Create the page "N^4 4" on this wiki! See also the search results found.

• 1989 AIME Problems
  <center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center> == Problem 4 ==
  7 KB (1,045 words) - 20:47, 14 December 2023
• 1990 AIME Problems
  == Problem 4 == [[1990 AIME Problems/Problem 4|Solution]]
  6 KB (870 words) - 10:14, 19 June 2021
• 1991 AIME Problems
  ...le <math>ABCD_{}^{}</math> has sides <math>\overline {AB}</math> of length 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</m == Problem 4 ==
  7 KB (1,106 words) - 22:05, 7 June 2021
• 1992 AIME Problems
  == Problem 4 == \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt}
  8 KB (1,117 words) - 05:32, 11 November 2023
• 1993 AIME Problems
  ...etc. If the candidate went <math>\frac{n^{2}}{2}</math> miles on the <math>n^{\mbox{th}}_{}</math> day of this tour, how many miles was he from his star ...many contestants caught <math>n\,</math> fish for various values of <math>n\,</math>.
  8 KB (1,275 words) - 06:55, 2 September 2021
• 1994 AIME Problems
  ..., where <math>m\,</math> and <math>n\,</math> are integers. Find <math>m + n\,</math>. == Problem 4 ==
  7 KB (1,141 words) - 07:37, 7 September 2018
• 1995 AIME Problems
  ...> and <math>n</math> are relatively prime positive integers. Find <math>m-n.</math> ...h> and <math>n</math> are relatively prime positive integers, find <math>m+n.</math>
  6 KB (1,000 words) - 00:25, 27 March 2024
• 1996 AIME Problems
  .../math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even integer? ...tive 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.
  6 KB (931 words) - 17:49, 21 December 2018
• 1997 AIME Problems
  ...and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> == Problem 4 ==
  7 KB (1,098 words) - 17:08, 25 June 2020
• 1998 AIME Problems
  == Problem 4 == ...ath>n</math> are [[relatively prime]] [[positive integer]]s. Find <math>m+n.</math>
  7 KB (1,084 words) - 02:01, 28 November 2023
• 1999 AIME Problems
  ...nd <math>n_{}</math> are relatively prime positive integers. Find <math>m+n.</math> Find the sum of all positive integers <math>n</math> for which <math>n^2-19n+99</math> is a perfect square.
  7 KB (1,094 words) - 13:39, 16 August 2020
• 2000 AIME I Problems
  {{AIME Problems|year=2000|n=I}} ...he least positive integer <math>n</math> such that no matter how <math>10^{n}</math> is expressed as the product of any two positive integers, at least
  7 KB (1,204 words) - 03:40, 4 January 2023
• 2001 AIME I Problems
  {{AIME Problems|year=2001|n=I}} == Problem 4 ==
  7 KB (1,212 words) - 22:16, 17 December 2023
• 2002 AIME I Problems
  {{AIME Problems|year=2002|n=I}} ...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  8 KB (1,374 words) - 21:09, 27 July 2023
• 2003 AIME I Problems
  {{AIME Problems|year=2003|n=I}} <center><math> \frac{((3!)!)!}{3!} = k \cdot n!, </math></center>
  6 KB (965 words) - 16:36, 8 September 2019
• 2000 AIME II Problems
  {{AIME Problems|year=2000|n=II}} ...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  6 KB (947 words) - 21:11, 19 February 2019
• 2001 AIME II Problems
  {{AIME Problems|year=2001|n=II}} .../math> forms a perfect square. What are the leftmost three digits of <math>N</math>?
  8 KB (1,282 words) - 21:12, 19 February 2019
• 2002 AIME II Problems
  {{AIME Problems|year=2002|n=II}} == Problem 4 ==
  7 KB (1,177 words) - 15:42, 11 August 2023
• 2003 AIME II Problems
  {{AIME Problems|year=2003|n=II}} ...is the sum of the other two. Find the sum of all possible values of <math>N</math>.
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1983 AIME Problems/Problem 3
  ...he roots of this equation are real, since its discriminant is <math>18^2 - 4 \cdot 1 \cdot 20 = 244</math>, which is positive. Thus by [[Vieta's formula ...he square root of a real number can't be negative, the only possible <math>n</math> is <math>5</math>.
