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  • B = (4, 3); C = (4, 0);
    5 KB (886 words) - 21:12, 22 January 2024
  • Isaac Newton was born on January 4, 1643, in Lincolnshire, England. Newton was born very shortly after the dea ...places a force on the matter with the same mass <math>n</math>, then <math>n</math> will put an equivalent force in the opposite direction.
    9 KB (1,355 words) - 07:29, 29 September 2021
  • ...e series: <center><math>3+\frac{11}4+\frac 94 + \cdots + \frac{n^2+2n+3}{2^n}+\cdots</math>.</center> ...^{\infty} \left(\frac{2n}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)</math>
    1 KB (193 words) - 21:13, 18 May 2021
  • ...gers <math>n\geq 3</math>, there exists a balanced set consisting of <math>n</math> points. </li> ...ath> for which there exists a balanced centre-free set consisting of <math>n</math> points. </li>
    4 KB (692 words) - 22:33, 15 February 2021
  • ...emainder 1. Show that there is an integer <math>{n}</math> such that <math>n^2 + 1</math> is divisible by <math>{p}</math>.
    4 KB (639 words) - 01:53, 2 February 2023
  • *Show that <math>\sum_{k=1}^{n}a_k^2 \geq a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1</math>. [[Inequality_Introductory_Problem_2|Solution]] *Show that <math>x^2+y^4\geq 2x+4y^2-5</math> for all real <math>x</math> and <math>y</math>.
    3 KB (560 words) - 22:51, 13 January 2024
  • <center><math>AM=\frac{x_1+x_2+\cdots+x_n}{n}</math></center> is the arithmetic mean of the <math>{n}</math> numbers <math>x_1,x_2,\ldots,x_n</math>.
    699 bytes (110 words) - 12:44, 20 September 2015
  • * <math>5x^4 - 2x^2 + 9</math>, in the variable <math>x</math> A polynomial in one variable is a function <math>P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0</math>. Here, <math>a_i</math> is the <m
    6 KB (1,100 words) - 01:44, 17 January 2024
  • ...integers <math>(x,y)</math> that are solutions to the equation <math>\frac{4}{x}+\frac{5}{y}=1</math>. (2021 CEMC Galois #4b) ...ice how I artificially grouped up the <math>y</math> terms by adding <math>4*5=20</math>).
    7 KB (1,107 words) - 07:35, 26 March 2024
  • <cmath>a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})</cmath> If <math>n=2</math>, this creates the difference of squares factorization, <cmath>a^2-
    3 KB (532 words) - 22:00, 13 January 2024
  • ...um of the [[series]] <math>\frac11 + \frac14 + \frac19 + \cdots + \frac{1}{n^2} + \cdots</math><br> ...}+\frac{x^4}{5!}-\cdots=\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{4\pi^2}\right)\left(1-\frac{x^2}{9\pi^2}\right)\cdots</math><br>
    2 KB (314 words) - 06:45, 1 May 2014
  • ...e''' states that if <math>n+1</math> or more pigeons are placed into <math>n</math> holes, one hole must contain two or more pigeons. This seemingly tri ...if <math>n</math> balls are to be placed in <math>k</math> boxes and <math>n>k</math>, then at least one box must contain more than one ball.
    11 KB (1,985 words) - 21:03, 5 August 2023
  • ...^4 + 6x^3 + 11x^2 + 3x + 31</math> is the square of an integer. Then <math>n</math> is: <math>\textbf{(A)}\ 4 \qquad
    3 KB (571 words) - 00:42, 22 October 2021
  • ...numbers. Note that if <math>n</math> is even, we take the positive <math>n</math>th root. It is analogous to the [[arithmetic mean]] (with addition r ...1 and 2 is <math>\sqrt[4]{6\cdot 4\cdot 1 \cdot 2} = \sqrt[4]{48} = 2\sqrt[4]{3}</math>.
    2 KB (282 words) - 22:04, 11 July 2008
  • ...four guys in order. By the same logic as above, this is <math>2!\binom{6}{4}=30</math>. Again, <math>|A\cap C|</math> would be putting five guys in ord If <math>(A_i)_{1\leq i\leq n}</math> are finite sets, then:
    9 KB (1,703 words) - 07:25, 24 March 2024
  • ...th>t</math> such that <math>a_i = t b_i</math> for all <math>1 \leq i \leq n</math>, or if one list consists of only zeroes. Along with the [[AM-GM Ineq ...ghtarrow{v}</math> and <math>\overrightarrow{w}</math> in <math>\mathbb{R}^n</math>, where <math>\overrightarrow{v} \cdot \overrightarrow{w}</math> is t
    13 KB (2,048 words) - 15:28, 22 February 2024
  • ...[[recursion|recursive definition]] for the factorial is <math>n!=n \cdot (n-1)!</math>. * <math>4! = 24</math>
    10 KB (809 words) - 16:40, 17 March 2024
  • ==Discriminant of polynomials of degree n== .../math> with all the coefficients being real. But for polynomials of degree 4 or higher it can be difficult to use it.
