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• 1993 AIME Problems/Problem 2
  ...^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|</math>. Applying [[difference of squares]], we see that ...)^2 - (4i+3)^2}{2}\right| &= \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right|\\ &= \left|\sum_{i=0}^9 -(8i+4) \right|.
  2 KB (241 words) - 11:56, 13 March 2015
• 1993 AIME Problems/Problem 1
  The thousands digit is <math>\in \{4,5,6\}</math>. Case <math>1</math>: Thousands digit is even
  3 KB (440 words) - 21:20, 22 July 2021
• 1994 AIME Problems/Problem 15
  ...rs and <math>s\,</math> is not divisible by the square of any prime. What is <math>q+r+s\,</math>? ...th>; the [[locus]] of each of the respective conditions for <math>P</math> is the region inside the (semi)circles with diameters <math>\overline{AB}, \ov
  4 KB (717 words) - 22:20, 3 June 2021
• 1994 AIME Problems/Problem 13
  ...\text{cis}\left(\frac {(2k + 1)\pi}{10}\right)</math> where <math>k</math> is an integer between <math>0</math> and <math>9</math>. The expression to find is <math>\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}</
  3 KB (375 words) - 23:46, 6 August 2021
• 1994 AIME Problems/Problem 12
  ...the sides of the squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can be partiti ...ac {13}6n</math> squares in every row. Then <math>6|n</math>, and our goal is to maximize the value of <math>n</math>.
  3 KB (473 words) - 17:06, 1 January 2024
• 1994 AIME Problems/Problem 11
  ...multiples of <math>3</math>, the change will always be a multiple of <math>3</math>, so we just need to find the number of changes we can get from <math ...r <math>m,n</math> being [[positive integer]]s is <math>5 \times 2 - 5 - 2=3</math>.
  4 KB (645 words) - 15:12, 15 July 2019
• 1994 AIME Problems/Problem 10
  ...ths of the sides of <math>\triangle ABC\,</math> are integers, <math>BD=29^3,\,</math> and <math>\cos B=m/n\,</math>, where <math>m\,</math> and <math>n ...he form <math>29^2 x</math> and <math>29 x^2</math>, respectively, where x is an integer.
  3 KB (534 words) - 16:23, 26 August 2018
• 1994 AIME Problems/Problem 9
  ...nues until the bag is empty. The probability that the bag will be emptied is <math>p/q,\,</math> where <math>p\,</math> and <math>q\,</math> are relativ *Case 1. We draw a pair on the first two cards. The second card is the same as the first with probability <math>\frac {1}{2k - 1}</math>, then
  3 KB (589 words) - 14:18, 21 July 2019
• 1994 AIME Problems/Problem 8
  Consider the points on the [[complex plane]]. The point <math>b+37i</math> is then a rotation of <math>60</math> degrees of <math>a+11i</math> about the ...)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.</cmath>
  5 KB (788 words) - 13:53, 8 July 2023
• 1994 AIME Problems/Problem 7
  has at least one solution, and each solution is an ordered pair <math>(x,y)\,</math> of integers. How many such ordered pa ...h> where the signs are all independent of each other, for a total of <math>3\cdot 2\cdot 2=12</math> lattice points. They are indicated by the blue dots
  3 KB (442 words) - 19:51, 8 January 2024
• 1994 AIME Problems/Problem 6
  <center><math>y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,</math></center> ...the plane into equilateral triangles of side length <math>\tfrac{2}{\sqrt{3}}.\,</math> How many such triangles are formed?
  4 KB (721 words) - 16:14, 8 March 2021
• 1994 AIME Problems/Problem 5
  ...math>. (If <math>n\,</math> has only one digits, then <math>p(n)\,</math> is equal to that digit.) Let <center><math>S=p(1)+p(2)+p(3)+\cdots+p(999)</math></center>.
  2 KB (275 words) - 19:27, 4 July 2013
• 1994 AIME Problems/Problem 4
  \lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994 (For real <math>x\,</math>, <math>\lfloor x\rfloor\,</math> is the greatest integer <math>\le x.\,</math>)
  2 KB (264 words) - 13:33, 11 August 2018
• 1994 AIME Problems/Problem 2
  ...ngent at <math>P</math> to a circle of radius 20. Square <math>ABCD</math> is constructed with <math>A</math> and <math>B</math> on the larger circle, <m ...that <math>OE = AB - OQ = AB - 10</math>. The other leg, <math>AE</math>, is just <math>\frac 12 AB</math>.
  2 KB (272 words) - 03:53, 23 January 2023
• 1994 AIME Problems/Problem 1
  ...square]]. What is the [[remainder]] when the 1994th term of the sequence is divided by 1000? ...= 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
  ...ld start with a block of <tt>H</tt>'s, the total probability is that <math>3/2</math> of it has to start with an <tt>H</tt>. ...ose sum is <math>\frac{3/64}{1-(15/32)}=\frac{3}{34}</math>, so the answer is <math>\boxed{037}</math>.
  6 KB (979 words) - 13:20, 11 April 2022
• 1995 AIME Problems/Problem 14
  ...,</math> and <math>d_{}</math> are positive integers and <math>d_{}</math> is not divisible by the square of any prime number. Find <math>m+n+d.</math> ...math>, and <math>EF = \sqrt{OE^2 - OF^2} = 9</math>. Then <math>OEF</math> is a <math>30-60-90</math> [[right triangle]], so <math>\angle OEB = \angle OE
  3 KB (484 words) - 13:11, 14 January 2023
• 1995 AIME Problems/Problem 13
  ...- \left(k^4 - 2k^3 + \frac 32k^2 - \frac 12k + \frac 1{16}\right)\\ &= 4k^3 + k. \end{align*}</cmath> ...^3 + k</math> times, and the sum for each <math>k</math> is then <math>(4k^3 + k) \cdot \frac{1}{k} = 4k^2 + 1</math>. From <math>k = 1</math> to <math>
  2 KB (287 words) - 01:25, 12 December 2019
• 1995 AIME Problems/Problem 12
  ...ity]], let <math>AP = 1.</math> It follows that <math>\triangle OPA</math> is a <math>45-45-90</math> [[right triangle]], so <math>OP = AP = 1,</math> <m ...= 1 + 1 - 2\cos \theta \Longrightarrow \cos \theta = - 3 + 2\sqrt {2} = - 3 + \sqrt{8}.</cmath>
  8 KB (1,172 words) - 21:57, 22 September 2022
• 1995 AIME Problems/Problem 11
  ...of <math>P_{}</math> cuts <math>P_{}</math> into two prisms, one of which is [[similar]] to <math>P_{},</math> and both of which have nonzero volume. G ...nce <math>x < a, y < b, z < c</math>, it follows that the only possibility is <math>y=a,z=b=1995</math>. Then,
  2 KB (292 words) - 19:30, 4 July 2013
• 1995 AIME Problems/Problem 10
  What is the largest positive integer that is not the sum of a positive integral multiple of <math>42</math> and a positi ...h>\mod {42}</math> must be a [[composite]] number. Also, every number that is a multiple of <math>42</math> greater than that prime number must also be p
  3 KB (436 words) - 19:26, 2 September 2023
• 1995 AIME Problems/Problem 9
  ...on <math>\overline{AM}</math> with <math>AD=10</math> and <math>\angle BDC=3\angle BAC.</math> Then the perimeter of <math>\triangle ABC</math> may be ...defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3;
  7 KB (1,181 words) - 13:47, 3 February 2023
• 1995 AIME Problems/Problem 8
  ...tory positive integers for all integers <math>y \le 100</math>. The answer is ...1}^{99} \left\lfloor\frac{100-y}{y(y+1)} \right\rfloor = 49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.</cmath>
  4 KB (646 words) - 17:37, 1 January 2024
• 1995 AIME Problems/Problem 7
  ...s t - \sin t - \cos t + 1 = \frac{13}{4} - \sqrt{10}</math>, so the answer is <math>13 + 4 + 10 = \boxed{027}</math>. ...ath>. Therefore, <math>x = \frac{13}{4} - \sqrt{10}</math>, and the answer is <math> 13 + 4 + 10 = \boxed{027}</math>.
  3 KB (427 words) - 09:23, 13 December 2023
• 1995 AIME Problems/Problem 6
  Let <math>n=2^{31}3^{19}.</math> How many positive [[integer]] [[divisor]]s of <math>n^2</math ...h>) into pairs that multiply to <math>n^2</math>, then one factor per pair is less than <math>n</math>, and so there are <math>\frac{63\times 39-1}{2} =
  2 KB (407 words) - 08:14, 4 November 2022
• 1995 AIME Problems/Problem 5
  ...ese roots is <math>13+i</math> and the sum of the other two roots is <math>3+4i,</math> where <math>i=\sqrt{-1}.</math> Find <math>b.</math> ...pairs. Let the first two roots be <math>m,n</math>. Since <math>m+n</math> is not real, <math>m,n</math> are not conjugates, so the other pair of roots m
  3 KB (451 words) - 15:02, 6 September 2021
• 1995 AIME Problems/Problem 4
  ...adius <math>9</math>. The circle of radius <math>9</math> has a chord that is a common external tangent of the other two circles. Find the square of the ...i(acos(1/3)), F=B+3*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9));
  3 KB (605 words) - 11:30, 5 May 2024
• 1995 AIME Problems/Problem 3
  ...coordinate plane]] via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let <math>p</math> be t ...reach <math>(2,2)</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 2
  ...2}</math>. Thus, the product of the two roots (both of which are positive) is <math>1995^{1+\sqrt{2}/2} \cdot 1995^{1 - \sqrt{2}/2} = 1995^2</math>, maki ...od{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}</math>, so our answer is <math>\boxed{025}</math>.
