB (1,514 words) - 14:35, 29 March 2024
• 1997 AIME Problems/Problem 4
  ...and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> [[Image:1997_AIME-4.png]]
  2 KB (354 words) - 22:33, 2 February 2021
• 1997 AIME Problems/Problem 3
  ...also <math>0\pmod{10+b}</math>. Rewrite <math>(100x+10y+z)</math> as <math>n\times(10a+b)</math>. <math>(90a+9b-1)\times n(10a+b)= 1000(10a + b)</math>
  2 KB (375 words) - 19:34, 4 August 2021
• 1998 AIME Problems/Problem 15
  ...is thereby represented by a directed segment from one vertex of the <math>n</math> -gon to another, and a proper sequence is represented as a path that ...it can be arranged that <math>n-2</math> segments will emanate from <math>n-2</math> of the vertices and that an odd number of segments will emanate fr
  9 KB (1,671 words) - 22:10, 15 March 2024
• 1998 AIME Problems/Problem 14
  ..., where <math>m, n,</math> and <math>p</math> are integers, and <math>m\le n\le p.</math> What is the largest possible value of <math>p</math>? <cmath>2mnp = (m+2)(n+2)(p+2)</cmath>
  2 KB (390 words) - 21:05, 29 May 2023
• 1998 AIME Problems/Problem 11
  label("\((15,20,20)\)",Pa,N); We can find the lengths of the sides of the polygons now. There are 4 [[right triangle]]s with legs of length 5 and 10, so their [[hypotenuse]]s
  7 KB (1,084 words) - 11:48, 13 August 2023
• 1998 AIME Problems/Problem 8
  | 0 || 1 || 2 || 3 || 4 || 5 || 6 *<math>n \equiv 0 \pmod{2}\quad\quad F_{n-1}\cdot 1000-F_n\cdot x</math>
  2 KB (354 words) - 19:37, 24 September 2023
• 1998 AIME Problems/Problem 7
  ...er]]s that satisfy <math>\sum_{i = 1}^4 x_i = 98.</math> Find <math>\frac n{100}.</math> ..._i + 1</math> will be odd. Substituting we get <cmath>2y_1+2y_2+2y_3+2y_4 +4 = 98 \implies y_1+y_2+y_3+y_4 =47</cmath>
  5 KB (684 words) - 11:41, 13 August 2023
• 1998 AIME Problems/Problem 5
  ...dd. <math>\frac {k(k-1)}2</math> will be even if <math>4|k</math> or <math>4|k-1</math>, and odd otherwise. ...c{(n)(n-1)}{2} + \frac{(n+1)(n)}{2} = \left(\frac n2\right)(n+1 - (n-1)) = n</math>. So the first two fractions add up to <math>19</math>, the next two
  1 KB (225 words) - 02:20, 16 September 2017
• 1998 AIME Problems/Problem 4
  ...ath>n</math> are [[relatively prime]] [[positive integer]]s. Find <math>m+n.</math> ...and an odd tile. Thus, since there are <math>5</math> odd tiles and <math>4</math> even tiles, the only possibility is that one player gets <math>3</ma
  5 KB (917 words) - 02:37, 12 December 2022

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