  3 KB (532 words) - 05:18, 21 July 2022
• 1983 AIME Problems/Problem 4
  <cmath> AC = \sqrt{AB^2 + BC^2} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}. </cmath> string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"};
  11 KB (1,741 words) - 22:40, 23 November 2023
• 1983 AIME Problems/Problem 6
  Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <mat ...ively, we could have noted that <math>a^b\equiv a^{b\pmod{\phi{(n)}}}\pmod n</math>. This way, we have <math>6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\
  3 KB (361 words) - 20:20, 14 January 2023
• 1983 AIME Problems/Problem 11
  label("E",E,N); label("F",F,N);
  5 KB (865 words) - 21:11, 6 February 2023
• 1983 AIME Problems/Problem 13
  ...simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<!-- don't remove the following tag, for PoTW on the Wiki front pa ...<math>S</math> be a non-[[empty set | empty]] [[subset]] of <math>\{1,2,3,4,5,6\}</math>.
  5 KB (894 words) - 22:02, 5 April 2024
• 1983 AIME Problems/Problem 14
  ...abel("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);</asy> ...,b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0);
  13 KB (2,149 words) - 18:44, 5 February 2024
• 1983 AIME Problems/Problem 15
  ...rational number. If this number is expressed as a fraction <math>\frac{m}{n}</math> in lowest terms, what is the product <math>mn</math>? label("$D$",D,N);
  19 KB (3,221 words) - 01:05, 7 February 2023
• 1984 AIME Problems/Problem 1
  A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>. We can substi == Solution 4 ==
  4 KB (576 words) - 21:03, 23 December 2023
• 1984 AIME Problems/Problem 3
  ...h>t_{2}</math>, and <math>t_{3}</math> in the figure, have [[area]]s <math>4</math>, <math>9</math>, and <math>49</math>, respectively. Find the area of pair A=(0,0),B=(12,0),C=(4,5);
  4 KB (726 words) - 13:39, 13 August 2023
• 1984 AIME Problems/Problem 5
  ...</math>. If we multiply the two equations together, we get that <math>a^4b^4 = 2^{36}</math>, so taking the fourth root of that, <math>ab = 2^9 = \boxed ...ac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2</math>. This means that <math>\frac{\ln ab}{\ln 2} = 9</math
  6 KB (863 words) - 16:10, 16 May 2024
• 1984 AIME Problems/Problem 7
  ...nction]] f is defined on the [[set]] of [[integer]]s and satisfies <math>f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\
  4 KB (617 words) - 22:09, 15 May 2024
• 1984 AIME Problems/Problem 9
  triple M,N,P[],Q[]; int n=s.length;
  6 KB (947 words) - 20:44, 26 November 2021
• 1984 AIME Problems/Problem 11
  ...robability]] that no two birch trees are next to one another. Find <math>m+n</math>. ...e out one tree between each pair of birch trees. So you would remove <math>4</math> trees that aren't birch. What you are left with is a unique arrangem
  7 KB (1,115 words) - 00:52, 7 September 2023
• 1984 AIME Problems/Problem 14
  ...>21</math>, ... , and in general <math>9 + 6n</math> for nonnegative <math>n</math> are odd composites. We now have 3 cases: ...th> can be expressed as <math>9 + (9+6n)</math> for some nonnegative <math>n</math>. Note that <math>9</math> and <math>9+6n</math> are both odd composi
  8 KB (1,346 words) - 01:16, 9 January 2024
• 1985 AIME Problems/Problem 14
  ...tal, for an average of 9. Thus we must have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = \boxed{25}</math>. ...nts vs them, which is a contradiction since it must be larger. Thus, <math>n=\boxed{25}</math>.
  5 KB (772 words) - 22:14, 18 June 2020
• 1985 AIME Problems/Problem 13
  ...<math>a_{n+1}</math>. Find the maximum value of <math>d_n</math> as <math>n</math> ranges through the [[positive integer]]s. ...h>2n+1</math> if it is going to divide the entire [[expression]] <math>100+n^2+2n+1</math>.