  2 KB (362 words) - 00:40, 29 January 2021
• 1995 AIME Problems/Problem 1
  ...</math> The total area enclosed by at least one of <math>S_{1}, S_{2}, S_{3}, S_{4}, S_{5}</math> can be written in the form <math>m/n,</math> where <m The sum of the areas of the [[square]]s if they were not interconnected is a [[geometric sequence]]:
  2 KB (302 words) - 19:29, 4 July 2013
• 1996 AIME Problems/Problem 15
  ...each twice as large as angle <math>DBA</math>, and angle <math>ACB</math> is <math>r</math> times as large as angle <math>AOB</math>. Find <math>\lfloor 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);
  5 KB (710 words) - 21:04, 14 September 2020
• 1996 AIME Problems/Problem 14
  A <math>150\times 324\times 375</math> [[rectangle|rectangular]] [[solid]] is made by gluing together <math>1\times 1\times 1</math> cubes. An internal [ ...point on the diagonal with coordinates <math>(ma, mb, mc)</math>. We have 3 key observations as this point moves from <math>(0,0,0)</math> towards <mat
  5 KB (923 words) - 21:21, 22 September 2023
• 1996 AIME Problems/Problem 13
  .../math> [[bisect]]s <math>\overline{BC}</math>, and <math>\angle ADB</math> is a right angle. The ratio <math>\frac{[ADB]}{[ABC]}</math> can be written in ...he problem asks for a ratio, we can divide each side length by <math>\sqrt{3}</math> to make things simpler. We now have a triangle with sides <math>\sq
  3 KB (521 words) - 01:18, 25 February 2016
• 1996 AIME Problems/Problem 12
  ...utation]] <math>a_1,a_2,a_3,\cdots,a_{10}</math> of the integers <math>1,2,3,\cdots,10</math>, form the sum + |2 - 10| + \ldots + |2 - 3| + |2 - 1|\\
  5 KB (879 words) - 11:23, 5 September 2021
• 1996 AIME Problems/Problem 11
  ...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 that <m 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 10
  ...ent|tangent]] function is <math>180^\circ</math>, and the tangent function is [[one-to-one]] over each period of its domain. Therefore, the smallest positive solution is <math>x = \boxed{159}</math>.
  4 KB (503 words) - 15:46, 3 August 2022
• 1996 AIME Problems/Problem 9
  ...s wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens? ...>, leaving lockers <math>86, 342, 598</math>, and <math>854</math>, and he is at where he started again. He then opens <math>86</math> and <math>598</mat
  3 KB (525 words) - 23:51, 6 September 2023
• 1996 AIME Problems/Problem 8
  ...rdered pairs of positive integers <math>(x,y)</math> with <math>x<y</math> is the harmonic mean of <math>x</math> and <math>y</math> equal to <math>6^{20 ...h>x<y</math>, the answer is half of the remaining number of factors, which is <math>\frac{1599-1}{2}= \boxed{799}</math>.
  1 KB (155 words) - 19:32, 4 July 2013
• 1996 AIME Problems/Problem 7
  pathpen = black; pair O = (3.5,3.5); D(O); fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4));
  4 KB (551 words) - 11:44, 26 June 2020
• 1996 AIME Problems/Problem 6
  The probability that one team wins all games is <math>5\cdot \left(\frac{1}{2}\right)^4=\frac{5}{16}</math>. Similarity, the probability that one team loses all games is <math>\frac{5}{16}</math>.
  3 KB (461 words) - 00:33, 16 May 2024
• 1996 AIME Problems/Problem 5
  ...</math>, <math>b</math>, and <math>c</math>, and that the roots of <math>x^3+rx^2+sx+t=0</math> are <math>a+b</math>, <math>b+c</math>, and <math>c+a</m ...^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0</math>, we have <math>a + b + c = s = -3</math>, <math>ab + bc + ca = 4</math>, and <math>abc = 11</math>. Then
  3 KB (585 words) - 22:08, 19 November 2022
• 1996 AIME Problems/Problem 4
  .... The area of the shadow, which does not include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed <math (Figure not to scale) The area of the square shadow base is <math>48 + 1 = 49</math>, and so the sides of the shadow are <math>7</math>
  2 KB (257 words) - 17:50, 4 January 2016
• 1996 AIME Problems/Problem 3
  ...square after <math>1996</math> is <math>2025 = 45^2</math>, so our answer is <math>45 - 1 = \boxed{044}</math>. ...</math> terms. Therefore, we need to find the smallest perfect square that is greater than <math>1996</math>. From trial and error, we get <math>44^2 = 1
  3 KB (515 words) - 04:29, 27 November 2023
• 1996 AIME Problems/Problem 2
  ...that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even integer? ...4</math> for the third, and <math>256</math> for the fourth, so the answer is <math>4+16+64+256=\boxed{340}</math>.
  1 KB (163 words) - 19:31, 4 July 2013
• 1996 AIME Problems/Problem 1
  ...c [[square]], the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Fin \multicolumn{3}{c}{\text{Table}}\\\hline
  2 KB (332 words) - 11:28, 4 August 2021
• 1997 AIME Problems/Problem 15
  ...ath>q</math>, and <math>r</math> are positive integers, and <math>q</math> is not divisible by the square of any prime number. Find <math>p+q+r</math>. ...\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i</math>.
  4 KB (609 words) - 22:49, 17 July 2023
• 1997 AIME Problems/Problem 14
  ...t <math>\frac{m}{n}</math> be the [[probability]] that <math>\sqrt{2+\sqrt{3}}\le\left|v+w\right|</math>, where <math>m</math> and <math>n</math> are [[ ...rt{3}</math>, which simplifies to <cmath>\cos((m-n)\theta) \ge \frac{\sqrt{3}}{2}</cmath>Thus, <cmath>|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi}
  5 KB (874 words) - 22:30, 1 April 2022
• 1997 AIME Problems/Problem 13
  ...<math>a</math> and <math>b</math> are positive integers and <math>b</math> is not divisible by the square of any prime number. Find <math>a+b</math>. | <math>f(3) = 0</math> || <math>f(1.1) = 0.1</math>
  7 KB (1,225 words) - 19:56, 4 August 2021
• 1997 AIME Problems/Problem 12
  ...r all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>. The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the tw
  11 KB (2,063 words) - 22:59, 21 October 2023
• 1997 AIME Problems/Problem 11
  ...{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</math>? We want to pair up <math>[1, 44]</math>, <math>[2, 43]</math>, <math>[3, 42]</math>, etc. from the numerator and <math>[46, 89]</math>, <math>[47,
  10 KB (1,514 words) - 14:35, 29 March 2024
• 1997 AIME Problems/Problem 10
  ...pe-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true: *'''Case 1''': All three attributes are the same. This is impossible since sets contain distinct cards.
  3 KB (585 words) - 19:37, 25 April 2022
• 1997 AIME Problems/Problem 9
  ...ve, <math>\langle a^{-1}\rangle=\langle a^2\rangle</math>, and <math>2<a^2<3</math>. Find the value of <math>a^{12}-144a^{-1}</math>. ...+\sqrt{5}}2</math> (the [[phi|golden ratio]]) is the answer. The following is the way to derive that:
  4 KB (586 words) - 21:53, 30 December 2023
• 1997 AIME Problems/Problem 8
  ...of the entries in each row is 0 and the sum of the entries in each column is 0? The problem is asking us for all configurations of <math>4\times 4</math> grids with 2 1's
  4 KB (638 words) - 16:41, 22 January 2024
• 1997 AIME Problems/Problem 7
  ...</math> mile per minute. At time <math>t=0</math>, the center of the storm is <math>110</math> miles due north of the car. At time <math>t=t_1</math> min ...ar is at <math>\left(\frac 23t,0\right)</math> and the center of the storm is at <math>\left(\frac{t}{2}, 110 - \frac{t}{2}\right)</math>. Using the dist
  4 KB (617 words) - 18:47, 17 July 2022
• 1997 AIME Problems/Problem 6
  ...A_n</math>, and <math>A_1A_2B</math> is an [[equilateral triangle]]. What is the largest value of <math>n</math> for which <math>A_1</math>, <math>A_n</ Clearly <math>n</math> is maximized when <math>m = 7, n = \boxed{042}</math>.
  3 KB (497 words) - 00:39, 22 December 2018
• 1997 AIME Problems/Problem 4
  ...<math>5, x, 5+r</math> and <math>5, 8+r+x, 13</math>, wher <math>x</math> is the distance between the center of the circle in question and the segment c NOTE: It can be seen that there is no apparent need to use the variable x as a 5,12,13 right triangle has been
  2 KB (354 words) - 22:33, 2 February 2021
• 1997 AIME Problems/Problem 3
  ...number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number? ...and <math>y=112</math>, which satisifies the conditions. Hence the answer is <math>112 + 14 = \boxed{126}</math>.
  2 KB (375 words) - 19:34, 4 August 2021
• 1997 AIME Problems/Problem 2
  ...is pattern can be easily generalized and we see that the number of squares is just <math>\sum^8_{i=1}{i^2}</math>. This can be simplified by using the we ...{i=1}{i}}</math>. This gets us <math>{(\frac{9\cdot8}{2})}^2,</math> which is just <math>1296.</math>
  3 KB (416 words) - 21:09, 27 October 2022
• 1998 AIME Problems/Problem 15
  ...n ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after ...the domino a set of 40 points (labeled 1 through 40) in which every point is connected with every other point. The connections represent the dominoes.
  9 KB (1,671 words) - 22:10, 15 March 2024
• 1998 AIME Problems/Problem 14
  ...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>? ...the second pair gives <math>98</math>. We now check that <math>130</math> is optimal, setting <math>a=m-2</math>, <math>b=n-2</math> in order to simplif
  2 KB (390 words) - 21:05, 29 May 2023
• 1998 AIME Problems/Problem 13
  ...n,</math> its complex power sum is defined to be <math>a_1i + a_2i^2+ a_3i^3 + \cdots + a_ni^n,</math> where <math>i^2 = - 1.</math> Let <math>S_n</mat ...we will just define to have a power sum of zero) with <math>9</math> in it is equal to the number of subsets without a <math>9</math>. To easily see this
  2 KB (384 words) - 19:02, 20 October 2023
• 1998 AIME Problems/Problem 12
  ...and <math>c</math> is not divisible by the square of any [[prime]]. What is <math>a^{2} + b^{2} + c^{2}</math>? ...math> gives <math>x = \frac {\sqrt {5} - 1}{2}</math> and <math>y = \frac {3 - \sqrt {5}}{2}</math>.
  5 KB (876 words) - 20:27, 9 June 2022
• 1998 AIME Problems/Problem 9
  ...math> and <math>c</math> are [[positive]] [[integer]]s, and <math>c</math> is not divisible by the square of any [[prime]]. Find <math>a + b + c.</math> ...e the area of the unshaded region over the area of the total region, which is the probability that the mathematicians do not meet:
  4 KB (624 words) - 18:34, 18 February 2018
• 1998 AIME Problems/Problem 8
  ...he preceding term from the one before that. The last term of the sequence is the first [[negative]] term encounted. What positive integer <math>x</math The best way to start is to just write out some terms.