    4 KB (734 words) - 19:19, 10 October 2023
  • * [[2006 AIME II Problems/Problem 4]] {{AIME box|year=2006|n=II|before=[[2006 AIME I]]|after=[[2007 AIME I]], [[2007 AIME II|II]]}}
    1 KB (133 words) - 12:32, 22 March 2011
  • ...s, then <math>a^{\varphi(n)} \equiv 1 \pmod{n}</math>, where <math>\varphi(n)</math> denotes [[Euler's totient function]]. In particular, <math>\varphi( ...let <math>S = \{\text{natural numbers relatively prime to and less than}\ n\}</math> <math>\square</math>
    16 KB (2,675 words) - 10:57, 7 March 2024
  • ...range <math>\{1,2,3\cdots{,n}\}</math> which are relatively prime to <math>n</math>. If <math>{a}</math> is an integer and <math>m</math> is a positive ...{k}\equiv 1\pmod{n}</math>, and by [[Lagrange's Theorem]] <math>bk|\varphi(n)</math> which means
    3 KB (542 words) - 17:45, 21 March 2023
  • ...[[area]] <math>A</math> and [[perimeter]] <math>P</math>, then <math>\frac{4\pi A}{P^2} \le 1</math>. This means that given a perimeter <math>P</math> f <b>Proof of Lemma: </b> Let <math>M</math> and <math>N</math> be the projections of <math>E</math> and <math>F</math> onto line <m
    7 KB (1,296 words) - 14:22, 22 October 2023
  • ...f those numbers (<math>1 \leq k \leq n</math>). For example, if <math>n = 4</math>, and our set of numbers is <math>\{a, b, c, d\}</math>, then: ...um_{sym}f(x)</math>. The <math>n</math>th can be written <math>\sum_{sym}^{n}f(x)</math>
    2 KB (275 words) - 12:51, 26 July 2023
  • <cmath>f(z)=\sum_{n\ge 0}a_nq^n.</cmath> ...n series <math>G_4</math> and <math>G_6</math> are modular forms of weight 4 and 6 respectively.
    5 KB (849 words) - 16:14, 18 May 2021
  • ...nly if its units digit is divisible by 2, i.e. if the number ends in 0, 2, 4, 6 or 8. A number is divisible by <math>5^n</math> if and only if the last <math>n</math> digits are divisible by that power of 5.
    8 KB (1,315 words) - 18:18, 2 March 2024
  • ...it works for <math>n=1+1=2</math>, which in turn means it works for <math>n=2+1=3</math>, and so on. ...nd show that if <math>{n=k}</math> gives the desired result, so does <math>n=k+2</math>. If you wish, you can similarly induct over the powers of 2.
    5 KB (768 words) - 20:45, 1 September 2022
  • <math>270=2\cdot3^3\cdot5</math> and <math>144=2^4\cdot3^2</math>. The common factors are 2 and <math>3^2</math>, so <math>GCD ...an use the recursive formula <math>GCD(a_1,\dots,a_n)=GCD(GCD(a_1,\dots,a_{n-1}),a_n)</math>.
    2 KB (288 words) - 22:40, 26 January 2021
  • "How many numbers less than or equal to 100 are divisible by 2 or 3 but not 4?". ...Suppose we know the total number of people invited to the party, say <math>n</math>.
    4 KB (635 words) - 12:19, 2 January 2022
  • ...uare \square \square \square,</cmath> which we have to populate with <math>4</math> <math>A</math>s and <math>3</math> <math>B</math>s. Using constructi ...ath>B</math>s. Starting with the <math>A</math>s, we must choose the <math>4</math> boxes of their placement; because all the <math>A</math>s are indist
    12 KB (1,896 words) - 23:55, 27 December 2023
  • <math>r_{n-1} \pmod {r_n} \equiv 0</math><br> <math>r_{n-1} = r_n \cdot q_{n+1} +0</math><br>
    6 KB (924 words) - 21:50, 8 May 2022
  • ...se 0}+{n \choose 1}x + {n \choose 2}x^2+\cdots+</math><math>{n \choose n}x^n</math>. ...s the number of ways we can get <math>{k}</math> heads when flipping <math>n</math> different coins.
    4 KB (659 words) - 12:54, 7 March 2022
  • ...x]] <math>a</math>, <math>b</math>, and [[non-negative]] [[integer]] <math>n</math>, <center><math>(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k</math></center>
    5 KB (935 words) - 13:11, 20 February 2024
  • ...N = p_1p_2\cdots p_n + 1</math> is not divisible by any of them, but <math>N</math> must [[#Importance of Primes|have]] a prime factor, which leads to a ...ividing larger numbers would result in a quotient smaller than <math>\sqrt{n}</math>.