  4 KB (671 words) - 20:04, 6 March 2024
• 1985 AIME Problems/Problem 12
  ...when it has crawled exactly <math>7</math> meters. Find the value of <math>n</math>. P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\
  17 KB (2,837 words) - 13:34, 4 April 2024
• 1985 AIME Problems/Problem 10
  ...o <math>\frac n2</math> to <math>\frac {n+1}2</math> for any integer <math>n</math> (same reasoning as above). So now we only need to test every 10 numb *<math>4</math>: We can partition as <math>1+1+2</math>, and from the previous case
  12 KB (1,859 words) - 18:16, 28 March 2022
• 1985 AIME Problems/Problem 5
  ...is chosen so that <math>a_n = a_{n - 1} - a_{n - 2}</math> for each <math>n \ge 3</math>. What is the sum of the first 2001 terms of this sequence if t ...h>n</math> times, <math>a_{j + 6n} = a_j</math> for all [[integer]]s <math>n</math> and <math>j</math>.
  2 KB (410 words) - 13:37, 1 May 2022
• 1985 AIME Problems/Problem 4
  ...division points closest to the opposite vertices. Find the value of <math>n</math> if the the [[area]] of the small square is exactly <math>\frac1{1985 ...2 - 2n + 1 = 1985</math>. Solving this [[quadratic equation]] gives <math>n = \boxed{32}</math>.
  3 KB (484 words) - 21:40, 2 March 2020
• 1986 AIME Problems/Problem 15
  ...> and <math>B</math> lie along the lines <math>y=x+3</math> and <math>y=2x+4</math> respectively, find the area of triangle <math>ABC</math>. label("$B$", B, N);
  11 KB (1,722 words) - 09:49, 13 September 2023
• 1986 AIME Problems/Problem 13
  ...s this just becomes the ball-and-urn argument. We want to add 5 balls into 4 urns, which is the same as 3 dividers; hence this gives <math>{{5+3}\choose .../math> and multiplication, the answer is <math>{{2+4-1}\choose2} \cdot{{5+4-1}\choose5}=560</math> ~Slight edits in LaTeX by EthanSpoon
  4 KB (772 words) - 21:09, 7 May 2024
• 1986 AIME Problems/Problem 11
  === Solution 4(calculus) === ...ix} ,</math> where the term <math>\dbinom{n}{k}</math> is negated if <math>n+k</math> is odd.
  6 KB (872 words) - 16:51, 9 June 2023
• 1986 AIME Problems/Problem 10
  ...>. Play the role of the magician and determine <math>(abc)</math> if <math>N= 3194</math>. Let <math>n=abc</math> then
  3 KB (565 words) - 16:51, 1 October 2023
• 1986 AIME Problems/Problem 9
  D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle); pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N));
  11 KB (1,850 words) - 18:07, 11 October 2023
• 1986 AIME Problems/Problem 8
  ...So the overall power of <math>2</math> and <math>5</math> is <math>7(1+2+3+4+5+6) = 7 \cdot 21 = 147</math>. However, since the question asks for proper ...d(n))/2}</math>, where <math>d(n)</math> is the number of divisor of <math>n</math>.
  3 KB (487 words) - 20:52, 16 September 2020
• 1986 AIME Problems/Problem 7
  The increasing [[sequence]] <math>1,3,4,9,10,12,13\cdots</math> consists of all those positive [[integer]]s which a ...o determine the 100th number. <math>100</math> is equal to <math>64 + 32 + 4</math>, so in binary form we get <math>1100100</math>. However, we must cha
  5 KB (866 words) - 00:00, 22 December 2022
• 1986 AIME Problems/Problem 5
  ...]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>? ...<math>900</math>. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides <math>900</math> is <math>\boxed{890}</math>; we can doub
  2 KB (338 words) - 19:56, 15 October 2023
• 1987 AIME Problems/Problem 15
  == Solution 4 (Algebra) == label("$A$",A,N);
  5 KB (838 words) - 18:05, 19 February 2022
• 1987 AIME Problems/Problem 14
  ...(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.</cmath> ...^2 + 2b^2 + 2ab\right).</math> Each of the terms is in the form of <math>x^4 + 324.</math> Using Sophie Germain, we get that
  7 KB (965 words) - 10:42, 12 April 2024
• 1987 AIME Problems/Problem 12
  ...s a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>. In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.
  4 KB (673 words) - 19:48, 28 December 2023
• 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

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