  2 KB (354 words) - 19:37, 24 September 2023
• 1998 AIME Problems/Problem 7
  Note that this is an algebraic bijection, we have simplified the problem and essentially remo ...hrough 50; thus the answer is <math>n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600</math>, and <math>\frac n{100} = \boxed{196}</math>.
  5 KB (684 words) - 11:41, 13 August 2023
• 1998 AIME Problems/Problem 4
  ...values. The [[probability]] that all three players obtain an [[odd]] sum is <math>m/n,</math> where <math>m</math> and <math>n</math> are [[relatively ...that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.)
  5 KB (917 words) - 02:37, 12 December 2022
• 1998 AIME Problems/Problem 2
  ...ac 12</math> of the 30 by 30 [[square]] it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of th ...positive integers, so <math>(0,0)</math> doesn't count; hence, the answer is <math>480</math>.
  6 KB (913 words) - 16:34, 6 August 2020
• 1998 AIME Problems/Problem 1
  For how many values of <math>k</math> is <math>12^{12}</math> the [[least common multiple]] of the positive integers It is evident that <math>k</math> has only 2s and 3s in its prime factorization,
  2 KB (289 words) - 22:50, 23 April 2024
• 1999 AIME Problems/Problem 15
  ...ed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid? ...om <math>P</math> to <math>\triangle ABC</math>. The crux of this problem is the following lemma.
  7 KB (1,169 words) - 15:28, 13 May 2024
• 1999 AIME Problems/Problem 14
  ...ath> and <math>CA=15,</math> and the [[tangent]] of angle <math>PAB</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relativ /* constructing P, C is there as check */
  7 KB (1,184 words) - 13:25, 22 December 2022
• 1999 AIME Problems/Problem 12
  ...]] to <math>\overline{AB}</math> at <math>P_{},</math> and its [[radius]] is <math>21</math>. Given that <math>AP=23</math> and <math>PB=27,</math> fin We want the perimeter, which is <math>2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}</math>.
  3 KB (472 words) - 15:59, 25 February 2022
• 1999 AIME Problems/Problem 11
  ...s equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to [[telescope]] the sum. Using the ...}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},</cmath> and our answer is <math>\boxed{177}</math>.
  4 KB (614 words) - 04:38, 8 December 2023
• 1999 AIME Problems/Problem 10
  ...segments form a [[triangle]] whose vertices are among the ten given points is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relat ...e picked. Since the triangle accounts for 3 segments, there are <math>45 - 3 = 42</math> segments remaining.
  3 KB (524 words) - 17:25, 17 July 2023
• 1999 AIME Problems/Problem 9
  ...ction]] has the property that the image of each point in the complex plane is [[equidistant]] from that point and the [[origin]]. Given that <math>|a+bi ...h passes through <math>\left(\frac 12, \frac12\right)</math>, so its graph is <math>x + y = 1</math>. Substituting <math>x = (a-b)</math> and <math>y = (
  6 KB (1,010 words) - 19:01, 24 May 2023
• 1999 AIME Problems/Problem 7
  ...9</math>. At step i of a 1000-step process, the <math>i</math>-th switch is advanced one step, and so are all the other switches whose labels divide th ...f <math>\frac{N}{d}</math> must be a multiple of 4 to ensure that a switch is in position A:
  3 KB (475 words) - 13:33, 4 July 2016
• 1999 AIME Problems/Problem 6
  ...er manipulation <math>y = \frac {x}{\sqrt {3}}</math> and <math>y = \sqrt {3}x</math>, respectively, which are still linear functions. Basically the squ ...2\pi) = \frac {1}{12}(2400\pi - 1200\pi) = 100\pi</math>. Hence the answer is <math>\boxed{314}</math>.
  2 KB (354 words) - 16:42, 20 July 2021
• 1999 AIME Problems/Problem 4
  ...h> is <math>43/99</math> and the [[area]] of octagon <math>ABCDEFGH</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relat ...of all <math>8</math> of them is <math>\frac{86}{99}</math> and the answer is <math>\boxed{185}</math>.
  3 KB (398 words) - 13:27, 12 December 2020
• 1999 AIME Problems/Problem 3
  ...all [[positive integer]]s <math>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]]. ...math>. This gives <math>n=1, 9, 10,</math> or <math>18</math>, and the sum is <math>1+9+10+18=\boxed{38}</math>.
  2 KB (296 words) - 01:18, 29 January 2021
• 1999 AIME Problems/Problem 2
  ...his figure into two [[congruent]] [[polygon]]s. The [[slope]] of the line is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relat ...th>\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}</math>, and the solution is <math>m + n = \boxed{118}</math>.
  3 KB (423 words) - 11:06, 27 April 2023
• 1999 AIME Problems/Problem 1
  Find the smallest prime that is the fifth term of an increasing [[arithmetic sequence]], all four preceding ...th>, and <math>29</math> form an [[arithmetic sequence]]. Thus, the answer is <math>029</math>.
  2 KB (332 words) - 13:22, 3 August 2020
• 1999 AIME Problems/Problem 8
  ...f <math>\mathcal{S}</math> divided by the area of <math>\mathcal{T}</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relati ...of the above diagram, of <math>y \ge \frac{1}{3}, z \ge \frac{1}{6}</math> is the triangle at the right, and <math>x \ge \frac 12, z \ge \frac 16</math>
  3 KB (445 words) - 19:40, 4 July 2013
• 2000 AIME I Problems/Problem 15
  ...ft to right, the labels on the cards are now in ascending order: <math>1,2,3,\ldots,1999,2000.</math> In the original stack of cards, how many cards wer ...4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there
  15 KB (2,673 words) - 19:16, 6 January 2024
• 2000 AIME I Problems/Problem 14
  ...th>r</math> times as large as angle <math>APQ,</math> where <math>r</math> is a positive real number. Find <math>\lfloor 1000r \rfloor</math>. ...ath>which implies that <math>1000r = 571 + \tfrac 37</math>. So the answer is <math>\boxed{571}</math>.
  8 KB (1,275 words) - 03:04, 27 February 2022
• 2000 AIME I Problems/Problem 13
  ...an be reached by the firetruck within six minutes. The area of this region is <math>m/n</math> square miles, where <math>m</math> and <math>n</math> are ...Arrows(4)); D((0,-6)--(0,6),Arrows(4)); truck((1,0)); truck((2,0)); truck((3,0)); truck((4,0));
  3 KB (571 words) - 00:38, 13 March 2014
• 2000 AIME I Problems/Problem 11
  ...are [[relatively prime]] positive [[divisor]]s of <math>1000.</math> What is the [[floor function|greatest integer]] that does not exceed <math>S/10</ma ...pressed in the form of <math>2^{x}5^{y}</math>, where <math>-3 \le x,y \le 3</math>. Thus every number in the form of <math>a/b</math> will be expressed
  4 KB (667 words) - 13:58, 31 July 2020
• 2000 AIME I Problems/Problem 10
  ...th>1</math> and <math>100,</math> inclusive, the number <math>x_{k}</math> is <math>k</math> less than the sum of the other <math>99</math> numbers. Give ...=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}</math>, and the answer is <math>75+98=\boxed{173}</math>.
  2 KB (319 words) - 22:26, 29 December 2022
• 2000 AIME I Problems/Problem 9
  <cmath>\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &=& 3 - \log 2000\\ Small note from different author: <math>-(3 - \log 2000) = \log 2000 - 3 = \log 2000 - \log 1000 = \log 2.</math>
  4 KB (623 words) - 15:56, 8 May 2021
• 2000 AIME I Problems/Problem 8
  ...math>n,</math> and <math>p</math> are positive integers and <math>p</math> is not divisible by the cube of any prime number. Find <math>m + n + p</math>. ...rac{3}{4}\right)^{3}\right)^{1/3}}{1}</math> of the height when the vertex is at the top.
  4 KB (677 words) - 16:33, 30 December 2023
• 2000 AIME I Problems/Problem 7
  note: this is the type of problem that makes you think symmetry, but actually can be solv == Solution 3 ==
  5 KB (781 words) - 15:02, 20 April 2024
• 2000 AIME I Problems/Problem 6
  ...ath> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the [[geometric mean]] of <math>x</math> a ...er of pairs of <math>(\sqrt{x},\sqrt{y})</math>, and so <math>(x,y)</math> is then <math>\boxed{997}</math>.
  6 KB (966 words) - 21:48, 29 January 2024
• 2000 AIME I Problems/Problem 5
  .../math> and <math>n</math> are [[relatively prime]] positive integers. What is <math>m + n</math>? If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate <math>m/n</math>. The [[Principle
  7 KB (1,011 words) - 20:09, 4 January 2024
• 2000 AIME I Problems/Problem 4
  Expressing all terms 3 to 9 in terms of <math>a_1</math> and <math>a_2</math> and substituting the ...=69</math>. These numbers are relatively prime, as desired. The perimeter is <math>2(61)+2(69)=\boxed{260}</math>.
  3 KB (485 words) - 00:31, 19 January 2024
• 2000 AIME I Problems/Problem 2
  ...th>D</math> across the y-axis. The area of [[pentagon]] <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>. ...)</math> or <math>(11,10)</math>. Since <math>v < u</math> the latter case is the answer, and <math>u+v = \boxed{021}</math>.
  3 KB (434 words) - 22:43, 16 May 2021
• 2000 AIME I Problems/Problem 1
  ...ositive 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 one of thes <math>n = 3:</math> <cmath>2^3 = 8 , 5 ^3 = 125</cmath>
  1 KB (163 words) - 17:44, 16 December 2020
• 2001 AIME I Problems/Problem 15
  ...are considered to be consecutive, are written on faces that share an edge is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively pr ...resent adjacent octahedral faces. Each assignment of the numbers <math>1,2,3,4,5,6,7</math>, and <math>8</math> to the faces of the octahedron correspon
  11 KB (1,837 words) - 18:53, 22 January 2024
• 2001 AIME I Problems/Problem 14
  ...and <math>1</math> represent a house that does receive mail. This problem is now asking for the number of <math>19</math>-digit strings of <math>0</math n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\hline
  13 KB (2,298 words) - 19:46, 9 July 2020
• 2001 AIME I Problems/Problem 13
  ...h>d < 120.</math> The length of the chord of a <math>3d</math>-degree arc is <math>- m + \sqrt {n}</math> centimeters, where <math>m</math> and <math>n< ...5}}{2},</math> which equals <math>-9 + \sqrt{165}.</math> Thus, the answer is <math>9 + 165 = \boxed{174}</math>.