    6 KB (985 words) - 12:38, 25 February 2024
  • ...in the second. For instance, one function may map 1 to 1, 2 to 4, 3 to 9, 4 to 16, and so on. This function has the rule that it takes its input value ...ve from <math>\mathbb{R} \rightarrow \mathbb{R}</math> (since <math>f(2) = 4 = f(-2)</math>) nor surjective from <math>\mathbb{R} \rightarrow \mathbb{R}
    10 KB (1,761 words) - 03:16, 12 May 2023
  • ...ructive]] approach, the first digit can be one of seven integers; <math>1, 4, 5, 6, 7, 8,</math> and <math>9</math>. Note that the first digit cannot be ...use can be, four options for the second, and so on. Hence, there are <math>4^7 = 16384</math> ways she can color the four houses.
    8 KB (1,192 words) - 17:20, 16 June 2023
  • ...the '''prime factorization''' of <math>n</math> is an expression for <math>n</math> as a product of powers of [[prime number]]s. An important theorem o <math>n = {p_1}^{e_1} \cdot {p_2}^{e_2}\cdot{p_3}^{e_3}\cdots{p_k}^{e_k}</math>
    3 KB (496 words) - 22:14, 5 January 2024
  • ...me composite numbers are <math>4=2^2</math> and <math>12=2\times 6=3\times 4</math>. Composite numbers '''atleast have 2 distinct [[prime]] [[divisors]] 4 6 8 9 10 12 14 15 16 18 20 21 22 24 25 26 27 28 30 32 33 34 35 36 38 39 40
    6 KB (350 words) - 12:58, 26 September 2023
  • ...tegers (sometimes called [[whole number]]s). In particular, <math>\mathbb{N}</math> usually includes zero in the contexts of [[set theory]] and [[abstr
    1 KB (162 words) - 21:44, 13 March 2022
  • ...ts are perpendicular. Drawing all four semi-axes divides the ellipse into 4 [[congruent (geometry)|congruent]] quarters. pair P=(3,12/5), F1=(-4,0), F2=(4,0);
    5 KB (892 words) - 21:52, 1 May 2021
  • ...46. This number can be rewritten as <math>2746_{10}=2\cdot10^3+7\cdot10^2+4\cdot10^1+6\cdot10^0.</math> ...</math>, spits out <math>P(n)</math>, the value of the polynomial at <math>n</math>. However, the oracle charges a fee for each such computation, so you
    4 KB (547 words) - 17:23, 30 December 2020
  • \text{\textbullet}&&x^{n}-y^{n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots +xy^{n-1}+y^n) ...^2 \\\phantom{\text{\textbullet}}&&- b^4 + 2 b^2 d^2 - 4 b c^2 d + c^4 - d^4&=\det\begin{bmatrix}a&b&c&d\\d&a&b&c\\c&d&a&b\\b&c&d&a\end{bmatrix}\\&&&=(a
    2 KB (327 words) - 13:13, 6 July 2023
  • ...ost surprising places, such as in the sum <math>\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}</math>. Some common [[fraction]]al approximations for p ...approximates <math>\frac{\pi}{4}</math>. This can simply be multiplied by 4 to approximate <math>\pi</math>.
    8 KB (1,469 words) - 21:11, 16 September 2022
  • ...<math>F_1 = F_2 = 1</math> and <math>F_n=F_{n-1}+F_{n-2}</math> for <math>n \geq 3</math>. This is the simplest nontrivial example of a [[linear recur ...>n \geq 2</math>. In general, one can show that <math>F_n = (-1)^{n+1}F_{-n}</math>.
    6 KB (957 words) - 23:49, 7 March 2024
  • ...equence <math>(5,1)</math> majorizes <math>(4,2)</math> (as <math>5>4, 5+1=4+2</math>), Muirhead's inequality states that for any positive <math>x,y</ma x^5y^1+y^5x^1&=\frac{3}{4}\left(x^5y^1+y^5x^1\right)+\frac{1}{4}\left(x^5y^1+y^5x^1\right)\\
    8 KB (1,346 words) - 12:53, 8 October 2023
  • ...t]]s in that set, i.e. the size of the set. The cardinality of <math>\{3, 4\}</math> is 2, the cardinality of <math>\{1, \{2, 3\}, \{1, 2, 3\}\}</math> ...4\}</math> is <math>|\{3, 4\}| = 2</math>. Sometimes, the notations <math>n(A)</math> and <math>\# (A)</math> are used.
