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1987 AIME Problems/Problem 8
...is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>? <cmath>\begin{align*}104(n+k) &< 195n< 105(n+k)\\

2 KB (393 words) - 16:59, 16 December 2020
1987 AIME Problems/Problem 7
...3, 4), (2, 3, 4), (3, 3, 4), (3, 2, 4), (3, 1, 4)</math> and <math>(3, 0, 4)</math>. ...ach of the forms <math>(3, 3, n)</math>, <math>(3, n, 3)</math> and <math>(n, 3, 3)</math>.

3 KB (547 words) - 22:54, 4 April 2016
1987 AIME Problems/Problem 3
...product of the distinct proper divisors of <math>n</math>. A number <math>n</math> is ''nice'' in one of two instances: ...visors are <math>p</math> and <math>q</math>, and <math>p(n) = p \cdot q = n</math>.

3 KB (511 words) - 09:29, 9 January 2023
1987 AIME Problems/Problem 1
...non-negative]] [[integer]]s is called "simple" if the [[addition]] <math>m+n</math> in base <math>10</math> requires no carrying. Find the number of sim ...e then fixed). Thus, the number of [[ordered pair]]s will be <math>(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}</math>.

1 KB (191 words) - 14:42, 17 September 2016
1988 AIME Problems/Problem 13
Let <math>F_n</math> represent the <math>n</math>th number in the Fibonacci sequence. Therefore, x^2 - x - 1 = 0&\Longrightarrow x^n = F_n(x), \ n\in N \\

10 KB (1,585 words) - 03:58, 1 May 2023
1988 AIME Problems/Problem 10
...ly one vertex of a square/hexagon/octagon, we have that <math>V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48</math>. ...ron must be a diagonal of that face. Each square contributes <math>\frac{n(n-3)}{2} = 2</math> diagonals, each hexagon <math>9</math>, and each octagon

5 KB (811 words) - 19:10, 25 January 2021
1988 AIME Problems/Problem 9
...equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework: ...0}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution.

6 KB (893 words) - 08:15, 2 February 2023
1988 AIME Problems/Problem 5
...99}</math> is an integer multiple of <math>10^{88}</math>. Find <math>m + n</math>. ...math>\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}</math>, and <math>m + n = \boxed{634}</math>.

822 bytes (108 words) - 22:21, 6 November 2016
1988 AIME Problems/Problem 4
Suppose that <math>|x_i| < 1</math> for <math>i = 1, 2, \dots, n</math>. Suppose further that What is the smallest possible value of <math>n</math>?

2 KB (394 words) - 10:21, 27 January 2024
1988 AIME Problems/Problem 3
1) <math>\log_a b^n=n\log_a b</math>. 2) <math>\log_{a^n} b=\frac{1}{n}\log_a b</math>.

3 KB (481 words) - 21:52, 18 November 2020
1988 AIME Problems/Problem 2
...ts of <math>k</math>. For <math>n \ge 2</math>, let <math>f_n(k) = f_1(f_{n - 1}(k))</math>. Find <math>f_{1988}(11)</math>. We see that <math>f_{1}(11)=4</math>

696 bytes (103 words) - 19:16, 27 February 2018
1989 AIME Problems/Problem 15
real x = 0.4, y = 0.2, z = 1-x-y; label("$X$", X, N);

13 KB (2,091 words) - 00:20, 26 October 2023
1989 AIME Problems/Problem 14
...ressed in the base <math>-n+i</math> using the integers <math>0,1,2,\ldots,n^2</math> as digits. That is, the equation <center><math>r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0</math></center>

2 KB (408 words) - 17:28, 16 September 2023
1989 AIME Problems/Problem 12
C=origin; B=(8,0); D=IP(CR(C,6.5),CR(B,8)); A=(4,-3); P=midpoint(A--B); Q=midpoint(C--D); ...p); dot("$C$",C,left,p); dot("$D$",D,up,p); dot("$M$",P,dir(-45),p); dot("$N$",Q,0.2*(Q-P),p);

2 KB (376 words) - 13:49, 1 August 2022
1989 AIME Problems/Problem 11
Let the mode be <math>x</math>, which we let appear <math>n > 1</math> times. We let the arithmetic mean be <math>M</math>, and the sum ...t| = \left|\frac{S+xn}{121}-x\right| = \left|\frac{S}{121}-\left(\frac{121-n}{121}\right)x\right|

5 KB (851 words) - 18:01, 28 December 2022
1989 AIME Problems/Problem 10
pair A = (0,0), B = (3, 0), C = (1, 4); draw(rightanglemark(C,P, B, 4));

8 KB (1,401 words) - 21:41, 20 January 2024
1989 AIME Problems/Problem 9
...h that <cmath>133^5+110^5+84^5+27^5=n^{5}.</cmath> Find the value of <math>n</math>. n^5&\equiv0\pmod{2}, \\

6 KB (874 words) - 15:50, 20 January 2024
1989 AIME Problems/Problem 8
...tion is of the form <cmath>f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7</cmath> for some <math>k\in\{1,2,3\}.</math> and we wish to find <math>f(4).</math>

8 KB (1,146 words) - 04:15, 20 November 2023
1989 AIME Problems/Problem 6
D(B--A); D(A--C); D(B--C,dashed); MP("A",A,SW);MP("B",B,SE);MP("C",C,N);MP("60^{\circ}",A+(0.3,0),NE);MP("100",(A+B)/2);MP("8t",(A+C)/2,NW);MP("7t t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}.\end{align*}</cmath>

5 KB (864 words) - 19:55, 2 July 2023
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) - 19:31, 30 July 2021
1990 AIME Problems/Problem 7
pair P=(-8,5),Q=(-15,-19),R=(1,-7),S=(7,-15),T=(-4,-17); MP("P",P,N,f);MP("Q",Q,W,f);MP("R",R,E,f);

8 KB (1,319 words) - 11:34, 22 November 2023
1990 AIME Problems/Problem 6
...otal number of fish in September is <math>125 \%</math>, or <math>\frac{5}{4}</math> times the total number of fish in May. ...s the number of fish in May. Solving for <math>n</math>, we see that <math>n = \boxed{840}</math>

2 KB (325 words) - 13:16, 26 June 2022
1990 AIME Problems/Problem 5
...gral divisors, including <math>1_{}^{}</math> and itself. Find <math>\frac{n}{75}</math>. ...to the least power. Therefore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}</math>.

1 KB (175 words) - 03:45, 21 January 2023
1990 AIME Problems/Problem 3
...the interior angle of a regular sided [[polygon]] is <math>\frac{(n-2)180}{n}</math>. {{AIME box|year=1990|num-b=2|num-a=4}}

3 KB (516 words) - 19:18, 16 April 2024
1991 AIME Problems/Problem 15
<math>\sum_{k=1}^n \sqrt{(2k-1)^2+a_k^2},</math> ...}</math> for which <math>S_n^{}</math> is also an integer. Find this <math>n^{}_{}</math>.

4 KB (658 words) - 16:58, 10 November 2023
1991 AIME Problems/Problem 13
Solving the resulting quadratic equation <math>r^{2}-rt+t(t-1)/4=0</math>, for <math>r</math> in terms of <math>t</math>, one obtains that ...e present case <math>t\leq 1991</math>, and so one easily finds that <math>n=44</math> is the largest possible integer satisfying the problem conditions

7 KB (1,328 words) - 20:24, 5 February 2024
1991 AIME Problems/Problem 12
...t terms, denote the [[perimeter]] of <math>ABCD^{}_{}</math>. Find <math>m+n^{}_{}</math>. ...$Q$",Q,E);label("$R$",R,SW);label("$S$",S,W); label("$15$",B/2+P/2,N);label("$20$",B/2+Q/2,E);label("$O$",O,SW); </asy></center>

8 KB (1,270 words) - 23:36, 27 August 2023
1991 AIME Problems/Problem 11
...es. The area of one circle is thus <math>\pi(2 - \sqrt {3})^{2} = \pi (7 - 4 \sqrt {3})</math>, so the area of all <math>12</math> circles is <math>\pi ...nce, the radius <math>r_{}^{}=R\sin(\pi/n)</math>. The total area <math>A_{n}^{}</math> of the <math>n_{}^{}</math> circles is thus given by

4 KB (740 words) - 19:33, 28 December 2022
1991 AIME Problems/Problem 10
...es the partial sums of <math>P_b</math> (in other words, <math>S_b = \sum_{n=1}^{b} P_b</math>): aab & 4 & 2 & 3 \\

5 KB (813 words) - 06:10, 25 February 2024
1991 AIME Problems/Problem 9
...c mn,</math> where <math>\frac mn</math> is in lowest terms. Find <math>m+n^{}_{}.</math> ...ot will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = \boxed{044}</math>.

10 KB (1,590 words) - 14:04, 20 January 2023
1991 AIME Problems/Problem 3
...may write <math>A_{k}^{}={N\choose k}x^{k}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>. Taking logarithms in both sides of this last equati ...ft[\prod_{j=1}^{k}\frac{(N-j+1)x}{j}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)x}{j}\right]\, .

