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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
  6 KB (902 words) - 17:40, 19 May 2024
• 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

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