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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) - 17:46, 24 May 2024
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

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