5 KB (865 words) - 12:13, 21 May 2020
1991 AIME Problems/Problem 2
...}^{}</math> has [[edge | sides]] <math>\overline {AB}</math> of [[length]] 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</m pair A=(0,0),B=(4,0),C=(4,3),D=(0,3);

4 KB (595 words) - 12:51, 17 June 2021
1992 AIME Problems/Problem 15
...that the decimal representation of <math>m!</math> ends with exactly <math>n</math> zeroes. How many positive integers less than <math>1992</math> are n ...a multiple of <math>5</math>, <math>f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)</math>.

2 KB (358 words) - 01:54, 2 October 2020
1992 AIME Problems/Problem 13
...h> then plugging in <math>a+n</math> for <math>x</math> gives us <math>{(a+n)^2 \ge 0}</math>). ===Solution 4 (Involves Basic Calculus)===

4 KB (703 words) - 02:40, 29 December 2023
1992 AIME Problems/Problem 9
...<math>n^{}_{}</math> are relatively prime positive integers, find <math>m+n^{}_{}</math>. ...to find that <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>.

5 KB (874 words) - 10:27, 22 August 2021
1992 AIME Problems/Problem 8
...4-a_3,\ldots)</math>, whose <math>n^{\mbox{th}}_{}</math> term is <math>a_{n+1}-a_n^{}</math>. Suppose that all of the terms of the sequence <math>\Delt <cmath> a_{n} = \frac{1}{2}(n-19)(n-92) </cmath>

5 KB (778 words) - 21:36, 3 December 2022
1992 AIME Problems/Problem 4
\text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} ...iangle]] do three consecutive entries occur that are in the ratio <math>3 :4 :5</math>?

3 KB (476 words) - 14:13, 20 April 2024
1992 AIME Problems/Problem 3
...on, so that <math>\frac{n}{2n}=\frac{1}{2}</math>, and <math>\frac{n+3}{2n+4}>\frac{503}{1000}</math>. ..., <math>1000n+3000>1006n+2012</math>, so <math>n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}</math>. Thus, the answer is <math>\boxed{164}</math>.

2 KB (251 words) - 08:05, 2 January 2024
1992 AIME Problems/Problem 2
...> potential ascending numbers, one for each [[subset]] of <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9\}</math>. ...count each case individually: <math>\binom{n}{0}+\binom{n}{1}+...\binom{n}{n}</math> so the 2 statements are equivalent. Therfore we have <math>2^9-\bin

2 KB (336 words) - 05:18, 4 November 2022
2006 AMC 10B Problems/Problem 22
...math>, <math>J</math>, and <math>N</math> are positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? ...h>, so the peanut butter and jam for <math>N</math> sandwiches costs <math>N(4B+5J)\cent</math>.

2 KB (394 words) - 00:51, 25 November 2023
1993 AIME Problems/Problem 15
...</math> and <math>n\,</math> are relatively prime integers. Find <math>m + n\,</math>. A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H);

3 KB (449 words) - 21:39, 21 September 2023
1993 AIME Problems/Problem 14
...ath>\sqrt{N}\,</math>, for a positive integer <math>N\,</math>. Find <math>N\,</math>. ...s of this rectangle be <math>A(4,y)</math>, <math>B(-x,3)</math>, <math>C(-4,-y)</math> and <math>D(x,-3)</math> for nonnegative <math>x,y</math>. Then

3 KB (601 words) - 09:25, 19 November 2023
1993 AIME Problems/Problem 13
...0</math>. So, <math>\tan(\angle OXP)=\frac{OP}{PX}=\frac{50}{200}=\frac{1}{4}</math>. ...n^2(\angle OXP)} = \frac{2\cdot \left(\frac{1}{4}\right)}{1-\left(\frac{1}{4}\right)^2}=\frac{8}{15}</math>.

8 KB (1,231 words) - 20:06, 26 November 2023
1993 AIME Problems/Problem 12
...and <math>P_n\,</math> is the most recently obtained point, then <math>P_{n + 1}^{}</math> is the midpoint of <math>\overline{P_n L}</math>. Given tha ...e coordinates stay within the triangle. We have <cmath>P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\bmod{560},\ 2y_n\bmod{420})</cmath>

4 KB (611 words) - 13:59, 15 July 2023
1993 AIME Problems/Problem 11
...atively prime positive integers. What are the last three digits of <math>m+n\,</math>? ...frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\cdots = \frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{2}{3}</math>, and the probability of the second person winning is <

7 KB (1,058 words) - 20:57, 22 December 2020
1993 AIME Problems/Problem 8
...ion is double counted, except the case where both <math>m</math> and <math>n</math> contain all <math>6</math> elements of <math>S.</math> So our final ...possibilities; this is because it must contain all of the "missing" <math>n - k</math> elements and thus has a choice over the remaining <math>k.</math

9 KB (1,400 words) - 14:09, 12 January 2024
1993 AIME Problems/Problem 7
...ot6^2}{C(1000,6)\cdot6!}=\frac14,</cmath> from which the answer is <math>1+4=\boxed{005}.</math> ...ath>p = \frac{3 + 2}{20} = \frac{1}{4}</math>, and the answer is <math>1 + 4 = \boxed{005}</math>.

5 KB (772 words) - 09:04, 7 January 2022
1993 AIME Problems/Problem 6
...series of consecutive integers as <math>a,\ b,\ c</math>. Therefore, <math>n = a + (a + 1) \ldots (a + 8) = 9a + 36 = 10b + 45 = 11c + 55</math>. Simpli Let the desired integer be <math>n</math>. From the information given, it can be determined that, for positive

3 KB (524 words) - 18:06, 9 December 2023
1993 AIME Problems/Problem 5
...r [[integer]]s <math>n \ge 1\,</math>, define <math>P_n(x) = P_{n - 1}(x - n)\,</math>. What is the [[coefficient]] of <math>x\,</math> in <math>P_{20} Using the formula for the sum of the first <math>n</math> numbers, <math>1 + 2 + \cdots + 20 = \frac{20(20+1)}{2} = 210</math>

2 KB (355 words) - 13:25, 31 December 2018
1993 AIME Problems/Problem 4
...n</math>. From <math>a + d = b + c</math>, <math>d = b + c - a = a + 2m + n</math>. ...+ m</math>, <math>c = a + m + n</math>, and <math>d = b + c - a = a + 2m + n</math> into <math>bc - ad = 93</math>,

8 KB (1,343 words) - 16:27, 19 December 2023
1993 AIME Problems/Problem 3
...many contestants caught <math>n\,</math> fish for various values of <math>n\,</math>. <center><math>\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\

2 KB (364 words) - 00:05, 9 July 2022
1993 AIME Problems/Problem 2
...ast, etc. If the candidate went <math>n^{2}_{}/2</math> miles on the <math>n^{\mbox{th}}_{}</math> day of this tour, how many miles was he from his star ...,\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}</math>, and so on. The E/W displacement is thus <math>1^

2 KB (241 words) - 11:56, 13 March 2015
1994 AIME Problems/Problem 15
...0.7), dashes = linetype("8 6")+linewidth(0.7)+blue, bluedots = linetype("1 4") + linewidth(0.7) + blue; D(D(MP("A",A)) -- D(MP("B",B)) -- D(MP("C",C,N)) -- cycle);

4 KB (717 words) - 22:20, 3 June 2021
1994 AIME Problems/Problem 14
...), C = MP("C",D((1,0))), A = MP("A",expi(alpha * pi/180),N); path r = C + .4 * expi(beta * pi/180) -- C - 2*expi(beta * pi/180); ...,C+.4*expi(beta*pi/180)));MP("\beta",B,(5,1.2),fontsize(9));MP("\alpha",C,(4,1.2),fontsize(9));

2 KB (303 words) - 00:03, 28 December 2017
1994 AIME Problems/Problem 13
...i}{10}\right)</math> after expanding. Here <math>k</math> ranges from 0 to 4 because two angles which sum to <math>2\pi</math> are involved in the produ The expression to find is <math>\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}</math>.

3 KB (375 words) - 23:46, 6 August 2021
1994 AIME Problems/Problem 12
...row. Then <math>6|n</math>, and our goal is to maximize the value of <math>n</math>. ...th>, which we can easily verify works, and the answer is <math>\frac{13}{6}n^2 = \boxed{702}</math>.

3 KB (473 words) - 17:06, 1 January 2024
1994 AIME Problems/Problem 11
...tower 94 bricks tall. Each brick can be oriented so it contributes <math>4''\,</math> or <math>10''\,</math> or <math>19''\,</math> to the total heigh We have the smallest stack, which has a height of <math>94 \times 4</math> inches. Now when we change the height of one of the bricks, we eithe

4 KB (645 words) - 15:12, 15 July 2019
1994 AIME Problems/Problem 10
...and <math>n\,</math> are relatively prime positive integers. Find <math>m+n.\,</math> ...\frac{BC}{AB} = \frac{29^2 x}{29x^2} = \frac{29}{421}</math>, and <math>m+n = \boxed{450}</math>.