    2 KB (263 words) - 00:54, 17 November 2019
  • * Evaluate: <math>\int_2^5 x^3 dx</math> and <math>\int_{.2}^{.4} \cos(x) dx</math>. (The next few questions are meant as hints for how to ...ctly how far the object moved between times <math>t=.2</math> and <math>t=.4</math>. Interpret the distance that the object moved geometrically, as an
    11 KB (2,082 words) - 15:23, 2 January 2022
  • ...et]] of [[vertex|vertices]], <math>\{A_1, A_2, \ldots, A_n\}</math>, <math>n \geq 3</math>, with [[edge]]s <math>\{\overline{A_1A_2}, \overline{A_2A_3} ...ewer -- it will have "degenerated" from an <math>n</math>-gon to an <math>(n - 1)</math>-gon. (In the case of triangles, this will result in either a l
    2 KB (372 words) - 19:04, 30 May 2015
  • <math> \textbf{(A)}\ 5\sqrt{2} - 7 \qquad\textbf{(B)}\ 7 - 4\sqrt{3} \qquad\textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad\textbf{(D)}\ \frac{ ...s$",(W--Z),E,red); label("$s$",(X--Y),W,red); label("$s\sqrt{2}$",(W--X),N,red);
    4 KB (691 words) - 18:38, 19 September 2021
  • ...rected line segment. In many situations, a vector is best considered as an n-tuple of numbers (often real or complex). Most generally, but also most abs ...sional vector can be described in this coordinate form as an ordered <math>n</math>-tuple of numbers within angle brackets or parentheses, <math>(x\,\,y
    7 KB (1,265 words) - 13:22, 14 July 2021
  • ...2.285669651531203956336043826\ldots=x</cmath> such that:<cmath>(^24)^x=4^{4^x}\approx(3^5)^6</cmath> # Evaluate <math>(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)</math>.
    4 KB (680 words) - 12:54, 16 October 2023
  • draw(D--(30,4)--(34,4)--(34,0)--D); 1. If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
    6 KB (1,003 words) - 09:11, 7 June 2023
  • A=(4,2); D=(3,4);
    3 KB (575 words) - 15:27, 19 March 2023
  • pair A=(-1,5), B=(-4,-1), C=(4,-1), D, O; *If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
    4 KB (658 words) - 00:37, 8 September 2018
  • ...ath>\mathbb{N}</math> corresponds to the sequence <math>X = (x_n) = (0, 1, 4, 9, 16, \ldots)</math>. ...> is called the [[limit]] of <math>(x_n)</math> and is written <math>\lim_{n \to \infty} x_n</math>. The statement that <math>(x_n)</math> converges to
    2 KB (413 words) - 21:18, 13 November 2022
  • For example, <math>1, 2, 4, 8</math> is a geometric sequence with common ratio <math>2</math> and <mat ...progression if and only if <math>a_2 / a_1 = a_3 / a_2 = \cdots = a_n / a_{n-1}</math>. A similar definition holds for infinite geometric sequences. It
    4 KB (644 words) - 12:55, 7 March 2022
  • ...on difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic sequences, as the difference ...progression if and only if <math>a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}</math>. A similar definition holds for infinite arithmetic sequences. It
    4 KB (736 words) - 02:00, 7 March 2024
  • ...\geq 3</math>, there are no solutions to the equation <math>a^n + b^n = c^n</math>. ...he never published it, though he did publish a proof for the case <math>n=4</math>. It seems unlikely that he would have circulated a proof for the sp
    3 KB (453 words) - 11:13, 9 June 2023
  • ...ece of length <math>k_i</math> from the end of leg <math>L_i \; (i = 1,2,3,4)</math> and still have a stable table? For <math>0 \le x \le n</math>, it is easy to see that the number of stable tables is <math>(x+1)^2
    7 KB (1,276 words) - 20:51, 6 January 2024
  • If <math>n>1</math>, <math>2n, n^2 - 1, n^2 + 1</math> is a Pythagorean triple. ...ny <math>m,n</math>(<math>m>n</math>), we have <math>m^2 - n^2, 2mn, m^2 + n^2</math> is a Pythagorean triple.
    9 KB (1,434 words) - 13:10, 20 February 2024
  • ...s to the [[circumcenter]]. This creates a triangle that is <math>\frac{1}{n},</math> of the total area (consider the regular [[octagon]] below as an ex ...ound using [[trigonometry]] to be of length <math>\frac s2 \cot \frac{180}{n}^{\circ}</math>.
    6 KB (1,181 words) - 22:37, 22 January 2023
  • ...r+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor</cmath> *<math>\lfloor -3.2 \rfloor = -4</math>
    3 KB (508 words) - 21:05, 26 February 2024
  • ...he sum of the values on row <math>n</math> of Pascal's Triangle is <math>2^n</math>. ...ved from the combinatorics identity <math>{n \choose k}+{n \choose k+1} = {n+1 \choose k+1}</math>. Thus, any number in the interior of Pascal's Triang
    5 KB (838 words) - 17:20, 3 January 2023
  • Consider a polynomial <math>P(x)</math> of degree <math>n</math>, <center><math> P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center>
    4 KB (690 words) - 13:11, 20 February 2024
  • ...ath> and <math>BC</math> again at distinct points <math>K</math> and <math>N</math> respectively. Let <math>M</math> be the point of intersection of the {{IMO box|year=1985|num-b=4|num-a=6}}
    3 KB (496 words) - 13:35, 18 January 2023
  • ...=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \cdots</cmath>
    8 KB (1,217 words) - 20:15, 7 September 2023
  • ...</cmath> where <math>n_1>1,~~0<n_2<1,~~-1<n_3<0,~~n_4<-1</math>, and <math>n</math> is the root mean power. ...1}{a}}</math> and the harmonic mean's root mean power is -1 as <math>\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}=\sqrt[-1]{\frac{x_1^{-1}+\cdots+x_a^{-
    5 KB (912 words) - 20:06, 14 March 2023
  • <math>4 = 2 + 2</math> ...might expect the total number of ways to write a large even integer <math>n</math> as the sum of two odd primes to be roughly
    7 KB (1,201 words) - 16:59, 19 February 2024
  • ...function, it is easy to see that <math>\zeta(s)=0</math> when <math>s=-2,-4,-6,\ldots</math>. These are called the trivial zeros. This hypothesis is o ...uld hold. The Riemann Hypothesis would also follow if <math>M(n)\le C\sqrt{n}</math> for any constant <math>C</math>.