  3 KB (561 words) - 19:25, 27 November 2022
• 2001 AIME I Problems/Problem 12
  ...0,0,2),</math> and <math>D = (0,0,0).</math> The [[radius]] of the sphere is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively pr ...F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0);
  6 KB (1,050 words) - 18:44, 27 September 2023
• 2014 USAJMO Problems
  ===Problem 3=== [[2014 USAJMO Problems/Problem 3|Solution]]
  3 KB (600 words) - 16:42, 5 August 2023
• 2001 AIME I Problems/Problem 11
  ...be the number associated with <math>P_i</math> after the renumbering. It is found that <math>x_1 = y_2,</math> <math>x_2 = y_1,</math> <math>x_3 = y_4, <cmath>\begin{align}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\\
  3 KB (493 words) - 13:51, 22 July 2020
• 2001 AIME I Problems/Problem 10
  ...[[midpoint]] of the segment they determine also belongs to <math>S</math> is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively pr ...h>, <math>(0,2)</math>, <math>(2,0)</math>, <math>(1,3)</math>, and <math>(3,1)</math>, <math>8</math> possibilities.
  8 KB (1,187 words) - 02:40, 28 November 2020
• 2001 AIME I Problems/Problem 9
  ...<math>q</math>, and <math>r</math> are positive and satisfy <math>p+q+r=2/3</math> and <math>p^2+q^2+r^2=2/5</math>. The ratio of the area of triangle We let <math>[\ldots]</math> denote area; then the desired value is
  4 KB (673 words) - 20:15, 21 February 2024
• 2001 AIME I Problems/Problem 8
  ...e because its base-<math>7</math> representation is <math>102</math>. What is the largest 7-10 double? <cmath>2(a_na_{n-1}\cdots a_0)_7 = (a_na_{n-1}\cdots a_0)_{10}</cmath> (This is because the digits in <math>N</math> ' s base 7 representation make a numbe
  3 KB (502 words) - 11:28, 9 December 2023
• 2001 AIME I Problems/Problem 7
  ...h>\overline{AC}</math>, respectively, such that <math>\overline{DE}</math> is [[parallel]] to <math>\overline{BC}</math> and contains the center of the [ .../math> and <math>\triangle ABC</math> is <math>\frac{43}{63}</math>, which is the scale factor between the two similar triangles, and thus <math>DE = \fr
  9 KB (1,540 words) - 08:31, 1 December 2022
• 2001 AIME I Problems/Problem 6
  ...rolled four times. The [[probability]] that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form <ma The red path corresponds to the sequence of rolls <math>2, 3, 5, 5</math>. This establishes a [[bijection]] between valid dice roll seq
  11 KB (1,729 words) - 20:50, 28 November 2023
• 2001 AIME I Problems/Problem 5
  ...tude is contained in the y-axis, and the square of the length of each side is <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relat path e = xscale(2)*unitcircle; real x = -8/13*3^.5;
  6 KB (1,043 words) - 10:09, 15 January 2024
• 2001 AIME I Problems/Problem 4
  ...ath>b</math>, and <math>c</math> are positive integers, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c</math>. ...and <math>\angle TCA=75^{\circ}</math>, meaning <math>\triangle TAC</math> is an isosceles triangle and <math>AC=24</math>.
  3 KB (534 words) - 03:22, 23 January 2023
• 2001 AIME I Problems/Problem 3
  ...nx^n + a_{n-1}x^{n-1} + \cdots + a_0 = 0</math>, then the sum of the roots is <math>\frac{-a_{n-1}}{a_n}</math>. ...but <math>x^{2001}+-x^{2001}=0</math>, so the term with the largest degree is <math>x^{2000}</math>. So we need the coefficient of that term, as well as
  2 KB (335 words) - 18:38, 9 February 2023
• 2001 AIME I Problems/Problem 2
  ...h>\mathcal{S}</math>, and the mean of <math>\mathcal{S}\cup\{2001\}</math> is <math>27</math> more than the mean of <math>\mathcal{S}</math>. Find the me ...1</math> as it is from <math>2001</math>. Thus, the mean of <math>S</math> is
  1 KB (225 words) - 07:57, 4 November 2022
• 2001 AIME I Problems/Problem 1
  ...ore the case <math>b = 0</math> as we have been doing so far, then the sum is <math>495 + 120 + 15 = \boxed{630}</math>. == Solution 3 ==
  4 KB (687 words) - 18:37, 27 November 2022
• 2002 AIME I Problems/Problem 15
  ...h>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime, find <math>p + q + r.</math> ...h>. Since <math>ABFG</math> is an isosceles trapezoid and <math>CDE</math> is an isosceles triangle, we have symmetry about the <math>xz</math>-plane.
  7 KB (1,181 words) - 20:32, 8 January 2024
• 2002 AIME I Problems/Problem 14
  ...th> and that 2002 is the largest element of <math>\mathcal{S},</math> what is the greatest number of elements that <math>\mathcal{S}</math> can have? ...mod <math>n</math>. Since they are positive integers, the largest element is at least <math>n^2+1</math>, the <math>(n+1)</math>th positive integer cong
  2 KB (267 words) - 19:18, 21 June 2021
• 2002 AIME I Problems/Problem 13
  ...<math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. ...= black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C));
  6 KB (974 words) - 13:01, 29 September 2023
• 2002 AIME I Problems/Problem 10
  ...{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>AE=3</math> and <math>AF=10</math>. Given that <math>EB=9</math> and <math>FC=27 ...ey share a common side and angle, so the area of triangle <math>AGF</math> is <math>10/13</math> the area of triangle <math>AEF</math>.
  4 KB (643 words) - 22:44, 8 August 2023
• 2002 AIME I Problems/Problem 9
  Note that it is impossible for any of <math>h,t,u</math> to be <math>1</math>, since then e ...t be <math>3</math> as well. This configuration works, so <math>333</math> is paintable.
  4 KB (749 words) - 19:44, 25 April 2024
• 2002 AIME I Problems/Problem 8
  (1) <math>a_1,a_2,a_3\cdots</math> is a nondecreasing sequence of positive integers (3) <math>a_9=k</math>
  1 KB (205 words) - 19:54, 4 July 2013
• 2002 AIME I Problems/Problem 7
  ...[[Binomial Expansion]] is valid for exponents that are not integers. That is, for all real numbers <math>x,y</math> and <math>r</math> with <math>|x|>|y ...=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots</cmath>
  2 KB (316 words) - 19:54, 4 July 2013
• 2002 AIME I Problems/Problem 5
  ...12 combinations of two distinct vertices that form a square side only form 3 squares, and all 12 combinations of two vertices that form a square diagona ...onals, meaning we counted them 6 times. Therefore, our answer is <math>198-3(6-1)=198-15=\boxed{183}.</math>
  1 KB (220 words) - 20:50, 12 November 2022
• 2002 AIME I Problems/Problem 4
  Since <math>m</math> is an integer, <math>t+1 = 29</math>, so <math>t=28</math>. It quickly follows ...=29(28)</math> and <math>n=28</math>, which implies <math>m>n</math>. This is impossible since <math>n-m>0</math>.
  2 KB (320 words) - 07:55, 4 November 2022
• 2002 AIME I Problems/Problem 3
  ...will both be two-digit numbers and will have the property that Jane's age is obtained by interchanging the digits of Dick's age. Let <math>d</math> be D <center><math>(1,2), (1,3), (2,3), (1,4), (2,4), (3,4), \dots , (8,9)</math></center>
  2 KB (246 words) - 17:02, 21 May 2023
• 2002 AIME I Problems/Problem 2
  ...orean Theorem]], we find that the shorter dimension is <math>2r\left(\sqrt{3}+1\right)</math>. ...\cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)</math>. Thus we have <math
  2 KB (287 words) - 19:54, 4 July 2013
• 2002 AIME I Problems/Problem 1
  ...it arrangement that reads the same left-to-right as it does right-to-left) is <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are rela ...is <math>\frac{10 \times 10}{10^3} = \frac 1{10}</math>. Similarly, there is a <math>\frac 1{26}</math> probability of picking the three-letter palindro
  3 KB (369 words) - 23:36, 6 January 2024
• 2003 AIME I Problems/Problem 15
  pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A As <math>BD</math> is an angle bisector of both triangles <math>BAC</math> and <math>BF'F</math>,
  8 KB (1,382 words) - 14:23, 29 December 2022
• 2003 AIME I Problems/Problem 14
  ...in that order. Find the smallest value of <math> n </math> for which this is possible. ...th>10^k m - nX</math> is an integer and <math>\frac{10^k m - nX}{n}</math> is a fraction between <math>0</math> and <math>1</math>, we can rewrite this a
  3 KB (477 words) - 14:23, 4 January 2024
• 2003 AIME I Problems/Problem 13
  ...ath>'s than <math>0</math>'s. Find the [[remainder]] when <math> N </math> is divided by <math>1000</math>. ...are in the form <math>{2i \choose i}</math>, so the sum of these elements is <math>\sum_{i=0}^{5} {2i \choose i} = 1 + 2 +6 + 20 + 70 + 252 = 351</math>
  4 KB (651 words) - 19:42, 7 October 2023
• 2003 AIME I Problems/Problem 10
  ...</math> and <math> \angle ACB = 106^\circ. </math> Point <math> M </math> is in the interior of the triangle so that <math> \angle MAC = 7^\circ </math> ...congruent (by ASA), <math>CM = CN</math>. Hence <math>\triangle CMN</math> is an [[equilateral triangle]], so <math>\angle CNM = 60^\circ</math>.