3 KB (534 words) - 16:23, 26 August 2018
1994 AIME Problems/Problem 5
...<math>n\,</math>. (If <math>n\,</math> has only one digits, then <math>p(n)\,</math> is equal to that digit.) Let ...a three-digit number (so <math>5 \equiv 005</math>), and since our <math>p(n)</math> ignores all of the zero-digits, replace all of the <math>0</math>s

2 KB (275 words) - 19:27, 4 July 2013
1994 AIME Problems/Problem 3
<cmath>T_{n-1} + T_n = n^2,</cmath> where <math>T_n = 1+2+...+n = \frac{n(n+1)}{2}</math> is the <math>n</math>th triangular number.

2 KB (252 words) - 11:12, 3 July 2023
1994 AIME Problems/Problem 2
...ath>, where <math>m</math> and <math>n</math> are integers. Find <math>m + n</math>. ...adratic formula]] shows that the answer is <math>\frac{16 \pm \sqrt{16^2 + 4 \cdot 240}}{2} = 8 \pm \sqrt{304}</math>. Discard the negative root, so our

2 KB (272 words) - 03:53, 23 January 2023
1994 AIME Problems/Problem 1
...h term, so <math>n = 4 + (997-1) \cdot 3 = 2992</math>. The value of <math>n^2 - 1 = 2992^2 - 1 \pmod{1000}</math> is <math>\boxed{063}</math>.

946 bytes (139 words) - 21:05, 1 September 2023
1995 AIME Problems/Problem 15
...and <math>n_{}</math> are relatively prime positive integers, find <math>m+n</math>. ...s can occur in a row, so the sequence is blocks of <math>1</math> to <math>4</math> <tt>H</tt>'s separated by <tt>T</tt>'s and ending in <math>5</math>

6 KB (979 words) - 13:20, 11 April 2022
1995 AIME Problems/Problem 14
...}</math> is not divisible by the square of any prime number. Find <math>m+n+d.</math> ...; D(MP("A",A,NW)--MP("B",B,SE)); D(MP("C",C,NE)--MP("D",D,SW)); D(MP("E",E,N)); D(C--B--O--E,d);D(O--D(MP("F",F,NE)),d); MP("39",(B+F)/2,NE);MP("30",(C+

3 KB (484 words) - 13:11, 14 January 2023
1995 AIME Problems/Problem 13
Let <math>f(n)</math> be the integer closest to <math>\sqrt[4]{n}.</math> Find <math>\sum_{k=1}^{1995}\frac 1{f(k)}.</math> ...}\right)^4 \right\rfloor</math> values of <math>n</math> for which <math>f(n) = k</math>. Expanding using the [[binomial theorem]],

2 KB (287 words) - 01:25, 12 December 2019
1995 AIME Problems/Problem 12
...> where <math>m_{}</math> and <math>n_{}</math> are integers, find <math>m+n.</math> // n = normal to plane

8 KB (1,172 words) - 21:57, 22 September 2022
1995 AIME Problems/Problem 10
Let our answer be <math>n</math>. Write <math> n = 42a + b </math>, where <math>a, b</math> are positive integers and <math> ...math>5</math> is the only prime divisible by <math>5</math>. We get <math> n = 215</math> as our largest possibility in this case.

3 KB (436 words) - 19:26, 2 September 2023
1995 AIME Problems/Problem 9
...ntsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE

7 KB (1,181 words) - 13:47, 3 February 2023
1995 AIME Problems/Problem 8
..._{y=1}^{99} \left\lfloor\frac{100-y}{y(y+1)} \right\rfloor = 49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.</cmath> ...+1)<100</math>. This yields the equations <math>x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9</math>.

4 KB (646 words) - 17:37, 1 January 2024
1995 AIME Problems/Problem 7
Given that <math>(1+\sin t)(1+\cos t)=5/4</math> and ...ath>m_{}</math> and <math>n_{}</math> [[relatively prime]], find <math>k+m+n.</math>

3 KB (427 words) - 09:23, 13 December 2023
1995 AIME Problems/Problem 5
...lues of <math>a, b, c,</math> and <math>d_{},</math> the equation <math>x^4+ax^3+bx^2+cx+d=0</math> has four non-real roots. The product of two of the ...conjugate of <math>m</math>, and <math>n'</math> be the conjugate of <math>n</math>. Then,

3 KB (451 words) - 15:02, 6 September 2021
1995 AIME Problems/Problem 4
<cmath>PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}</cmath> D(A--MP("A_9",G,N)); D(B--MP("A_3",F,N)); D(C--MP("A_6",D,N)); D(A--P); D(rightanglemark(A,G,P,12));

2 KB (339 words) - 19:29, 4 July 2013
1995 AIME Problems/Problem 3
...h> and <math>n</math> are relatively prime positive integers, find <math>m+n.</math> ...)</math>, so the number of steps the object may have taken is either <math>4</math> or <math>6</math>.

3 KB (602 words) - 23:15, 16 June 2019
1995 AIME Problems/Problem 1
...> and <math>n</math> are relatively prime positive integers. Find <math>m-n.</math> :<math>1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 + \left(\frac{1}{16}\right)^2</math>

2 KB (302 words) - 19:29, 4 July 2013
1996 AIME Problems/Problem 15
pair B=(0,0), A=expi(pi/4), C=IP(A--A + 2*expi(17*pi/12), B--(3,0)), D=A+C, O=IP(A--C,B--D); D(MP("A",A,N)--MP("B",B)--MP("C",C)--MP("D",D,N)--cycle); D(B--D); D(A--C); D(MP("O",O,SE));

5 KB (710 words) - 21:04, 14 September 2020
1996 AIME Problems/Problem 13
...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. BD^2 + DE^2 &= \frac{15}{4} \\

3 KB (521 words) - 01:18, 25 February 2016
1996 AIME Problems/Problem 12
...imes because there are <math>5</math> places <math>a_n</math> and <math>a_{n + 1}</math> can be. To find all possible values for <math>|a_n - a_{n - 1}|</math> we have to compute

5 KB (879 words) - 11:23, 5 September 2021
1996 AIME Problems/Problem 11
Let <math>\mathrm {P}</math> be the product of the [[root]]s of <math>z^6+z^4+z^3+z^2+1=0</math> that have a positive [[imaginary]] part, and suppose tha 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\

6 KB (1,022 words) - 20:23, 17 April 2021
1996 AIME Problems/Problem 9
...even lockers closed. Then he opens the lockers that are multiples of <math>4</math>, leaving only lockers <math>2 \pmod{8}</math> and <math>6 \pmod{8}</ .../math> and <math>6 \pmod{8}</math> are just lockers that are <math>2 \pmod{4}</math>. Edit by [[User: Yiyj1|Yiyj1]]

3 KB (525 words) - 23:51, 6 September 2023
1996 AIME Problems/Problem 7
fill(shift(4,3)*unitsquare,rgb(1,1,.4));fill(shift(4,5)*unitsquare,rgb(1,1,.4)); fill(shift(3,4)*unitsquare,rgb(.8,.8,.5));fill(shift(1,4)*unitsquare,rgb(.8,.8,.5));

4 KB (551 words) - 11:44, 26 June 2020
1996 AIME Problems/Problem 6
...th>m</math> and <math>n</math> are relatively prime integers. Find <math>m+n</math>. ...lity that one team wins all games is <math>5\cdot \left(\frac{1}{2}\right)^4=\frac{5}{16}</math>.

3 KB (461 words) - 01:00, 19 June 2019
1996 AIME Problems/Problem 3
...[[integer]] <math>n</math> for which the expansion of <math>(xy-3x+7y-21)^n</math>, after like terms have been collected, has at least 1996 terms. ...>y</math> and so none of the terms will need to be collected. Hence <math>(n+1)^2 \ge 1996</math>, the smallest square after <math>1996</math> is <math>

3 KB (515 words) - 04:29, 27 November 2023
1996 AIME Problems/Problem 2
.../math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even integer? ...<math>n</math> must satisfy these [[inequality|inequalities]] (since <math>n < 1000</math>):

1 KB (163 words) - 19:31, 4 July 2013
1997 AIME Problems/Problem 14
...n</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m+n</math>. ...\pi/1997</math>, and let <math>w</math> be the root corresponding to <math>n\theta=2n\pi/ 1997</math>. Then

5 KB (874 words) - 22:30, 1 April 2022
1997 AIME Problems/Problem 11
...h>x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}</math>. What is the greatest integer that does not exceed <math>100x ...m_{n=1}^{44} \sin n} = \frac{\sum_{n=46}^{89} \sin n}{\sum_{n=1}^{44} \sin n} = \frac {\sin 89 + \sin 88 + \dots + \sin 46}{\sin 1 + \sin 2 + \dots + \s

10 KB (1,514 words) - 14:35, 29 March 2024
1997 AIME Problems/Problem 4
...and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> [[Image:1997_AIME-4.png]]

2 KB (354 words) - 22:33, 2 February 2021
1997 AIME Problems/Problem 3
...also <math>0\pmod{10+b}</math>. Rewrite <math>(100x+10y+z)</math> as <math>n\times(10a+b)</math>. <math>(90a+9b-1)\times n(10a+b)= 1000(10a + b)</math>