    2 KB (425 words) - 12:01, 20 October 2016
  • ...ulo]] <math>m</math> if there is some integer <math>n</math> so that <math>n^2-a</math> is [[divisibility | divisible]] by <math>m</math>. ...p-1}{2}}</math>, so <math>\left(\frac{-1}{p}\right)=1 \iff p \equiv 1 \mod 4</math>
    5 KB (778 words) - 13:10, 29 November 2017
  • Let <math>P</math> be a point, and let <math>S</math> be an <math>n</math>-sphere. Let two arbitrary lines passing through <math>P</math> inter ...th> intersect at <math>R</math>. If <math>AR:BR=1:4</math> and <math>CR:DR=4:9</math>, find the ratio <math>AB:CD</math>.
    5 KB (827 words) - 17:30, 21 February 2024
  • * [[2004 AIME I Problems/Problem 4]] {{AIME box|year=2004|n=I|before=[[2003 AIME I]], [[2003 AIME II|II]]|after=[[2004 AIME II]]}}
    1 KB (135 words) - 18:15, 19 April 2021
  • ...sitive integer]] <math>n</math>, the sequence <math>\{n,f(n),f(f(n)),f(f(f(n))),\ldots\}</math> contains 1. This conjecture is still open. Some people h ==Properties of <math>f(n)</math> ==
    1 KB (231 words) - 19:45, 24 February 2020
  • * [[2004 AIME II Problems/Problem 4]] {{AIME box|year=2004|n=II|before=[[2004 AIME I]]|after=[[2005 AIME I]], [[2005 AIME II|II]]}}
    1 KB (135 words) - 12:24, 22 March 2011
  • * [[2005 AIME I Problems/Problem 4 | Problem 4]] {{AIME box|year=2005|n=I|before=[[2004 AIME I|2004 AIME I]], [[2004 AIME II|II]]|after=[[2005 AIME
    1 KB (154 words) - 12:30, 22 March 2011
  • * [[2006 AIME I Problems/Problem 4]] {{AIME box|year=2006|n=I|before=[[2005 AIME I]], [[2005 AIME II|II]]|after=[[2006 AIME II]]}}
    1 KB (135 words) - 12:31, 22 March 2011
  • * [[2005 AIME II Problems/Problem 4]] {{AIME box|year=2005|n=II|before=[[2005 AIME I]]|after=[[2006 AIME I]], [[2006 AIME II|II]]}}
    1 KB (135 words) - 12:30, 22 March 2011
  • | n/a | n/a
    51 KB (6,175 words) - 20:58, 6 December 2023
  • ...em. A more widely known version states that there is a prime between <math>n</math> and <math>2n</math>. ...closer look at the [[combinations|binomial coefficient]] <math>\binom{2n}{n}</math>. Assuming that the reader is familiar with that proof, the Bertrand
    2 KB (309 words) - 21:43, 11 January 2010
  • <cmath>\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}= 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots</cmath>
    9 KB (1,547 words) - 03:04, 13 January 2021
  • draw((0,0), linewidth(4)); <math>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots </math>
    15 KB (2,396 words) - 20:24, 21 February 2024
  • ...so that <math>m^2=n</math>. The first few perfect squares are <math>0, 1, 4, 9, 16, 25, 36</math>. ...th>n</math> square numbers (starting with <math>1</math>) is <math>\frac{n(n+1)(2n+1)}{6}</math>
    954 bytes (155 words) - 01:14, 29 November 2023
  • ...th>n</math>), if the difference <math>{a - b}</math> is divisible by <math>n</math>. ...ntegers modulo <math>n</math>''' (usually known as "the integers mod <math>n</math>," or <math>\mathbb{Z}_n</math> for short). This structure gives us
    14 KB (2,317 words) - 19:01, 29 October 2021
  • {{AIME Problems|year=2006|n=I}} == Problem 4 ==
    7 KB (1,173 words) - 03:31, 4 January 2023
  • === Solution 4 === {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}
    6 KB (910 words) - 19:31, 24 October 2023
  • ...> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the triple M=(B+C)/2,S=(4*A+T)/5;
    6 KB (980 words) - 21:45, 31 March 2020
  • ...h> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 such that <math> S_n </math> is a [[perfect square]] .../math> but not by <math>2^n, \ldots,</math> and <math>2^{n-1}-2^{n-2} = 2^{n-2}</math> elements of <math>S</math> that are divisible by <math>2^1</math>
    10 KB (1,702 words) - 00:45, 16 November 2023
  • ...ough the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra {{AIME box|year=2006|n=I|num-b=9|num-a=11}}
    4 KB (731 words) - 17:59, 4 January 2022
  • pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1, label("$\mathcal{Q}$",(4.2,-1),NW);