  7 KB (1,058 words) - 01:41, 6 December 2022
• 2003 AIME I Problems/Problem 9
  An [[integer]] between <math>1000</math> and <math>9999</math>, inclusive, is called ''balanced'' if the sum of its two leftmost [[digit]]s equals the su ...balanced numbers. If the common sum of the first two and last two digits is <math>n</math>, such that <math>10 \leq n \leq 18</math>, there are <math>1
  4 KB (696 words) - 11:55, 10 September 2023
• 2003 AIME I Problems/Problem 7
  ...h <math> AB = 9 </math> and <math> BC = 21. </math> Point <math> D </math> is not on <math> \overline{AC} </math> so that <math> AD = CD, </math> and <ma ...<math>y^2 - 36 = x^2 - 225 \Longrightarrow x^2 - y^2 = 189</math>. The LHS is [[difference of squares]], so <math>(x + y)(x - y) = 189</math>. As both <m
  3 KB (490 words) - 18:13, 13 February 2021
• 2003 AIME I Problems/Problem 6
  ...also vertices of a <math>1</math> by <math>1</math> by <math>1</math> cube is <math> m + \sqrt{n} + \sqrt{p}, </math> where <math> m, n, </math> and <mat ...x | vertices]] of a [[cube (geometry) | cube]], there are <math>{8 \choose 3} = 56</math> total [[triangle]]s to consider. They fall into three categor
  3 KB (477 words) - 18:35, 27 December 2021
• 2003 AIME I Problems/Problem 5
  ...th>4</math> by <math>5</math> units. Given that the [[volume]] of this set is <math>\frac{m + n\pi}{p}, </math> where <math> m, n, </math> and <math> p < import three; currentprojection = perspective(5,4,3); defaultpen(linetype("8 8")+linewidth(0.6));
  2 KB (288 words) - 19:58, 4 July 2013
• 2003 AIME I Problems/Problem 4
  ...tity, <math>\sin^2 x + \cos^2 x = 1</math> and <math>2\sin x \cos x</math> is simply <math>\frac{1}{5}</math>, via substitution. Thus, substituting these ==Solution 3==
  3 KB (516 words) - 21:59, 22 October 2020
• 2003 AIME I Problems/Problem 3
  Let the [[set]] <math> \mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}. </math> Susan makes a list as follows: for each two-element subse ...the set from greatest to least to reduce error: <math>\{34, 21, 13, 8, 5, 3, 2, 1\}.</math>
  2 KB (317 words) - 00:09, 9 January 2024
• 2003 AIME I Problems/Problem 2
  ...th>1</math> is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. T ...>, while the total area is given by <math>100^{2} \pi</math>, so the ratio is
  4 KB (523 words) - 15:49, 8 March 2021
• 2024 AMC 8 Problems/Problem 8
  ...day. How many different dollar amounts could Taye have on Thursday, <math>3</math> days later? <math>\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf
  2 KB (384 words) - 22:57, 17 February 2024
• 2003 AIME II Problems/Problem 15
  where <math>m, n,</math> and <math>p</math> are integers and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p.</math> The five-element sum is just <math>\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \sin 120^\circ +
  4 KB (675 words) - 17:23, 30 July 2022
• 2003 AIME II Problems/Problem 14
  .../math> where <math>m</math> and <math>n</math> are positive integers and n is not divisible by the square of any prime. Find <math>m + n.</math> ...rac{8}{\sqrt{3}}, 4\right)</math> and that <math>B = \left(\frac{10}{\sqrt{3}}, 2\right)</math>.
  9 KB (1,461 words) - 15:09, 18 August 2023
• 2003 AIME II Problems/Problem 13
  ...he probability that the bug moves to its starting vertex on its tenth move is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively pr ...t the bug is at its starting vertex after <math>n</math> moves. If the bug is on its starting vertex after <math>n</math> moves, then it must be not on i
  15 KB (2,406 words) - 23:56, 23 November 2023
• 2003 AIME II Problems/Problem 12
  ...ath>s=v_1+\cdots+v_{27}</math> be the total number of votes cast. Our goal is to determine the smallest possible <math>s</math>. ...is means "For all <math>i</math>, <math>\frac{100v_i}s + 1 \leq v_i</math> is true")
  4 KB (759 words) - 13:00, 11 December 2022
• 2003 AIME II Problems/Problem 11
  ...<math>m</math> and <math>p</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime, find <math>m + n + p.</math> Hence, the answer is <math>527+11+40=\boxed{578}.</math>
  5 KB (772 words) - 19:47, 1 August 2023
• 2003 AIME II Problems/Problem 9
  ...the roots of <math>Q(x) = 0,</math> find <math>P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).</math> ...g division to divide <math>P(x)</math> by <math>Q(x)</math>, the remainder is <math>x^2-x+1</math>.
  3 KB (475 words) - 21:53, 6 May 2024
• 2003 AIME II Problems/Problem 8
  ...ath> such that <math>f(1)=1440</math>, <math>f(2)=1716</math>, and <math>f(3)=1848</math>. Plugging in the values for x gives us a system of three equat ...gives <math>a=-72, b=492,</math> and <math>c=1020</math>. Thus, the answer is <math>-72(8)^2+492\cdot8+1020= \boxed{348}.</math>
  5 KB (793 words) - 15:18, 14 July 2023
• 2003 AIME II Problems/Problem 6
  ...tively, after a <math>180^\circ</math> rotation about <math>G.</math> What is the area of the union of the two regions enclosed by the triangles <math>AB Since a <math>13-14-15</math> triangle is a <math>5-12-13</math> triangle and a <math>9-12-15</math> triangle "glued"
  5 KB (787 words) - 17:38, 30 July 2022
• 2003 AIME II Problems/Problem 4
  .... The ratio of the volume of the smaller tetrahedron to that of the larger is <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively pr ...rac{1}{3}, \frac{1}{3})</math>,<math>(0,\frac{1}{3}, \frac{1}{3}, \frac{1}{3})</math>.
  3 KB (563 words) - 17:36, 30 July 2022
• 2003 AIME II Problems/Problem 3
  .../math> is never immediately followed by <math>C</math>, and <math>C</math> is never immediately followed by <math>A</math>. How many seven-letter good wo ...s is restricted. Therefore, the number of seven-letter good words is <math>3*2^6=192</math>
  2 KB (336 words) - 17:29, 30 July 2022
• 2003 AIME II Problems/Problem 2
  ...wo of whose digits are the same. What is the remainder when <math>N</math> is divided by 1000? ...<math>0,1,2</math> is also divisible by 8. The only arrangement that works is <math>120</math>.
  1,013 bytes (162 words) - 09:00, 11 July 2023
• 2003 AIME II Problems/Problem 1
  ...eger]]s is <math>6</math> times their [[sum]], and one of the [[integer]]s is the sum of the other two. Find the sum of all possible values of <math>N</m ...<math>13, 8, 7</math> so the sum of all possible values of <math>N</math> is <math>12 \cdot (13 + 8 + 7) = 12(28) = \boxed{336}</math>.
  1 KB (174 words) - 08:56, 11 July 2023
• 2002 AIME II Problems/Problem 15
  ..., <math>b</math>, and <math>c</math> are positive integers, <math>b</math> is not divisible by the square of any prime, and <math>a</math> and <math>c</m ...and <math>b</math> respectively. We know that the point <math>(9,6)</math> is a point on both circles, so we have that
  7 KB (1,182 words) - 09:56, 7 February 2022
• 2002 AIME II Problems/Problem 14
  ...th center <math>O</math> on <math>\overline{AP}</math> is drawn so that it is [[Tangent (geometry)|tangent]] to <math>\overline{AM}</math> and <math>\ove ...Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have
  4 KB (658 words) - 19:15, 19 December 2021
• 2002 AIME II Problems/Problem 13
  ...area of triangle <math>PQR</math> to the area of triangle <math>ABC</math> is <math>m/n,</math> where <math>m</math> and <math>n</math> are relatively pr C=(1.9375,3.4994);
  6 KB (935 words) - 13:23, 3 September 2021
• 2002 AIME II Problems/Problem 12
  ...h>a_n\le.4</math> for all <math>n</math> such that <math>1\le n\le9</math> is given to be <math>p^aq^br/\left(s^c\right)</math> where <math>p</math>, <ma ...h> to <math>(6,4)</math> that always stay below the line <math>y=\frac{2x}{3}</math>. We can find the number of such paths using a Pascal's Triangle typ
  7 KB (1,127 words) - 13:34, 19 June 2022
• 2002 AIME II Problems/Problem 11
  ...math>n</math>, and <math>p</math> are positive integers and <math>m</math> is not divisible by the square of any prime. Find <math>100m+10n+p</math>. ...>. Then, the common ratio is <math>\frac{1}{8x}</math>, and the first term is <math>8x^2</math>.
  4 KB (696 words) - 16:27, 22 March 2022
• 2002 AIME II Problems/Problem 9
  ...have no common elements.) Find the remainder obtained when <math>n</math> is divided by <math>1000</math>. ...</math>, <math>i \in B</math>, or <math>i \in C</math>. So there are <math>3^{10}</math> ways to organize the elements of <math>S</math> into disjoint <
  3 KB (404 words) - 23:07, 4 May 2024
• 2002 AIME II Problems/Problem 8
  ...<math>n(n+1)<2002</math>, which is easy to solve by trial, as the solution is obviously <math>\simeq \sqrt{2002}</math>.) ...2002}{n}\right\rfloor=k</math> has no integer solutions for <math>n</math> is <math>\boxed{049}</math>.
  6 KB (908 words) - 14:22, 14 July 2023
• 2002 AIME II Problems/Problem 7
  It is known that, for all positive integers <math>k</math>, <center><math>1^2+2^2+3^2+\ldots+k^{2}=\frac{k(k+1)(2k+1)}6</math>.</center>
  3 KB (403 words) - 12:10, 9 September 2023
• 2002 AIME II Problems/Problem 6
  Find the integer that is closest to <math>1000\sum_{n=3}^{10000}\frac1{n^2-4}</math>. ...1 = (n-2)A + (n+2)B</math> or <math>1 = n(A+B)+ 2(B-A)</math>. Since there is no n term on the left hand side, <math> A+B=0</math> and by inspection <mat
  2 KB (330 words) - 05:56, 23 August 2022
• 2002 AIME II Problems/Problem 5
  ...> and <math>m</math> are non-negative integers, for which <math>a^6</math> is not a divisor of <math>6^a</math>. ...l pairs of non-negative integers (n,m) for which <math>(2^n3^m)^{6}</math> is not a divisor of <math>6^{2^n3^m}</math>
  3 KB (515 words) - 14:46, 14 February 2021
• 2002 AIME II Problems/Problem 4
  ...ath>m</math> is a positive integer. Find the remainder when <math>m</math> is divided by <math>1000</math>. ...Thus, the area of the garden enclosed by the path when <math>n=202</math> is
  2 KB (268 words) - 07:28, 13 September 2020
• 2002 AIME II Problems/Problem 3
  ...r]]s that form an increasing [[geometric sequence]] and <math>b - a</math> is the [[Perfect square|square]] of an integer. Find <math>a + b + c.</math> ...the [[geometric mean]] of the [[product]] <math>abc</math>. <math>b=\sqrt[3]{abc}=6^2=36</math>.