2 KB (375 words) - 19:34, 4 August 2021
1998 AIME Problems/Problem 15
...is thereby represented by a directed segment from one vertex of the <math>n</math> -gon to another, and a proper sequence is represented as a path that ...it can be arranged that <math>n-2</math> segments will emanate from <math>n-2</math> of the vertices and that an odd number of segments will emanate fr

9 KB (1,671 words) - 22:10, 15 March 2024
1998 AIME Problems/Problem 14
..., where <math>m, n,</math> and <math>p</math> are integers, and <math>m\le n\le p.</math> What is the largest possible value of <math>p</math>? <cmath>2mnp = (m+2)(n+2)(p+2)</cmath>

2 KB (390 words) - 21:05, 29 May 2023
1998 AIME Problems/Problem 11
label("$(15,20,20)$",Pa,N); We can find the lengths of the sides of the polygons now. There are 4 [[right triangle]]s with legs of length 5 and 10, so their [[hypotenuse]]s

7 KB (1,084 words) - 11:48, 13 August 2023
1998 AIME Problems/Problem 8
| 0 || 1 || 2 || 3 || 4 || 5 || 6 *<math>n \equiv 0 \pmod{2}\quad\quad F_{n-1}\cdot 1000-F_n\cdot x</math>

2 KB (354 words) - 19:37, 24 September 2023
1998 AIME Problems/Problem 7
...er]]s that satisfy <math>\sum_{i = 1}^4 x_i = 98.</math> Find <math>\frac n{100}.</math> ..._i + 1</math> will be odd. Substituting we get <cmath>2y_1+2y_2+2y_3+2y_4 +4 = 98 \implies y_1+y_2+y_3+y_4 =47</cmath>

5 KB (684 words) - 11:41, 13 August 2023
1998 AIME Problems/Problem 5
...dd. <math>\frac {k(k-1)}2</math> will be even if <math>4|k</math> or <math>4|k-1</math>, and odd otherwise. ...c{(n)(n-1)}{2} + \frac{(n+1)(n)}{2} = \left(\frac n2\right)(n+1 - (n-1)) = n</math>. So the first two fractions add up to <math>19</math>, the next two

1 KB (225 words) - 02:20, 16 September 2017
1998 AIME Problems/Problem 4
...ath>n</math> are [[relatively prime]] [[positive integer]]s. Find <math>m+n.</math> ...and an odd tile. Thus, since there are <math>5</math> odd tiles and <math>4</math> even tiles, the only possibility is that one player gets <math>3</ma

5 KB (917 words) - 02:37, 12 December 2022
1998 AIME Problems/Problem 2
===Solution 4 - Unrigorous engineers induction solution=== ...was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480.

6 KB (913 words) - 16:34, 6 August 2020
1999 AIME Problems/Problem 14
...math>n_{}</math> are relatively [[prime]] positive integers. Find <math>m+n.</math> D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle);

7 KB (1,184 words) - 13:25, 22 December 2022
1999 AIME Problems/Problem 13
...n_{}</math> are [[relatively prime]] positive integers. Find <math>\log_2 n.</math> ...his into the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime. The only necessary step is to factor out all t

2 KB (329 words) - 01:38, 6 October 2015
1999 AIME Problems/Problem 10
...}</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m+n.</math> ...t 43 \cdot 42 \cdot 3!} = \frac{16}{473}</math>. The solution is <math>m + n = 489</math>.

3 KB (524 words) - 17:25, 17 July 2023
1999 AIME Problems/Problem 9
...nd <math>n_{}</math> are relatively prime positive integers, find <math>m+n.</math> ...th>\left(\frac{1}{2}\right)^2 + b^2 = 8^2 \Longrightarrow b^2 = \frac{255}{4}</math>, and the answer is <math>\boxed{259}</math>.

6 KB (1,010 words) - 19:01, 24 May 2023
1999 AIME Problems/Problem 7
...neral, the divisor-count of <math>\frac{N}{d}</math> must be a multiple of 4 to ensure that a switch is in position A: ...onsider the cases where the 3 factors above do not contribute multiples of 4.

3 KB (475 words) - 13:33, 4 July 2016
1999 AIME Problems/Problem 4
...}</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m+n.</math> ...h>. The area of the [[octagon]] (by [[subtraction]] of areas) is <math>1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy</math>.

3 KB (398 words) - 13:27, 12 December 2020
1999 AIME Problems/Problem 3
Find the sum of all [[positive integer]]s <math>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]]. ...h> for some positive integer <math>x</math>, then rearranging we get <math>n^2-19n+99-x^2=0</math>. Now from the quadratic formula,

2 KB (296 words) - 01:18, 29 January 2021
1999 AIME Problems/Problem 2
...}</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m+n</math>. ...\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 8
...and <math>n_{}</math> are relatively prime positive integers. Find <math>m+n.</math> ...3}}{2}} = \frac{14}{36} = \frac{7}{18}</math>, and the answer is <math>m + n = \boxed{025}</math>.

3 KB (445 words) - 19:40, 4 July 2013
2000 AIME I Problems/Problem 15
...eck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 ca ...mall number. If you don't want to do this, define sequence <math>a_n = 2a_{n-1} - 1</math>, and solve for the closed form, which is very easy). Conseque

15 KB (2,673 words) - 19:16, 6 January 2024
2000 AIME I Problems/Problem 14
...=80</math>, <math>\angle APQ = 180-2\cdot 20 = 140 \Longrightarrow r=\frac{4}{7}</math> ...math>, <math>\angle ACB = 80</math>, so <math>r = \frac {80}{140} = \frac {4}{7}</math>.

8 KB (1,275 words) - 03:04, 27 February 2022
2000 AIME I Problems/Problem 13
...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. ...t most <math>t=\frac{1}{10}-\frac{x}{50}</math> hours, or <math>d=rt=14t=1.4-\frac{7x}{25}</math> miles. It can end up anywhere off the highway in a cir

3 KB (571 words) - 00:38, 13 March 2014
2000 AIME I Problems/Problem 11
...of <math>1000 = 2^35^3</math> can be written in the form of <math>2^{m}5^{n}</math>, it follows that <math>\frac{a}{b}</math> can also be expressed in == Solution 4 (head on)==

4 KB (667 words) - 13:58, 31 July 2020
2000 AIME I Problems/Problem 9
<cmath>\begin{eqnarray*}\log_{10}(2000xy) - (\log_{10}x)(\log_{10}y) & = & 4 \\ ...h>y=10</math>. Substituting into the first equation yields <math>\log20000=4</math> which is not possible.

4 KB (623 words) - 15:56, 8 May 2021
2000 AIME I Problems/Problem 8
...>p</math> is not divisible by the cube of any prime number. Find <math>m + n + p</math>. ...{1}</math> of the volume, and therefore <math>\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}</math> of the height when the vertex is at the

4 KB (677 words) - 16:33, 30 December 2023
2000 AIME I Problems/Problem 7
...<math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>. ..., <math>z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}</math>, so <math>m+n=\boxed{005}</math>.

5 KB (781 words) - 15:02, 20 April 2024
2000 AIME I Problems/Problem 6
x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\

6 KB (966 words) - 21:48, 29 January 2024
2000 AIME I Problems/Problem 5
...ath>n</math> are [[relatively prime]] positive integers. What is <math>m + n</math>? ...quickly see that there is no direct combinatorics way to calculate <math>m/n</math>. The [[Principle of Inclusion-Exclusion]] still requires us to find

7 KB (1,011 words) - 20:09, 4 January 2024
2000 AIME I Problems/Problem 4
...h chasing" and get <math>4a - 4 = 2a + 5</math>. Solving, we get <math>a = 4.5</math> and the side lengths are <math>61</math> and <math>69</math>. Thus {{AIME box|year=2000|n=I|num-b=3|num-a=5}}

3 KB (485 words) - 00:31, 19 January 2024
2000 AIME I Problems/Problem 3
{{AIME box|year=2000|n=I|num-b=2|num-a=4}}

679 bytes (98 words) - 00:51, 2 November 2023
2000 AIME I Problems/Problem 1
...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 If a factor of <math>10^{n}</math> has a <math>2</math> and a <math>5</math> in its [[prime factorizat

1 KB (163 words) - 17:44, 16 December 2020
2001 AIME I Problems/Problem 15
...and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> ...sent 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 corresponds

11 KB (1,837 words) - 18:53, 22 January 2024
2001 AIME I Problems/Problem 14
The last two digits of any <math>n</math>-digit string can't be <math>11</math>, so the only possibilities are ...ngs ending in <math>01</math>, and <math>c_n</math> be the number of <math>n</math>-digit strings ending in <math>10</math>.

13 KB (2,298 words) - 19:46, 9 July 2020
2001 AIME I Problems/Problem 13
...e <math>m</math> and <math>n</math> are positive integers. Find <math>m + n.</math> <cmath>x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}</cmath>

3 KB (561 words) - 19:25, 27 November 2022
2001 AIME I Problems/Problem 12
...and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> currentprojection = perspective(-2,9,4);

6 KB (1,050 words) - 18:44, 27 September 2023
2014 USAJMO Problems
===Problem 4=== ...that cannot be represented in the form <math>n+s_b(n)</math>, where <math>n</math> is a positive integer.