    5 KB (730 words) - 15:05, 15 January 2024
  • for(int i=0; i<4; i=i+1) { pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10);
    4 KB (709 words) - 01:50, 10 January 2022
  • <math> b = 4 </math> <math>b=4</math>
    3 KB (439 words) - 18:24, 10 March 2015
  • ...oduct <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is divided by <math>1000</math>. ...<math>5</math> dividing it, for <math>76</math> extra; every n! for <math>n\geq 50</math> has one more in addition to that, for a total of <math>51</ma
    2 KB (278 words) - 08:33, 4 November 2022
  • ...math> so we take <math>N_0 = 25</math> and <math>n = 2.</math> Then <cmath>N = 7 \cdot 10^2 + 25 = \boxed{725},</cmath> and indeed, <math>725 = 29 \cdot ...ow that <math>N<1000</math> (because this is an AIME problem). Thus, <math>N</math> has <math>1,</math> <math>2</math> or <math>3</math> digits. Checkin
    4 KB (622 words) - 03:53, 10 December 2022
  • Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) ...]</math> from <math>5\cdot[(1)+(2)]</math> to get <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath>
    4 KB (670 words) - 13:03, 13 November 2023
  • ...bf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 7</math> ==Problem 4==
    12 KB (1,784 words) - 16:49, 1 April 2021
  • ...ne <math>x\spadesuit y = (x + y)(x - y)</math>. What is <math>3\spadesuit(4\spadesuit 5)</math>? == Problem 4 ==
    13 KB (2,058 words) - 12:36, 4 July 2023
  • == Problem 4 == [[2006 AMC 12A Problems/Problem 4|Solution]]
    15 KB (2,223 words) - 13:43, 28 December 2020
  • ...8 \qquad (\mathrm {B}) \ -4 \qquad (\mathrm {C})\ 2 \qquad (\mathrm {D}) \ 4 \qquad (\mathrm {E})\ 8 == Problem 4 ==
    13 KB (1,971 words) - 13:03, 19 February 2020
  • == Problem 4 == [[2004 AMC 12A Problems/Problem 4|Solution]]
    13 KB (1,953 words) - 00:31, 26 January 2023
  • Members of the Rockham Soccer League buy socks and T-shirts. Socks cost &#36;4 per pair and each T-shirt costs &#36;5 more than a pair of socks. Each memb <math> \mathrm{(A) \ } 4.5\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 18
    13 KB (1,955 words) - 21:06, 19 August 2023
  • <math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> <math> \mathrm{(A) \ } \frac{7}{2}\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 13 </
    12 KB (1,792 words) - 13:06, 19 February 2020
  • ...00^{4000} \qquad \textbf{(D)}\ 4,000,000^{2000} \qquad \textbf{(E)}\ 2000^{4,000,000}</math> == Problem 4 ==
    13 KB (1,948 words) - 12:26, 1 April 2022
  • Let <math>P(n)</math> and <math>S(n)</math> denote the product and the sum, respectively, of the digits ...example, <math>P(23) = 6</math> and <math>S(23) = 5</math>. Suppose <math>N</math> is a
    13 KB (1,957 words) - 12:53, 24 January 2024
  • \qquad\mathrm{(C)}\ 4 when <math>x=4</math>?
    10 KB (1,547 words) - 04:20, 9 October 2022
  • <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath> == Problem 4 ==
    13 KB (1,987 words) - 18:53, 10 December 2022
  • == Problem 4 == [[2004 AMC 12B Problems/Problem 4|Solution]]
    13 KB (2,049 words) - 13:03, 19 February 2020
  • ...x</math> has the property that <math>x\%</math> of <math>x</math> is <math>4</math>. What is <math>x</math>? \mathrm{(B)}\ 4 \qquad
    12 KB (1,781 words) - 12:38, 14 July 2022
  • ...h>, <math>J</math> and <math>N</math> are all positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? <math>253=N(4B+5J)</math>
    1 KB (227 words) - 17:21, 8 December 2013
  • ...e object only makes <math>1</math> move, it is obvious that there are only 4 possible points that the object can move to. At this point we can guess that for n moves, there are <math>(n + 1)^2</math> different endpoints. Thus, for 10 moves, there are <math>11^2
    2 KB (354 words) - 16:57, 28 December 2020
  • ...th hold at the same time if and only if <math>10^k \leq x < \frac{10^{k+1}}4</math>. ...math>k</math> the length of the interval <math>\left[ 10^k, \frac{10^{k+1}}4 \right)</math> is <math>\frac 32\cdot 10^k</math>.