  2 KB (263 words) - 22:50, 5 April 2024
• 2002 AIME II Problems/Problem 2
  ...7,12,10)</math>, <math>Q=(8,8,1)</math>, and <math>R=(11,3,9)</math>. What is the [[surface area]] of the cube? <math>PR=\sqrt{(11-7)^2+(3-12)^2+(9-10)^2}=\sqrt{98}</math>
  709 bytes (103 words) - 03:32, 6 December 2019
• 2002 AIME II Problems/Problem 1
  &(2)& y\text{ is the number formed by reversing the digits of }x\text{; and}\\ &(3)& z=|x-y|.
  971 bytes (150 words) - 21:56, 11 February 2020
• 2001 AIME II Problems/Problem 15
  ...math>n</math>, and <math>p</math> are positive integers and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p</math>. import three; currentprojection = orthographic(camera=(1/4,2,3/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype("10 2");
  4 KB (518 words) - 15:01, 31 December 2021
• 2001 AIME II Problems/Problem 14
  ...{2}i</math> and <math>\text{cis\,} {8}\theta = -\frac{1}{2}- \frac{\sqrt{3}}{2}i</math> ...}}{2}i</math> and <math>\text{cis\,} {8}\theta = -\frac{1}{2}+\frac{\sqrt{3}}{2}i</math>
  2 KB (380 words) - 15:03, 22 July 2018
• 2001 AIME II Problems/Problem 13
  ...Hence <math>\angle ADB = \angle DEC</math>, and <math>\triangle BDE</math> is [[isosceles triangle|isosceles]]. Then <math>BD = BE = 10</math>. The answer is <math>m+n = \boxed{069}</math>.
  4 KB (743 words) - 03:32, 23 January 2023
• 2001 AIME II Problems/Problem 12
  ...has the midpoint triangle as a face. The [[volume]] of <math>P_{3}</math> is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are rela ...ath>\left(\frac 12\right)^3 = \frac 18</math>. The total volume added here is then <math>\Delta P_1 = 4 \cdot \frac 18 = \frac 12</math>.
  2 KB (380 words) - 00:28, 5 June 2020
• 2001 AIME II Problems/Problem 11
  ...ty]] that Club Truncator will finish the season with more wins than losses is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are rela ...and losses. Thus, by the [[complement principle]], the desired probability is half the probability that Club Truncator does not have the same number of w
  3 KB (415 words) - 23:25, 20 February 2023
• 2001 AIME II Problems/Problem 10
  ...0^{j-i} - 1</math>. From the factorization <math>10^6 - 1 = (10^3 + 1)(10^{3} - 1)</math>, we see that <math>j-i = 6</math> works; also, <math>a-b | a^n ...^{0},\dots\implies 94 - 6 = 88</math>, and so forth. Therefore, the answer is <math>94 + 88 + 82 + \dots + 4\implies 16\left(\dfrac{98}{2}\right) = \boxe
  4 KB (549 words) - 23:16, 19 January 2024
• 2001 AIME II Problems/Problem 9
  ...[[probability]] of obtaining a grid that does not have a 2-by-2 red square is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are [[re ...easy: 4 ways to choose which the side the squares will be on, and <math>2^3</math> ways to color the rest of the squares, so 32 ways to do that. For th
  8 KB (1,207 words) - 20:04, 5 September 2023
• 2001 AIME II Problems/Problem 8
  ...<math>x</math>, and that <math>f(x) = 1-|x-2|</math> for <math>1\le x \le 3</math>. Find the smallest <math>x</math> for which <math>f(x) = f(2001)</ma ...h>f(2001) = 3^kf\left(\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6</math>. Indeed,
  3 KB (545 words) - 23:41, 14 June 2023
• 2001 AIME II Problems/Problem 7
  ...h> and <math>C_{3}</math> can be written as <math>\sqrt {10n}</math>. What is <math>n</math>? ...> and <math>b</math> are the legs of the right triangle and <math>c</math> is the hypotenuse. (This formula should be used ''only for right triangles''.)
  7 KB (1,112 words) - 02:15, 26 December 2022
• 2001 AIME II Problems/Problem 3
  x_{3}&=420,\\ x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}
  2 KB (300 words) - 01:28, 12 November 2022
• 2001 AIME II Problems/Problem 2
  ...5</math> percent of the school population, and the number who study French is between <math>30</math> percent and <math>40</math> percent. Let <math>m</m Therefore, the answer is <math>M - m = 499 - 201 = \boxed{298}</math>.
  2 KB (252 words) - 00:54, 10 January 2024
• 2000 AIME II Problems/Problem 15
  ...ac{1 - 0}{\sin 1} = \frac{1}{\sin 1^{\circ}}</math>. Therefore, the answer is <math>\boxed{001}</math>. The average term is around the 60's which gives <math>\frac{4}{3}</math>.
  3 KB (469 words) - 21:14, 7 July 2022
• 2000 AIME II Problems/Problem 14
  ..., and <math>0<f_m</math>. Given that <math>(f_1,f_2,f_3,\ldots,f_j)</math> is the factorial base expansion of <math>16!-32!+48!-64!+\cdots+1968!-1984!+20 ...k+1)\cdot k!- k!)} = 1+\sum_{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \cdots + (n! - (n-1)!)) = n!</math>.
  7 KB (1,131 words) - 14:49, 6 April 2023
• 2000 AIME II Problems/Problem 13
  ...2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math>\frac{m+\sqrt{n}}r</math>, where <math>m</math>, <math>n</math> and < 2000x^6+100x^5+10x^3+x-2&=0\\
  6 KB (1,060 words) - 17:36, 26 April 2024
• 2000 AIME II Problems/Problem 11
  ...the absolute values of all possible slopes for <math>\overline{AB}</math> is <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively pr ...ction of <math>D</math> across that perpendicular. Then <math>ABCD'</math> is a [[parallelogram]], and <math>\overrightarrow{AB} = \overrightarrow{D'C}</
  4 KB (750 words) - 22:55, 5 February 2024
• 2000 AIME II Problems/Problem 10
  A [[circle]] is [[inscribe]]d in [[quadrilateral]] <math>ABCD</math>, [[tangent]] to <math> == Solution 3 (Smart algebra to make 2 less annoying) ==
  2 KB (399 words) - 17:37, 2 January 2024
• 2000 AIME II Problems/Problem 9
  ...ch that <math>z+\frac 1z=2\cos 3^\circ</math>, find the least integer that is greater than <math>z^{2000}+\frac 1{z^{2000}}</math>. ...ac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}</math>.
  4 KB (675 words) - 13:42, 4 April 2024
• 2000 AIME II Problems/Problem 8
  In [[trapezoid]] <math>ABCD</math>, leg <math>\overline{BC}</math> is [[perpendicular]] to bases <math>\overline{AB}</math> and <math>\overline{C ...<math>\overline{CD}</math>. Then <math>AE = x</math>, and <math>ADE</math> is a [[right triangle]]. By the [[Pythagorean Theorem]],
  4 KB (584 words) - 19:35, 7 December 2019
• 2000 AIME II Problems/Problem 7
  ...N{1!18!}</math></center> find the [[floor function|greatest integer]] that is less than <math>\frac N{100}</math>. <cmath>\frac {19!}{2!17!}+\frac {19!}{3!16!}+\frac {19!}{4!15!}+\frac {19!}{5!14!}+\frac {19!}{6!13!}+\frac {19!}{7
  2 KB (281 words) - 12:09, 5 April 2024
• 2000 AIME II Problems/Problem 6
  .../math> be the length of the segment joining the legs of the trapezoid that is [[parallel]] to the bases and that divides the trapezoid into two regions o ...length of the midline of the trapezoid is the average of its bases, which is <math>\frac{b+b+100}{2} = b+50</math>. The two regions which the midline di
  3 KB (433 words) - 19:42, 20 December 2021
• 2000 AIME II Problems/Problem 4
  What is the smallest positive integer with six positive odd integer divisors and tw ...math>18 = 2 \cdot 3 \cdot 3</math> factors, then it can have at most <math>3</math> distinct primes in its factorization.
  2 KB (397 words) - 15:55, 11 May 2022
• 2000 AIME II Problems/Problem 2
  A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola <math> {{AIME box|year=2000|n=II|num-b=1|num-a=3}}
  804 bytes (126 words) - 20:30, 4 July 2013
• 2000 AIME II Problems/Problem 1
  <center><math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math></center> <math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math>
  2 KB (292 words) - 13:33, 4 April 2024
• 2005 AMC 10B Problems/Problem 3
  ...the remaining paint is used. What fraction of the original amount of paint is available to use on the third day? ...frac{1}{10} \qquad \textbf{(B) } \frac{1}{9} \qquad \textbf{(C) } \frac{1}{3} \qquad \textbf{(D) } \frac{4}{9} \qquad \textbf{(E) } \frac{5}{9} </math>
  1 KB (163 words) - 14:00, 14 December 2021
• 2005 AMC 10B Problems/Problem 7
  ...ed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smallest circle to the area of the largest squ ...rac{\pi}{16} \qquad \textbf{(B) } \frac{\pi}{8} \qquad \textbf{(C) } \frac{3\pi}{16} \qquad \textbf{(D) } \frac{\pi}{4} \qquad \textbf{(E) } \frac{\pi}{
  2 KB (381 words) - 14:28, 14 December 2021
• 2005 AMC 10B Problems/Problem 12
  .... What is the probability that the product of the numbers on the top faces is prime? ...e which die will have the prime number, so the probability is <math>\dfrac{3}{6}\times\left(\dfrac{1}{6}\right)^{11}\times\dbinom{12}{1} = \dfrac{1}{2}\
  3 KB (385 words) - 14:03, 16 June 2022
• 2005 AMC 10B Problems/Problem 13
  ...etween <math>1</math> and <math>2005</math> are integer multiples of <math>3</math> or <math>4</math> but not <math>12</math>? ...12</math> are <math>\frac{2005}{12} = 167\text{ }R1.</math> So, the answer is <math>668+501-167-167 = \boxed{\textbf{(C) } 835}</math>
  1 KB (212 words) - 14:44, 15 December 2021
• 2005 AMC 10B Problems/Problem 14
  ..., and <math> C</math> is the midpoint of <math> \overline{BD}</math>. What is the area of <math> \triangle CDM</math>? ...}}{2}\qquad \textbf{(B) }\ \frac {3}{4}\qquad \textbf{(C) }\ \frac {\sqrt {3}}{2}\qquad \textbf{(D) }\ 1\qquad \textbf{(E) }\ \sqrt {2}</math>
  5 KB (882 words) - 22:12, 30 April 2024
• 2005 AMC 10B Problems/Problem 17
  ...<math>5^b = 6</math>, <math>6^c = 7</math>, and <math>7^d = 8</math>. What is <math>a \cdot b\cdot c \cdot d</math>? ...quad \textbf{(C) } 2 \qquad \textbf{(D) } \frac{5}{2} \qquad \textbf{(E) } 3 </math>
  2 KB (324 words) - 15:30, 16 December 2021
• 2005 AMC 10B Problems/Problem 18
  ..., and <math>g</math> are distinct digits and in increasing order, and none is either <math>0</math> or <math>1</math>. How many different telephone numbe ...r of ways to choose <math>7</math> numbers from <math>8</math>. The answer is then <math>\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{\textbf{(D) } 8}</ma
  1 KB (195 words) - 15:33, 16 December 2021
• 2005 AMC 10B Problems/Problem 20
  ...) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? ...math> times. Therefore, the sum of all such numbers is <math> 24 \times (1+3+5+7+8) \times (11111) = 24 \times 24 \times 11111 = 6399936. </math> Since
  2 KB (356 words) - 19:15, 17 September 2023
• 2005 AMC 10B Problems/Problem 21
  ...<math>a</math> and the other two bear a number <math>b \neq a</math>. What is the value of <math>q/p</math>? ...ath> ways to pick the slips, so <math>q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}</math>.