3 KB (600 words) - 16:42, 5 August 2023
2001 AIME I Problems/Problem 11
...th> and <math>x_5 = y_3.</math> Find the smallest possible value of <math>N.</math> <cmath>\begin{align*}x_i &= (i - 1)N + c_i\\

3 KB (493 words) - 13:51, 22 July 2020
2001 AIME I Problems/Problem 10
...and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> .../math>, <math>(2,0)</math>, <math>(4,0)</math>, <math>(2,4)</math>, <math>(4,2)</math>, <math>(1,3)</math>, and <math>(3,1)</math>, <math>13</math> poss

8 KB (1,187 words) - 02:40, 28 November 2020
2001 AIME I Problems/Problem 9
...nd <math>n</math> are [[relatively prime]] positive integers. Find <math>m+n</math>. D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,N))--cycle);

4 KB (673 words) - 20:15, 21 February 2024
2001 AIME I Problems/Problem 7
...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. D(MP("A",A,N)--MP("B",B)--MP("C",C)--cycle); D(MP("I",I,NE)); D(MP("E",E,NE)--MP("D",D,N

9 KB (1,540 words) - 08:31, 1 December 2022
2001 AIME I Problems/Problem 6
...</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m + n</math>. The solution to this problem is therefore <math>\dfrac{\binom{9}{4}}{6^4} = {\dfrac{7}{72}}</math>. So the answer is <math>\boxed{079}</math>.

11 KB (1,729 words) - 20:50, 28 November 2023
2001 AIME I Problems/Problem 5
...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. ...th>A,B,</math> and <math>C,</math> where <math>B</math> is in [[quadrant]] 4 and <math>C</math> is in quadrant <math>3.</math>

6 KB (1,043 words) - 10:09, 15 January 2024
2001 AIME I Problems/Problem 4
<math>\sin(75)</math> <math>=</math> <math>\frac{\sqrt{6} + \sqrt{2}}{4}</math> Therefore, the area of the triangle is <math>\frac{\sqrt{6} + \sqrt{2}}{4}</math> <math>\cdot</math> <math>24</math> <math>\cdot</math> <math>12\sqrt

3 KB (534 words) - 03:22, 23 January 2023
2001 AIME I Problems/Problem 3
...1} + \cdots + a_0 = 0</math>, then the sum of the roots is <math>\frac{-a_{n-1}}{a_n}</math>. ...the roots is <cmath>2000(\sum_{n=1}^{2000} y+\frac{1}{4})=2000(0+\frac{1}{4})=\boxed{500}.</cmath>

2 KB (335 words) - 18:38, 9 February 2023
2001 AIME I Problems/Problem 1
*For <math>b = a</math>, we have <math>n = 11a</math> for nine possibilities, giving us a sum of <math>11 \cdot \fra ...er ones give <math>b > 9</math>), giving us a sum of <math>12 \cdot \frac {4(5)}{2} = 120</math>.

4 KB (687 words) - 18:37, 27 November 2022
2002 AIME I Problems/Problem 15
...over the square, so we must have <math>E(2,0,12)</math> so <math>CE=\sqrt{4^2+6^2+12^2}=\sqrt{196}=14</math> (the other solution, <math>E(10,0,12)</mat ...may also do this by vectors; <math>\vec{AD}\times\vec{DE}=(12,0,0)\times(-4,-6,12)=(0,-144,-72)=-72(0,2,1)</math>, so the plane is <math>2y+z=2\cdot6=1

7 KB (1,181 words) - 20:32, 8 January 2024
2002 AIME I Problems/Problem 13
...d <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. BC^2 + 4 \cdot 18^2 &= 2\left(24^2 + AC^2\right) \\

6 KB (974 words) - 13:01, 29 September 2023
2002 AIME I Problems/Problem 10
== Solution 4 No Trig == {{AIME box|year=2002|n=I|num-b=9|num-a=11}}

4 KB (643 words) - 22:44, 8 August 2023
2002 AIME I Problems/Problem 9
...h>, <math>t</math> must be <math>2</math> and <math>u</math> must be <math>4</math>, in order for <math>5,6</math> to be paint-able. Thus <math>424</mat <math>h</math> cannot be greater than <math>4</math>, since if that were the case then the answer would be greater than <

4 KB (749 words) - 19:44, 25 April 2024
2002 AIME I Problems/Problem 6
<center><math>\log_{225}x+\log_{64}y=4</math></center> From the first equation: <math>A+B=4 \Rightarrow B = 4-A</math>.

1 KB (194 words) - 19:55, 23 April 2016
2002 AIME I Problems/Problem 5
{{AIME box|year=2002|n=I|num-b=4|num-a=6}}

1 KB (220 words) - 20:50, 12 November 2022
2002 AIME I Problems/Problem 3
...e Dick's present age. How many ordered pairs of positive integers <math>(d,n)</math> are possible? Let Jane's age <math>n</math> years from now be <math>10a+b</math>, and let Dick's age be <math>10

2 KB (246 words) - 17:02, 21 May 2023
2002 AIME I Problems/Problem 2
...three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is <math>2\sqrt{3}</math>, and the sho {{AIME box|year=2002|n=I|num-b=1|num-a=3}}

2 KB (287 words) - 19:54, 4 July 2013
2003 AIME I Problems/Problem 15
...d <math> n </math> are relatively prime positive integers. Find <math> m + n. </math> ..."M",M)));D(A--MP("F'",Fprime,SW)--D); MP("E",E,NE); D(rightanglemark(F,D,B,4)); MP("390",(M+C)/2); MP("390",(M+C)/2); MP("360",(A+B)/2,NW); MP("507",(B+

8 KB (1,382 words) - 14:23, 29 December 2022
2003 AIME I Problems/Problem 14
...</math> consecutively and in that order. Find the smallest value of <math> n </math> for which this is possible. To find the smallest value of <math>n</math>, we consider when the first three digits after the decimal point are

3 KB (477 words) - 14:23, 4 January 2024
2003 AIME I Problems/Problem 13
...<math>1</math>'s than <math>0</math>'s. Find the [[remainder]] when <math> N </math> is divided by <math>1000</math>. ...</math>. Thus there are <math>{n \choose k}</math> numbers that have <math>n+1</math> digits in base <math>2</math> notation, with <math>k+1</math> of t

4 KB (651 words) - 19:42, 7 October 2023
2003 AIME I Problems/Problem 10
D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D(MP("M",M))--B); D(C--M);

7 KB (1,058 words) - 01:41, 6 December 2022
2003 AIME I Problems/Problem 9
...th pairs. This gives <math>\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^9 n^2 = 285</math> balanced numbers. Thus, there are in total <math>330 + 285 ...uare|sum of consecutive squares]], namely <math>\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}</math>.

4 KB (696 words) - 11:55, 10 September 2023
2003 AIME I Problems/Problem 8
<cmath>400=2^4 \cdot 5^2=(y+x)(y-x).</cmath> {{AIME box|year=2003|n=I|num-b=7|num-a=9}}

5 KB (921 words) - 23:21, 22 January 2023
2003 AIME I Problems/Problem 7
D(D(MP("B",B))--D); D((0,-4)--(0,12),linetype("4 4")+linewidth(0.7)); {{AIME box|year=2003|n=I|num-b=6|num-a=8}}

3 KB (490 words) - 18:13, 13 February 2021
2003 AIME I Problems/Problem 6
...math> m, n, </math> and <math> p </math> are [[integer]]s. Find <math> m + n + p. </math> Each face of the cube contains <math>{4\choose 3} = 4</math> triangles of the first type, and there are <math>6</math> faces, so

3 KB (477 words) - 18:35, 27 December 2021
2003 AIME I Problems/Problem 5
...> n </math> and <math> p </math> are [[relatively prime]], find <math> m + n + p. </math> 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 3
*<math>8</math> will be the greater number in <math>4</math> subsets. Note: Note that <math>7+6+5+4+3+2+1=\binom{8}{2}</math>, so we have counted all the possible cases.

2 KB (317 words) - 00:09, 9 January 2024
2003 AIME I Problems/Problem 2
...n </math> are [[relatively prime]] [[positive integer]]s. Find <math> m + n. </math> *For <math>2</math> circles, the ratio is <math>3/4</math>.

4 KB (523 words) - 15:49, 8 March 2021
2003 AIME II Problems/Problem 15
<center><math>\sum_{k = 1}^{r}|b_{k}| = m + n\sqrt {p},</math></center> ...<math>p</math> is not divisible by the square of any prime. Find <math>m + n + p.</math>

4 KB (675 words) - 17:23, 30 July 2022
2003 AIME II Problems/Problem 14
...integers and n is not divisible by the square of any prime. Find <math>m + n.</math> The y-coordinate of <math>F</math> must be <math>4</math>. All other cases yield non-convex and/or degenerate hexagons, which

9 KB (1,461 words) - 15:09, 18 August 2023
2003 AIME II Problems/Problem 13
...and <math>n</math> are relatively prime positive integers, find <math>m + n.</math> ...hing the starting vertex in the next move. Thus <math>P_n=\frac{1}{2}(1-P_{n-1})</math>.