    3 KB (485 words) - 14:09, 21 May 2021
  • ...divisible by <math>10</math>. What is the smallest possible value of <math>n</math>? n{5}\right\rfloor +
    5 KB (881 words) - 15:52, 23 June 2021
  • MP("B",D(B),plain.N,f); MP("B",D(B),plain.N,f);
    7 KB (1,169 words) - 14:04, 10 June 2022
  • pair f = (4.34, 74.58); label("F", f, N);
    6 KB (958 words) - 23:29, 28 September 2023
  • Given a finite sequence <math>S=(a_1,a_2,\ldots ,a_n)</math> of <math>n</math> real numbers, let <math>A(S)</math> be the sequence <math>\left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)</math>
    3 KB (466 words) - 22:40, 29 September 2023
  • <cmath>2(x+x^3+x^5\cdots)(1+x^2+x^4\cdots)(1+x+x^2+x^3\cdots) = \frac{2x}{(1-x)^3(1+x)^2}</cmath> ...n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)</math>, where <math>n=2006</math> (we may omit the coefficients, as we are seeking for the number
    8 KB (1,332 words) - 17:37, 17 September 2023
  • ...any ways are there to choose <math>k</math> elements from an ordered <math>n</math> element [[set]] without choosing two consecutive members? ...n with <math>k</math> elements where the largest possible element is <math>n-k+1</math>, with no restriction on consecutive numbers. Since this process
    8 KB (1,405 words) - 11:52, 27 September 2022
  • ...h>, and <math>f^{[n + 1]}(x) = f^{[n]}(f(x))</math> for each integer <math>n \geq 2</math>. For how many values of <math>x</math> in <math>[0,1]</math> ...h>\frac{5}{2^{n+1}}</math>, <math>\cdots</math> ,<math>\frac{2^{n+1}-1}{2^{n+1}}</math>.
    3 KB (437 words) - 23:49, 28 September 2022
  • ...f <math>m,n,</math> and <math>p</math> is zero. What is the value of <math>n/p</math>? ...xtbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}</math>
    2 KB (317 words) - 12:27, 16 December 2021
  • ...e greatest integer <math>k</math> such that <math>7^k</math> divides <math>n</math>? ...\mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}</math>
    888 bytes (140 words) - 20:04, 24 December 2020
  • <cmath>z_{n+1} = \frac {iz_{n}}{\overline {z_{n}}},</cmath> where <math>\overline {z_{n}}</math> is the [[complex conjugate]] of <math>z_{n}</math> and <math>i^{2}=-1</math>. Suppose that <math>|z_{0}|=1</math> and
    4 KB (660 words) - 17:40, 24 January 2021
  • ...th> are relatively prime positive integers. What is the value of <math>m + n</math>? draw((0,-0.5)--(0,4),Arrows);
    4 KB (761 words) - 09:10, 1 August 2023
  • ...osing which ant moves to <math>A</math>. Hence, there are <math>2 \times 2=4</math> ways the ants can move to different points. ...ath> can actually move to four different points, there is a total of <math>4 \times 20=80</math> ways the ants can move to different points.
    10 KB (1,840 words) - 21:35, 7 September 2023
  • ...sect at a right angle at <math>E</math> . Given that <math> BE = 16, DE = 4, </math> and <math> AD = 5 </math>, find <math> CE </math>. pair D = (0,4);
    1 KB (177 words) - 02:14, 26 November 2020
  • ...wice, triple roots three times, and so on, there are in fact exactly <math>n</math> complex roots of <math>P(x)</math>. ...1}x^{n-1}}{c_n} + \dots + \frac{c_1x}{c_n} + \frac{c_0}{c_n} = \sum_{j=0}^{n} \frac{c_jx^j}{c_n}.</cmath>
    8 KB (1,427 words) - 21:37, 13 March 2022
  • == Problem 4 == [[2006 AMC 10A Problems/Problem 4|Solution]]
    13 KB (2,028 words) - 16:32, 22 March 2022
  • MP('8', (16,-4), W); MP('8', (20,-8), N);
    3 KB (424 words) - 10:14, 17 December 2021
  • D((0,0)--(4*t,0)--(2*t,8)--cycle); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N);
    5 KB (732 words) - 23:19, 19 September 2023
  • D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);
    6 KB (1,066 words) - 00:21, 2 February 2023
  • ...multiple indistinct elements, such as the following: <math>\{1,4,5,3,24,4,4,5,6,2\}</math> Such an entity is actually called a multiset. ...t of size <math>n</math> then <math>\mathcal{P}(A)</math> has size <math>2^n</math>.
    11 KB (2,021 words) - 00:00, 17 July 2011
  • label("\Large{$\Gamma_-$}",(-.45,.4)); Now, let <math>N(R)</math> be an upper bound for the quantity
    6 KB (1,034 words) - 07:55, 12 August 2019
  • ...that there exist integers <math>m</math> and <math>n</math> with <math>0<m<n<p</math> and ...th>N </math> but every subset of size <math>k</math> has sum at most <math>N/2</math>.