  3 KB (398 words) - 19:17, 17 September 2023
• 2005 AMC 10B Problems/Problem 23
  ...The area of <math>ABEF</math> is twice the area of <math>FECD</math>. What is <math>AB/DC</math>? <math>\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8 </math
  3 KB (489 words) - 19:22, 17 September 2023
• 2005 AMC 10B Problems/Problem 24
  ...x</math> and <math>y</math> be two-digit integers such that <math>y</math> is obtained by reversing the digits What is <math>x + y + m</math>?
  5 KB (845 words) - 19:23, 17 September 2023
• 2005 AMC 10B Problems/Problem 25
  ...perty that no two elements of <math>B</math> sum to <math>125</math>. What is the maximum possible number of elements in <math>B</math>? ...nd at most one number from each pair can be included in the set. The total is <math>24 + 38 = \boxed{\textbf{(C)}\ 62}</math>.
  3 KB (517 words) - 19:15, 15 October 2023
• 2006 USAMO Problems/Problem 2
  ...s of <math>k </math>, the minimum value of <math>N </math> for which there is a set of <math>2k+1 </math> distinct positive integers that has sum greater & = 2k^3 + 3k^2 + 3k
  2 KB (398 words) - 09:48, 5 August 2014
• 2006 USAMO Problems/Problem 4
  ...ve a sum and product of <math>n</math>. For <math>p_1+p_2=n</math>, which is only possible in one case, <math>n=4</math>, we consider <math>p_1=p_2=2</m ...we need to check for <math>n=1,2,3,5,7</math>. One is included because it is neither prime nor composite.
  3 KB (486 words) - 22:43, 5 August 2014
• 2006 USAMO Problems/Problem 6
  Thus, <math>XAE\sim XBF</math> by AA similarity, and <math>X</math> is the center of spiral similarity for <math>A,E,B,</math> and <math>F</math>. Thus, <math>YED\sim YFC</math> by AA similarity, and <math>Y</math> is the center of spiral similarity for <math>E,D,F,</math> and <math>C</math>.
  5 KB (986 words) - 22:46, 18 May 2015
• 2006 AMC 12A Problems/Problem 2
  Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>? ...bf{(B)}\ 0\qquad\textbf{(C)}\ h\qquad\textbf{(D)}\ 2h\qquad\textbf{(E)}\ h^3</math>
  694 bytes (110 words) - 14:03, 27 July 2023
• 2006 AMC 12A Problems/Problem 3
  ...lems|2006 AMC 12A #3]] and [[2006 AMC 10A Problems/Problem 3|2006 AMC 10A #3]]}} ...ice's age is <math>3:5</math>. Alice is <math>30</math> years old. How old is Mary?
  1 KB (155 words) - 17:30, 16 December 2021
• 2006 AMC 12A Problems/Problem 6
  ...e two hexagons can be repositioned without overlap to form a square. What is <math>y</math>? label("$y$",(3,4),S);
  3 KB (528 words) - 18:29, 7 May 2024
• 2006 AMC 12A Problems/Problem 8
  <math> \textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math> ...e must be either <math>1, 3, </math> or <math>5</math>, and <math>1</math> is clearly not possible. The other two possibilities both work:
  3 KB (450 words) - 02:00, 13 January 2024
• 2006 AMC 12A Problems/Problem 9
  ...more than an eraser, and both items cost a [[whole number]] of cents. What is the total cost, in cents, of one pencil and one eraser? ...1 \pmod 3</math> so <math>p</math> leaves a remainder of 1 on division by 3.
  3 KB (429 words) - 18:14, 26 September 2020
• Diagonal
  A '''diagonal''' of a [[polygon]] is any segment joining two [[vertex|vertices]] other than an [[edge]]. ...als of a polygon with <math>n</math> vertices is given by <math>\frac{n(n-3)}{2}</math>.
  2 KB (374 words) - 00:37, 25 January 2015
• Equation
  An '''equation''' is a [[relation]] which states that two [[expression]]s are equal, identical, A unique aspect to equations is the ability to modify an original equation by performing operations (such a
  5 KB (932 words) - 12:57, 26 July 2023
• Harvard-MIT Mathematics Tournament
  ...998. The contest takes place alternatively at Harvard or MIT each year. It is composed of two tournaments, the ''February Tournament'' and the ''November The February Tournament is the more difficult of the two tournaments, with its problems ranging from m
  4 KB (539 words) - 16:58, 19 February 2023
• Radius
  ...ons hold for the [[sphere]] and [[hypersphere]]. The plural form of radius is radii. The radius of a circle is often denoted using R or r.
  929 bytes (156 words) - 22:49, 5 January 2023
• Intersection
  ...math>\{1, 2, 3\}</math> and <math>\{1, 3, 5\}</math> is the set <math>\{1, 3\}</math>. ....e. <math>\bigcap_{i = 1}^n A_i = A_1 \cap A_2 \cap \ldots \cap A_n</math> is the intersection of the <math>n</math> sets <math>A_1, A_2, \ldots, A_n</ma
  1 KB (221 words) - 22:41, 11 April 2019
• Word problem
  A '''word problem''' is a problem posed in plain language, as opposed to a problem with only "math" Because word problems are based on a contextualized scenario, the context is important in interpreting the results. For instance, when determining the
  1 KB (200 words) - 15:05, 10 April 2020
• Multinomial Theorem
  where <math>\binom{n}{j_1; j_2; \ldots ; j_k}</math> is the [[multinomial coefficient]] <math>\binom{n}{j_1; j_2; \ldots ; j_k}=\df Note that this is a direct generalization of the [[Binomial Theorem]], when <math>k = 2</math
  3 KB (476 words) - 19:37, 4 January 2023
• Quadratic reciprocity
  <cmath> \genfrac{(}{)}{}{}{a}{p} =\begin{cases} 1 & \text{if } a \text{ is a quadratic residue modulo } p, \\ We say that <math>a</math> is a '''quadratic residue''' modulo <math>p</math> if there exists an integer
  7 KB (1,182 words) - 16:46, 28 April 2016
• 2006 AMC 10B Problems
  What is <math> (-1)^{1} + (-1)^{2} + ... + (-1)^{2006} </math> ? ...define <math> x \mathop{\spadesuit} y = (x+y)(x-y) </math>. What is <math> 3 \mathop{\spadesuit} (4 \mathop{\spadesuit} 5) </math>?
  14 KB (2,059 words) - 01:17, 30 January 2024
• 2005 AMC 10B Problems
  ...e property that <math>x\%</math> of <math>x</math> is <math>4</math>. What is <math>x</math>? == Problem 3 ==
  12 KB (1,874 words) - 21:20, 23 December 2020
• Noncommutative
  More formally, if <math>\star</math> is some [[binary operation]] on a [[set]], and <math>x</math> and <math>y</mat ...commutative]]. For example, <math>4\cdot3=3\cdot4=12</math>, and <math>2+3=3+2=5</math>.
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• Platonic solid
  A '''Platonic solid''' is a [[polyhedron]], or 3 dimensional figure, in which all [[face]]s are [[congruent (geometry)|congr It is easy to verify that all five Platonic solids satisfy [[Euler's polyhedral f
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• Law of Tangents
  ...'''Law of Tangents''' is a rather obscure [[trigonometric identity]] that is sometimes used in place of its better-known counterparts, the [[law of sine ...n a triangle, any ratio of linear combinations applied to lengths of sides is equal to the ratio of the same linear combinations applied to the sines of
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• 2000 AMC 12 Problems/Problem 3
  ...0 AMC 12 Problems|2000 AMC 12 #3]] and [[2000 AMC 10 Problems|2000 AMC 10 #3]]}} ...h>. If she ate <math>20\%</math> of the jellybeans, then <math>80\%</math> is remaining. Hence, after day 1, there are:
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• 2000 AMC 12 Problems/Problem 8
  Figures <math>0</math>, <math>1</math>, <math>2</math>, and <math>3</math> consist of <math>1</math>, <math>5</math>, <math>13</math>, and <mat draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);
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• Angle bisector theorem
  ...[[triangle]] <math>\triangle ABC</math> and [[angle bisector]] AD, where D is on side BC, then <math> \frac cm = \frac bn </math>. It follows that <math First, because <math>\overline{AD}</math> is an angle bisector, we know that <math>m\angle BAD = m\angle CAD</math> and
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• 2000 AMC 12 Problems/Problem 4
  ...rwards is the sum of its two predecessors. Which one of the ten [[digit]]s is the last to appear in the units position of a number in the Fibonacci seque <math>1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....</math>
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• Sigma notation
  ...ctly. Besides being easier to write than the explicit sum, sigma notation is also useful in that it shows the general form of each addend. ...er limit of summation is written ''under'' the sigma and the ''upper'' one is written above the sigma.