15 KB (2,406 words) - 23:56, 23 November 2023
2003 AIME II Problems/Problem 12
And if for some <math>i</math> we have <math>v_i=4</math>, then <math>s\geq \frac{400}3 = 133\frac13</math>, and hence <math>s ...ath> candidates got <math>5</math> votes each, and one candidate got <math>4</math> votes. In this situation, the total number of votes is exactly <math

4 KB (759 words) - 13:00, 11 December 2022
2003 AIME II Problems/Problem 11
...<math>n</math> is not divisible by the square of any prime, find <math>m + n + p.</math> Let <math>N</math> be the orthogonal projection from <math>C</math> to <math>AB.</math>

5 KB (772 words) - 19:47, 1 August 2023
2003 AIME II Problems/Problem 9
...of <math>Q(x) = 0,</math> find <math>P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).</math> <math>a_ns_2+a_{n-1}s_1+2a_{n-2}=0</math>

2 KB (376 words) - 11:47, 28 April 2024
2003 AIME II Problems/Problem 8
...ath> term of sequence <math>A</math> is <math>a+d_1 n</math> and the <math>n^{\text{th}}</math> term of <math>B</math> is <math>b + d_2 n</math>. Thus, the <math>n^{\text{th}}</math> term of the given sequence is

5 KB (793 words) - 15:18, 14 July 2023
2003 AIME II Problems/Problem 7
...th>. Also, the area of <math>\triangle ABC</math> is <math>ab=2b(a^2+b^2)/(4\cdot25)</math>. Setting these two expressions equal to each other and simpl {{AIME box|year=2003|n=II|num-b=6|num-a=8}}

2 KB (323 words) - 09:56, 16 September 2022
2003 AIME II Problems/Problem 6
D=(7.6667,-4); G=(6.3333,4);

5 KB (787 words) - 17:38, 30 July 2022
2003 AIME II Problems/Problem 5
...expressed as <math>n\pi</math>, where n is a positive integer. Find <math>n</math>. ...ing one). Thus, <math>V=\dfrac{6^2\cdot 12\pi}{2}=216\pi</math>, so <math>n=\boxed{216}</math>.

1 KB (204 words) - 17:41, 30 July 2022
2003 AIME II Problems/Problem 4
...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. Embed the tetrahedron in 4-space to make calculations easier.

3 KB (563 words) - 17:36, 30 July 2022
2003 AIME II Problems/Problem 3
...here are <math>3\cdot2^{n-1}\ n</math>-letter good words. Substitute <math>n=7</math> to find there are <math>3\cdot2^6=\boxed{192}</math> seven-letter {{AIME box|year=2003|n=II|num-b=2|num-a=4}}

2 KB (336 words) - 17:29, 30 July 2022
2003 AIME II Problems/Problem 1
...is the sum of the other two. Find the sum of all possible values of <math>N</math>. ...is one of <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
r[1] = 4.096; label("$y = mx$", (8,12*sqrt(221)/49*8), N);

7 KB (1,182 words) - 09:56, 7 February 2022
2002 AIME II Problems/Problem 13
...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. X=(4.3636,0);

6 KB (935 words) - 13:23, 3 September 2021
2002 AIME II Problems/Problem 12
...ath> and <math>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 ...be represented by the number of paths from <math>(0,0)</math> to <math>(6,4)</math> that always stay below the line <math>y=\frac{2x}{3}</math>. We can

7 KB (1,127 words) - 13:34, 19 June 2022
2002 AIME II Problems/Problem 11
...n in the form <math>\frac{\sqrt{m}-n}p</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers and <math>m</math> is not .... Thus, <math>64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}</math>.

4 KB (696 words) - 16:27, 22 March 2022
2002 AIME II Problems/Problem 8
...oor\frac{2002}{n}\right\rfloor=k</math> has no integer solutions for <math>n</math>. (The notation <math>\lfloor x\rfloor</math> means the greatest inte ...ther <math>\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor</math>,

6 KB (908 words) - 14:22, 14 July 2023
2002 AIME II Problems/Problem 7
...th>200</math> if <math>k(k+1)(2k+1)</math> is a multiple of <math>1200 = 2^4 \cdot 3 \cdot 5^2</math>. <math>7, 23, 14, 5, 21, 12, 3, 19, 10, 1, 17, 8, 24, 15, 6, 22, 13, 4, 20, 11, 27, 18, 9, 0,</math> and then it loops.

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>. ...to two fractions: <math>\frac{1}{(n+2)(n-2)} = \frac{A}{(n+2)} + \frac{B}{(n-2)}</math> for some A and B.

2 KB (330 words) - 05:56, 23 August 2022
2002 AIME II Problems/Problem 5
Find the sum of all positive integers <math>a=2^n3^m</math> where <math>n</math> and <math>m</math> are non-negative integers, for which <math>a^6</m ...</math> and <math>6^a</math>, and find all 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}

3 KB (515 words) - 14:46, 14 February 2021
2002 AIME II Problems/Problem 3
...math>, from which we can deduce that <math>c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48</math>. ...error shown in solution 1, we get <math>a = 27</math>, and <math>r = \frac{4}{3}</math>. Now, <math>27*r^2= c = 48</math>. Therefore, the answer is <mat

2 KB (263 words) - 22:50, 5 April 2024
2001 AIME II Problems/Problem 15
...<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
...gles are measured in degrees. Find the value of <math>\theta_{2} + \theta_{4} + \ldots + \theta_{2n}</math>. Listing all of these values, we find that <math>\theta_{2} + \theta_{4} + \ldots + \theta_{2n}</math> is equal to <math>(75 + 165 + 255 + 345) ^\c

2 KB (380 words) - 15:03, 22 July 2018
2001 AIME II Problems/Problem 13
...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. ...rk(E,C,D,anglesize));D(anglemark(A,B,D,5/4*anglesize));D(anglemark(E,C,D,5/4*anglesize));

4 KB (743 words) - 03:32, 23 January 2023
2001 AIME II Problems/Problem 12
...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. ...= \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
...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. ...{1 - \frac{141}{729}}{2} = \frac{98}{243}</math>, so the answer is <math>m+n = \boxed{341}</math>.

3 KB (415 words) - 23:25, 20 February 2023
2001 AIME II Problems/Problem 10
...</math>, we see that <math>j-i = 6</math> works; also, <math>a-b | a^n - b^n</math> implies that <math>10^{6} - 1 | 10^{6k} - 1</math>, and so any <math ...math>, and so forth. Therefore, the answer is <math>94 + 88 + 82 + \dots + 4\implies 16\left(\dfrac{98}{2}\right) = \boxed{784}</math>.

4 KB (549 words) - 23:16, 19 January 2024
2001 AIME II Problems/Problem 9
...<math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>. ...en there are <math>2^5</math> ways to color the rest of the squares. <math>4*32=128</math>

8 KB (1,207 words) - 20:04, 5 September 2023
2001 AIME II Problems/Problem 7
...ath>C_{3}</math> can be written as <math>\sqrt {10n}</math>. What is <math>n</math>? ...MP("V",V,NE)); D(O2 -- O3, rgb(0.2,0.5,0.2)+ linewidth(0.7) + linetype("4 4"));

7 KB (1,112 words) - 02:15, 26 December 2022
2001 AIME II Problems/Problem 6
...h> and <math>n</math> are relatively prime positive integers and <math>m < n</math>. Find <math>10n + m</math>. ...50); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7);

4 KB (772 words) - 19:31, 6 December 2023
2001 AIME II Problems/Problem 5
...ts have the triangle property. What is the largest possible value of <math>n</math>? <cmath>\mathcal{S} = \{\, 4,\, 5,\, 4+5, \,5+(4+5),\, \ldots\,\} = \{4, 5, 9, 14, 23, 37, 60, 97, 157, 254\}</cmath>

2 KB (286 words) - 22:32, 5 January 2024
2001 AIME II Problems/Problem 4
...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. D((0,0)--MP("x",(13,0),E),EndArrow(6)); D((0,0)--MP("y",(0,10),N),EndArrow(6));

2 KB (240 words) - 20:34, 4 July 2013
2001 AIME II Problems/Problem 3
x_{4}&=523,\ \text{and}\\ 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
2000 AIME II Problems/Problem 15
...rc\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.</math></center> ...n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{align*}</cmath>

3 KB (469 words) - 21:14, 7 July 2022
2000 AIME II Problems/Problem 13
...nd <math>r</math> are relatively prime, and <math>r>0</math>. Find <math>m+n+r</math>. 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\

6 KB (1,060 words) - 17:36, 26 April 2024
2000 AIME II Problems/Problem 12
...d <math>n</math> is not divisible by the square of any prime. Find <math>m+n+k</math>. <cmath>R = \frac{abc}{4K} = \frac{(13)(14)(15)}{4(84)} = \frac{65}{8}</cmath>

3 KB (532 words) - 13:14, 22 August 2020
2000 AIME II Problems/Problem 11
...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. pair A=(0,0), D=(1,7), Da = MP("D'",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot(A,C,D);

4 KB (750 words) - 22:55, 5 February 2024
2000 AIME II Problems/Problem 9
...3 )z + 1 = 0</math>, we have <math>z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}</math>. ...than 2. However <math>(4\cos^2 3 - 2)^2 -2</math> is also less than <math>4\cos^2 3 - 2</math>. we can see that every time we square the equation, the

4 KB (675 words) - 13:42, 4 April 2024
2000 AIME II Problems/Problem 8
<cmath>x^2 + \left(y-\sqrt{11}\right)^2 = 1001 \Longrightarrow x^4 - 11x^2 - 11^2 \cdot 9 \cdot 10 = 0</cmath> ..."B",B,W)--MP("C",C)--MP("D",D)--cycle); D(A--C);D(B--D);D(A--E,linetype("4 4") + linewidth(0.7));

4 KB (584 words) - 19:35, 7 December 2019
2000 AIME II Problems/Problem 7
...find the [[floor function|greatest integer]] that is less than <math>\frac N{100}</math>. ...{6!13!}+\frac {19!}{7!12!}+\frac {19!}{8!11!}+\frac {19!}{9!10!}=\frac {19!N}{1!18!}.</cmath>

2 KB (281 words) - 12:09, 5 April 2024
2000 AIME II Problems/Problem 6
<center><asy>pathpen = linewidth(0.7); pen d = linetype("4 4") + linewidth(0.7); D(A--B--C--D--cycle); D((A+D)/2 -- (B+C)/2, d); MP("b",(C+D)/2,N);MP("b+100",(A+B)/2);

3 KB (433 words) - 19:42, 20 December 2021
2000 AIME II Problems/Problem 5
...t each finger have a ring. Find the leftmost three nonzero digits of <math>n</math>. ...g the fingers is equivalent the number of ways we can drop five balls into 4 urns, or similarly dropping five balls into four compartments split by thre

1 KB (184 words) - 21:13, 12 September 2020
2000 AIME II Problems/Problem 4
We use the fact that the number of divisors of a number <math>n = p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}</math> is <math>(e_1 + 1)(e_2 + 1) \c Dividing the greatest power of <math>2</math> from <math>n</math>, we have an odd integer with six positive divisors, which indicates

2 KB (397 words) - 15:55, 11 May 2022
2000 AIME II Problems/Problem 3
...<math>n</math> are [[relatively prime]] positive integers. Find <math>m + n.</math> ...the answer is <math>\frac{54+1}{703} = \frac{55}{703}</math>, and <math>m+n = \boxed{758}</math>.

1 KB (191 words) - 04:27, 4 November 2022
2000 AIME II Problems/Problem 1
...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. Therefore, <math> m+n=1+6=\boxed{007}</math>

2 KB (292 words) - 13:33, 4 April 2024
2005 AMC 10B Problems/Problem 7
...i}{8} \qquad \textbf{(C) } \frac{3\pi}{16} \qquad \textbf{(D) } \frac{\pi}{4} \qquad \textbf{(E) } \frac{\pi}{2} </math> ...rt{2}}{4}</math>. The area of this circle is <math>\left(\dfrac{x\sqrt{2}}{4}\right)^2\pi=\dfrac{2\pi x^2}{16}=\dfrac{x^2\pi}{8}</math>, and the area of

2 KB (381 words) - 14:28, 14 December 2021
2005 AMC 10B Problems/Problem 24
...h> gives us <math>2a=12 \Longrightarrow a=6, b=5</math>. Testing <math>a-b=4</math> gives us <math>2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}</ ...t need to square 11. So <math>3^2 \cdot 11^2</math> gives us 1089 as <math>n</math> and <math>m = \sqrt{1089} = 33.</math> We now get the equation <math

5 KB (845 words) - 19:23, 17 September 2023
2005 AMC 10B Problems/Problem 25
...+(n+1) = 2n+1 = 125</math>. When we solve for <math>n</math>, we get <math>n =\boxed{\textbf{(C)}\ 62}</math>. ==Solution 4==

3 KB (517 words) - 19:15, 15 October 2023
2006 USAMO Problems/Problem 3
...d above. (In particular, this requires <math>f(n^2)\neq 0</math> for <math>n\ge 0</math>.) Let <math>g(n) = f(n^2)</math>, then <math>g(n)</math> is a polynomial of degree <math>2</math> or

9 KB (1,699 words) - 13:48, 11 April 2020
2006 USAMO Problems/Problem 5
...e <math>2^{m_n}</math> is the largest power of 2 that is a factor of <math>n</math>. Show that if <math>k\ge 2</math> is a positive integer and <math>i< {{USAMO newbox|year=2006|num-b=4|num-a=6}}

7 KB (1,280 words) - 17:23, 26 March 2016
2006 USAMO Problems/Problem 4
...math> satisfying <math>a_1 + a_2 + \cdots + a_k = a_1\cdot a_2\cdots a_k = 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</math>.

3 KB (486 words) - 22:43, 5 August 2014
2006 AMC 12A Problems/Problem 6
dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);

3 KB (528 words) - 18:14, 16 December 2021
2006 AMC 12A Problems/Problem 8
...bf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math> *<math>1 + 2 + 3 + 4 + 5 = 15</math>

3 KB (450 words) - 02:00, 13 January 2024
Diagonal
...onals of a polygon with <math>n</math> vertices is given by <math>\frac{n(n-3)}{2}</math>. ...dpoint. So, we simply divide by 2 to get our final formula, <math>\frac{n(n-3)}{2}</math>.

2 KB (374 words) - 00:37, 25 January 2015
Inradius
pair A=(0,0),B=(4,0),C=(1.5,2),I=incenter(A,B,C),F=foot(I,A,B); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(CR(D(MP("I",I,SW)),inradius(A,B,C))); D(F--I--foot(I,B,C)--I--f

2 KB (336 words) - 00:44, 23 April 2024
Equation
<math>x^2 = 4</math> ...as the form <math>ax + by + cz + ... = n</math>, where <math>a, b, c, ..., n</math> are numbers and <math>x, y, z, ...</math> are the variables. In line

5 KB (932 words) - 12:57, 26 July 2023
Quadratic reciprocity
...ulo <math>p</math> if there exists an integer <math>n</math> so that <math>n^2\equiv a\pmod p</math>. * <math>\genfrac{(}{)}{}{}{p}{q} \genfrac{(}{)}{}{}{q}{p} = (-1)^{(p-1)(q-1)/4} .</math>

7 KB (1,182 words) - 16:46, 28 April 2016
2006 AMC 10B Problems
...{\spadesuit} y = (x+y)(x-y) </math>. What is <math> 3 \mathop{\spadesuit} (4 \mathop{\spadesuit} 5) </math>? == Problem 4 ==

14 KB (2,059 words) - 01:17, 30 January 2024
2005 AMC 10B Problems
...x</math> has the property that <math>x\%</math> of <math>x</math> is <math>4</math>. What is <math>x</math>? <math>\mathrm{(A)} 2 \qquad \mathrm{(B)} 4 \qquad \mathrm{(C)} 10 \qquad \mathrm{(D)} 20 \qquad \mathrm{(E)} 40</math>

12 KB (1,874 words) - 21:20, 23 December 2020
Noncommutative
...and [[multiplication]] of numbers, are [[commutative]]. For example, <math>4\cdot3=3\cdot4=12</math>, and <math>2+3=3+2=5</math>. If <math>A</math> and <math>B</math> are both <math>n\times n</math> [[matrix|matrices]], then usually, <math>AB\ne BA</math>. For examp

2 KB (257 words) - 15:30, 26 December 2017
2000 AMC 12 Problems/Problem 8
draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);

7 KB (988 words) - 15:14, 10 April 2024
Angle bisector theorem
...[[Stewart's theorem]] it can be shown that <math>AD^2 = b\cdot c - m \cdot n</math> ...C)/2,(1,0));label("$c$",(A+B)/2,(0,1));label("$m$",(B+D)/2,(0,-1));label("$n$",(D+C)/2,(0,-1)); </asy>

3 KB (438 words) - 14:20, 4 March 2023
Divisibility rules/Rule 1 for 13 proof
Truncate the last digit, multiply it by 4 and add it to the rest of the number. The result is divisible by 13 if and ...d {13}</math> if and only if <math>4N \equiv 0 \pmod {13}</math>, so <math>n \equiv 0 \pmod{13}</math> if and only if <math>4d_0 - 12k \equiv 0 \pmod{13

1 KB (178 words) - 14:20, 12 April 2021
Sigma notation
...m_{k=m}^n a(k)</math> for the sum <math>a(m)+a(m+1)+a(m+2)+\ldots+a(n-1)+a(n)</math>. ...ber k in the sequence. [[Factoring]], we get, <math>5+10+15+20+25=5(1+2+3+4+5)</math>. Now, we see that each term is five times it's number in the seq

2 KB (335 words) - 17:17, 8 February 2024
University of South Carolina High School Math Contest/1993 Exam/Problem 11
...4 cards labeled 1 to 4 are 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 This is the number of [[derangement]]s of 4 objects. We can know the formula for derangements or count in one of two w

2 KB (334 words) - 16:27, 25 October 2023
1977 Canadian MO Problems/Problem 3
....</math> Find the smallest positive integer <math>b</math> for which <math>N</math> is the fourth power of an integer. ...ecause <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> we check to see if <math>

713 bytes (114 words) - 01:45, 19 August 2012
University of South Carolina High School Math Contest/1993 Exam/Problems
<cmath> \mathrm{(A) \ } 2 \qquad \mathrm{(B) \ }3\qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6 </cmath> == Problem 4 ==

14 KB (2,102 words) - 22:03, 26 October 2018
Mock AIME 1 2006-2007 Problems/Problem 3
...}</math> where <math>n</math> is an integer, find the remainder when <math>n^{2007}</math> is divided by <math>1000</math>. *[[Mock AIME 1 2006-2007 Problems/Problem 4 | Next Problem]]

963 bytes (135 words) - 15:53, 3 April 2012
Mock AIME 1 2006-2007 Problems/Problem 5
For a [[prime number]] <math>p</math>, define the [[function]] <math>f_p(n)</math> as follows: set <math>f_p(n) = y</math>. Otherwise, set <math>f_p(n) = 0</math>. Compute the sum <math>f_{11}(1) + f_{11}(2) + \ldots + f_{11}

2 KB (340 words) - 15:52, 3 April 2012
Mock AIME 1 2006-2007 Problems/Problem 9
...um of the positive [[prime number | prime]] [[divisor | factors]] of <math>n</math>. ...tive integers, then compute the sum of the positive prime factors of <math>n</math>.