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• University of South Carolina High School Math Contest/1993 Exam/Problem 3
  If 3 circles of radius 1 are mutually tangent as shown, what is the area of the gap they enclose? <center>[[Image:Usc93.3.PNG]]</center>
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• University of South Carolina High School Math Contest/1993 Exam/Problem 4
  If <math>(1 + i)^{100}</math> is expanded and written in the form <math>a + bi</math> where <math>a</math> a * [[University of South Carolina High School Math Contest/1993 Exam/Problem 3|Previous Problem]]
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• University of South Carolina High School Math Contest/1993 Exam/Problem 11
  ...e placed randomly into 4 boxes also labeled 1 to 4, one card per box. What is the probability that no card gets placed into a box having the same label a <center><math> \mathrm{(A) \ } 1/3 \qquad \mathrm{(B) \ }3/8 \qquad \mathrm{(C) \ }5/12 \qquad \mathrm{(D) \ } 1/2 \qquad \mathrm{(E)
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• University of South Carolina High School Math Contest/1993 Exam/Problem 29
  If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle? ...>R = \frac{2\cdot3\cdot4}{4 (\frac{3}{4}\sqrt{15})} = \frac{6\cdot\frac{4}{3}}{\sqrt{15}} = \frac8{\sqrt{15}} \Longrightarrow \mathrm{(B)}</math>.
  2 KB (219 words) - 09:57, 31 August 2012
• University of South Carolina High School Math Contest/1993 Exam/Problem 20
  ...th> denote the product of all the elements in <math>A_i</math>. Then what is the value of <math>\pi(A_1)+\pi(A_2)+\cdots+\pi(A_{63})</math>? ...(1+5)(1+6)-1</math> (The <math>-1</math> since we have one less set). This is <math>7!-1=5039</math>.
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• Canadian Mathematical Olympiad
  ...nadian Mathematical Olympiad Qualifying Repechage]] (CMOQR). The Olympiad is part of the selection process for the Canadian [[IMO]] team. ...5 - 8|breakdown=<u>Problem 1</u>: 5.5<br><u>Problem 2</u>: 6<br><u>Problem 3</u>: 6.5<br><u>Problem 4</u>: 7-7.5<br><u>Problem 5</u>: 7.5-8}}
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• 1977 Canadian MO Problems/Problem 1
  ...rfect square when <math>b</math> is an integer. Hence, when <math>b</math> is a positive integer, <math>a</math> cannot be. ...Thus, <math>b^{2} + b + 1</math> is not a perfect square, and thus there is no <math>b</math> that satisfies <math>4f(a) = f(b)</math>.
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• 1977 Canadian MO Problems/Problem 3
  ...Find the smallest positive integer <math>b</math> for which <math>N</math> is the fourth power of an integer. ...integer <math>a.</math> Because <math>7\mid a^4</math> and <math>7</math> is prime, <math>a^4 \ge 7^4.</math> Since we want to minimize <math>b,</math>
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• University of South Carolina High School Math Contest/1993 Exam/Problems
  ...s doubled and the length is increased by 3, then the area is tripled. What is the length of the rectangle? <cmath> \mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 6 \qquad \mathrm{(E) \ } 9 </cmath>
  14 KB (2,102 words) - 22:03, 26 October 2018
• Mock AIME 1 2006-2007 Problems/Problem 2
  ...ath>\star (n)=12</math> and <math>0\le n< 10^{7}</math>. If <math>m</math> is the number of elements in <math>\mathcal{S}</math>, compute <math>\star(m)< ...h>m = 18564 - 7 - 42 - 42 - 105 = 18368</math> so <math>\star(m) = 1 + 8 + 3 + 6 + 8 = 026</math>.
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• Mock AIME 1 2006-2007 Problems/Problem 3
  ...math>n</math> is an integer, find the remainder when <math>n^{2007}</math> is divided by <math>1000</math>. ...so <math>3^{2000}\equiv 1 \pmod{1000}</math> and so <math>3^{2007} \equiv 3^7 \equiv 2187 \equiv 187 \pmod{1000}</math>
  963 bytes (135 words) - 15:53, 3 April 2012
• Mock AIME 1 2006-2007 Problems/Problem 4
  ...respectively. If the <math>x</math>-coordinate of the triangle's centroid is <math>1</math>, find the area of <math>\triangle ABC</math>. ...^2)</math>. Then we have by the centroid condition that <math>a + b + c = 3</math>. From the first [[slope]] condition we have <math>10 = \frac{b^2 -
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• Mock AIME 1 2006-2007 Problems/Problem 6
  ...as a [[rational number | rational]] [[slope]]. If <math>\mathcal{L}</math> is written in the form <math>ax+by=c</math> for [[positive integer]]s <math>a, ...: 5</math> and since the [[greatest common divisor]] of the three numbers is 1, <math>a = 1, b = 5, c = 5</math> and <math>a + b + c = 011</math>.
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• Mock AIME 1 2006-2007 Problems/Problem 7
  ...</math> have <math>AC=6</math> and <math>BC=3</math>. Point <math>E</math> is such that <math>CE=1</math> and <math>AE=5</math>. Construct point <math>F< ...\cdot 6 = CF\cdot 3</math>. <math>CF</math> is then 2, and <math>BF</math> is 1. We can now use Menelaus on line <math>DF</math> with respect to triangle
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• Mock AIME 1 2006-2007 Problems/Problem 9
  ....</math>. If the sum of all possible [[distinct]] values of <math>S</math> is <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are [[rela ...11}+a_{12}+...</math>. If the sum of all distinct values of <math>S</math> is <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relati
  5 KB (744 words) - 19:46, 20 October 2020
• Mock AIME 1 2006-2007 Problems/Problem 12
  ...3}</math>, <math>10^6 \equiv 1 \pmod{13}</math> and the order of 10 mod 13 is 6. Thus, we get one value of <math>m</math> each time <math>n = 6j + 1</ma
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• Mock AIME 1 2006-2007 Problems/Problem 13
  ...<math>n</math>. If <math>d_{1}=1</math>, <math>d_{2}=2</math>, <math>d_{3}=3</math>, <math>d_{4}=-7</math>, <math>d_{5}=13</math>, and <math>d_{6}=-16</ ...these variables, which does uniquely determine these variables - but there is no obvious way of computing them. We will show a different solution.
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• Mock AIME 1 2006-2007 Problems/Problem 15
  ...nd <math>n</math> are relatively prime positive integers. Compute the last 3 digits of <math>m+n</math> ...||||||</math> is the actual number of arrangements in which <math>A</math> is next to <math>1</math>. There are <math>\frac{12!}{10!}</math> <math>-</mat
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• 1977 Canadian MO Problems/Problem 2
  ...h> on the circumference of the circle such that the angle <math>OPA</math> is a maximum. ...ent that <math>O</math> is the midpoint of <math>AC</math>, <math>X</math> is the midpoint of <math>AB</math>, and hence <math>OX=\dfrac{BC}{2}</math>.
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• Mock AIME 1 2006-2007/Problems
  ...ath>\star (n)=12</math> and <math>0\le n< 10^{7}</math>. If <math>m</math> is the number of elements in <math>\mathcal{S}</math>, compute <math>\star(m)< ==Problem 3==
  8 KB (1,355 words) - 14:54, 21 August 2020
• Mock AIME 2 2006-2007 Problems/Problem 2
  ...any [[element]]s <math>n</math> in <math>S</math> is <math>f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer? <math>f(n) = \frac{2n^3+n^2-n-2}{n^2-1} = \frac{(n - 1)(2n^2 + 3n + 2)}{(n - 1)(n + 1)} = \frac{2n^
  531 bytes (90 words) - 10:49, 4 April 2012
• Mock AIME 2 2006-2007 Problems/Problem 3
  .../math> is a [[perfect square]]. Find the [[remainder]] when <math>S</math> is divided by <math>1000.</math> ...>, which can be split into two [[factor]]s in 3 ways, <math>2043 \cdot 1 = 3 \cdot 681 = 227 \cdot 9</math>. This gives us three pairs of [[equation]]s
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• Mock AIME 2 2006-2007 Problems/Problem 4
  ...ger]] such that <math>(a + bi)^n = (a - bi)^n</math>, where <math>n</math> is as small as possible and <math>i = \sqrt{-1}</math>. Compute <math>\frac{b ...r <math>b = 0</math> or <math>3a^2 = b^2</math>. Thus <math>n = b^2/a^2 = 3</math>.
  1 KB (240 words) - 10:50, 4 April 2012
• Mock AIME 2 2006-2007 Problems/Problem 5
  Given that <math> iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>z=n\pm \sqrt{-i},</math> find <math> <center><math>iz^3 = z + 2 + \frac{3}{z} + \frac{4}{z^2} + \cdots</math>,</center>
  912 bytes (145 words) - 10:51, 4 April 2012
• Mock AIME 2 2006-2007 Problems/Problem 8
  .../math> and <math>x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>n = 1, 2, 3, 4</math>. Find the last three [[digit]]s of <math>x_7</math>. This solution is rather long and unpleasant, so a nicer solution may exist:
  3 KB (470 words) - 00:33, 10 August 2019
• Mock AIME 2 2006-2007 Problems/Problem 7
  ...If the length of this path is <math>m\sqrt{n},</math> where <math>n</math> is squarefree, find <math>m+n.</math> ...ius <math>51</math>, we must have that <math>\angle AOB=\frac{360^{\circ}}{3}=120^{\circ}</math>. We know that <math>AO=OB=51</math>, so we can use the
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• Mock AIME 2 2006-2007 Problems/Problem 9
  ...ly such that <math>\frac{BX}{CX}=\frac23</math>, <math>\frac{AY}{CY}=\sqrt 3,</math> and <math>CY=CX-BX</math>. If <math>\tan \angle APB= -\frac{a+b\sqr pair A=(0,6),B=(5,0),C=origin,X=(3,0),Y=A/(sqrt(3)+1);
  2 KB (358 words) - 23:22, 3 May 2014
• Mock AIME 2 2006-2007 Problems/Problem 12
  ...math>BC=6</math>, <math>BO=1,</math> and the [[area]] of <math>ABCD</math> is <math>\frac{a\sqrt{b}}{c},</math> where <math>a,b,c</math> are [[relatively <math>m\angle DAC=m\angle DBC \Rightarrow ABCD</math> is a cylic quadrilateral.
  2 KB (311 words) - 10:53, 4 April 2012
• Mock AIME 2 2006-2007 Problems/Problem 15
  ...ubes are colored red. An arrangement of the cubes is "intriguing" if there is exactly <math>1</math> red unit cube in every <math>1\times1\times4</math> ...3 squares to paint red as we can't select a red square from a square that is on the same row as the first column.
  4 KB (739 words) - 17:04, 24 November 2023

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