5 KB (744 words) - 19:46, 20 October 2020
Mock AIME 1 2006-2007 Problems/Problem 10
...here <math>m</math> and <math>n</math> are positive integers, find <math>m+n</math>. ...r}{a}+\frac{r}{b}+\frac{r}{c}=\frac{m}{n} =\frac{20}{3}</math>, so <math>m+n=23</math>.

1 KB (236 words) - 23:58, 24 April 2013
Mock AIME 1 2006-2007 Problems/Problem 11
...n}</math> be the [[set]] of strings with only 0's or 1's with length <math>n</math> such that any 3 adjacent place numbers sum to at least 1. For exampl ...le 0 and <math>A_3(n)</math> be the number of such strings of length <math>n</math> ending in a double zero. Then <math>A_1(1) = 1, A_2(1) = 1, A_3(1)

2 KB (424 words) - 15:51, 3 April 2012
Mock AIME 1 2006-2007 Problems/Problem 12
Let <math>k</math> be a [[positive integer]] with first [[digit]] 4 such that after removing the first digit, you get another positive integer, ...<math>n = 6j + 1</math>. There are <math>335</math> such values of <math>n</math> which fall in the required range.

2 KB (249 words) - 18:14, 3 April 2012
Mock AIME 1 2006-2007 Problems/Problem 13
...<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</math>, find <math>d Now consider the recurrence <math>\forall n: d_n = pd_{n-1} + qd_{n-2} + rd_{n-3}</math>. (<math>P</math> is called the ''characteristic polynomial'' of t

3 KB (568 words) - 15:50, 3 April 2012
Mock AIME 1 2006-2007 Problems/Problem 14
...}\angle AMP</math> for fixed <math>m</math> and <math>n</math> where <math>n>m</math>. If <math>f(m,49)</math> is an integer, find the sum of all possib Now we try to find <math>f(m,n)</math>.

3 KB (541 words) - 17:32, 22 November 2023
Mock AIME 1 2006-2007/Problems
...th> and <math>n</math> are relatively prime positive intgers, find <math>m+n</math>. (Note that <math>[ABC]</math> denotes the area of <math>\triangle A ...>\mathcal{S}</math>, we have that <math>\star (n)=12</math> and <math>0\le n< 10^{7}</math>. If <math>m</math> is the number of elements in <math>\mathc

8 KB (1,355 words) - 14:54, 21 August 2020
Mock AIME 2 2006-2007 Problems/Problem 1
...tegers <math>a,b,c,</math> and <math>d</math> such that <math>a+4=b-4=4c=d/4.</math> Find the smallest dragon. ...ath> and <math>c = 2</math> so our desired number is <math>a + b + c + d = 4 + 12 + 2 + 32 = 050</math>.

1 KB (178 words) - 01:27, 11 April 2024
Mock AIME 2 2006-2007 Problems/Problem 3
...th> be the sum of all [[positive integer]]s <math>n</math> such that <math>n^2+12n-2007</math> is a [[perfect square]]. Find the [[remainder]] when <mat ...t 9</math>. This gives us three pairs of [[equation]]s to solve for <math>n</math>:

1 KB (198 words) - 10:50, 4 April 2012
Mock AIME 2 2006-2007 Problems/Problem 5
...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> \lfloor 100n \rfloor</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
...th> 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>. ...3(x_2 + x_1) + x_3) = x_3^2(x_3 + x_2)(x_2 + x_1)(x_2 + x_1 + 1) = 144 = 2^4\cdot 3^2</math>.

3 KB (470 words) - 00:33, 10 August 2019
Mock AIME 2 2006-2007 Problems/Problem 9
label("$A$",A,N); draw(D--P--E,linetype("4 4"));

2 KB (358 words) - 23:22, 3 May 2014
Mock AIME 2 2006-2007 Problems/Problem 10
...degrees, to the equation <math>\sin^{10}x + \cos^{10}x = \frac{29}{16}\cos^4 2x,</math> where <math>0^\circ \le x^\circ \le 2007^\circ.</math> So, <math>\cos^22x=\frac{10\pm \sqrt{10^2-4\cdot24\cdot(-1)}}{2\cdot24}=\frac12</math> or <math>-\frac1{12}</math>

1 KB (183 words) - 00:09, 2 February 2013
Mock AIME 2 2006-2007 Problems/Problem 12
...to be the intersection of the diagonals of <math>ABCD</math>. If <math>AD=4,</math> <math>BC=6</math>, <math>BO=1,</math> and the [[area]] of <math>ABC Notice <math>\frac{[AOD]}{[BOC]}=(\frac{2}{3})^2\Rightarrow [BOC]=\frac{9}{4}[AOD]</math>

2 KB (311 words) - 10:53, 4 April 2012
Mock AIME 2 2006-2007 Problems/Problem 15
...be in every <math>1\times1\times4</math> rectangular box composed of <math>4</math> unit cubes. Determine the number of "intriguing" colorings. ...to solve this we must first look at the 2D problem using a <math>4 \times 4</math> grid.

4 KB (739 words) - 17:04, 24 November 2023
Mock AIME 2 2006-2007 Problems
...tegers <math>a,b,c,</math> and <math>d</math> such that <math>a+4=b-4=4c=d/4.</math> Find the smallest dragon. ...ments <math>n</math> in <math>S</math> is <math>f(n) = \frac{2n^3+n^2-n-2}{n^2-1}</math> an integer?

5 KB (848 words) - 23:49, 25 February 2017
Mock AIME 2 2006-2007 Problems/Problem 13
...h> and <math>n</math> are relatively prime positive integers, find <math>m+n.</math> {{Mock AIME box|year=2006-2007|n=2|num-b=12|num-a=14}}

1 KB (219 words) - 19:35, 25 June 2021
Expected value
...h> and <math>n</math> are relatively prime positive integers, find <math>m+n.</math> ...the expected value of the number of full circles formed, in terms of <math>n</math>.

4 KB (719 words) - 19:41, 25 November 2020
Base numbers/Conversion
<center><math>4201_5 = (4\cdot 5^3 + 2\cdot 5^2 + 0\cdot 5^1 + 1\cdot 5^0)_{10}</math></center> <center><math>=4\cdot 125 + 2\cdot 25 + 1</math></center>

7 KB (1,177 words) - 15:56, 18 April 2020
Arithmetico-geometric series
...n=a_ng_n</math>, where <math>a_n</math> and <math>g_n</math> are the <math>n</math>th terms of arithmetic and geometric sequences, respectively. ...1-drS_g}{r-1}</math>, where <math>S_g</math> is the sum of the first <math>n</math> terms of <math>g_n</math>.

2 KB (477 words) - 19:39, 17 August 2020
IMO Problems and Solutions
* [[1959 IMO Problems/Problem 4 | Problem 4]] proposed by Hungary * [[1960 IMO Problems/Problem 4 | Problem 4]] proposed by Hungary

35 KB (4,009 words) - 20:25, 21 February 2024
1959 IMO Problems
...h>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>. ...he equations in <math>\cos{x}</math> and <math>\cos{2x}</math> for <math>a=4, b=2, c=-1</math>.

3 KB (480 words) - 11:57, 17 September 2012
1959 IMO Problems/Problem 1
...h>\frac{21n+4}{14n+3}</math> is irreducible for every natural number <math>n</math>. <cmath>(21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 1) = 1</cmath>

5 KB (767 words) - 10:59, 23 July 2023
1959 IMO Problems/Problem 3
...he equations in <math>\cos{x}</math> and <math>\cos{2x}</math> for <math>a=4, b=2, c=-1</math>. Let the original equation be satisfied only for <math>\cos{x}=m, \cos{x}=n </math>. Then we wish to construct a quadratic with roots <math>2m^2 -1, 2

2 KB (300 words) - 03:20, 1 August 2019
1977 Canadian MO Problems
....</math> Find the smallest positive integer <math>b</math> for which <math>N</math> is the fourth power of an integer. == Problem 4 ==

3 KB (560 words) - 19:23, 10 March 2015

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