    3 KB (520 words) - 09:24, 14 May 2021
  • <math>\frac{16+8}{4-2}=</math> <math>\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \te
    17 KB (2,246 words) - 13:37, 19 February 2020
  • * [[1997 I Problems/Problem 4|Problem 4]] * [[1997 II Problems/Problem 4|Problem 4]]
    856 bytes (98 words) - 14:53, 3 July 2009
  • ...<math>a_n-g_n</math> is divisible by <math>m</math> for all integers <math>n>1</math>; <cmath>g_{l-1}(r-1)^2\equiv0\pmod{m}\text{ (4)}.</cmath>
    2 KB (321 words) - 23:43, 29 December 2023
  • .../math> is a positive integer. Find the number of possible values for <math>n</math>. <math>\log_{10} 12 + \log_{10} n > \log_{10} 75 </math>
    1 KB (164 words) - 14:58, 14 April 2020
  • ...imes the number of possible sets of 3 cards that can be drawn. Find <math> n. </math> ...<math>{n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>.
    1 KB (239 words) - 11:54, 31 July 2023
  • ...and <math> n </math> are [[relatively prime]] [[integer]]s, find <math> m+n. </math> *Person 2: <math>\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25</math>
    3 KB (463 words) - 21:07, 15 February 2021
  • ...and <math> n </math> are [[relatively prime]] [[integer]]s. Find <math> m+n. </math> ...-r} = 2005</math>. Then we form a new series, <math>a^2 + a^2 r^2 + a^2 r^4 + \ldots</math>. We know this series has sum <math>20050 = \frac{a^2}{1 -
    3 KB (581 words) - 07:54, 4 November 2022
  • ...th power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).) {{AIME box|year=2005|n=II|num-b=4|num-a=6}}
    2 KB (404 words) - 14:27, 27 November 2019
  • {{AIME Problems|year=2005|n=II}} ...imes the number of possible sets of 3 cards that can be drawn. Find <math> n. </math>
    7 KB (1,119 words) - 21:12, 28 February 2020
  • Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math>(x+1)^{48}</math ...\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1 </math>.
    2 KB (279 words) - 12:33, 27 October 2019
  • ...> n </math> is not divisible by the square of any [[prime]], find <math> m+n+p. </math> pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7));
    4 KB (693 words) - 13:03, 28 December 2021
  • ...ath> less than or equal to <math>1000</math> is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all real <math> t </math>? ...</math> for all [[real number]]s <math>t</math> and all [[integer]]s <math>n</math>. So, we'd like to somehow convert our given expression into a form
    6 KB (1,154 words) - 03:30, 11 January 2024
  • ..., D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);draw(G--O); dot(A^^B^^C^^D^^E^^F^^G^^ ...90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linetype("4 4")); dot(A^^B^^C^^D^^E^^F^^G^^J^^O); label("\(A\)",A,(-1,1));label("\(B\)",B
    13 KB (2,080 words) - 21:20, 11 December 2022
  • ...</math> and <math> n </math> are relatively prime integers, find <math> m+n. </math> import three; currentprojection = perspective(4,-15,4); defaultpen(linewidth(0.7));
    3 KB (436 words) - 03:10, 23 September 2020
  • ...states that for any [[real number]] <math>\theta</math> and integer <math>n</math>, ...i\sin(\theta))^n = (e^{i\theta})^n = e^{in\theta} = \cos(n\theta) + i\sin(n\theta)</math>.
    3 KB (452 words) - 23:17, 4 January 2021
  • {{AIME Problems|year=2005|n=I}} ...<math> k. </math> For example, <math> S_3 </math> is the sequence <math> 1,4,7,10,\ldots. </math> For how many values of <math> k </math> does <math> S_
    6 KB (983 words) - 05:06, 20 February 2019
  • ...h> k</math>. For example, <math> S_3 </math> is the [[sequence]] <math> 1,4,7,10,\ldots. </math> For how many values of <math> k </math> does <math> S_ ...th>(12,167)</math>, <math>(167,12)</math>,<math>(334,6)</math>, <math>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math> and <math>(2004,1)</ma
    2 KB (303 words) - 01:31, 5 December 2022
  • ...s,, so <math>n</math> must be in the form <math>n=p\cdot q</math> or <math>n=p^3</math> for distinct [[prime number]]s <math>p</math> and <math>q</math> In the first case, the three proper divisors of <math>n</math> are <math>1</math>, <math>p</math> and <math>q</math>. Thus, we nee
    2 KB (249 words) - 09:37, 23 January 2024
  • ...<math>n \leq 14</math>. In fact, when <math>n = 14</math> we have <math>n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5</math>, so this number works and no larg ...</math>. The [[quadratic formula]] yields <math>r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}</math>. <math>\sqrt{4s
    8 KB (1,248 words) - 11:43, 16 August 2022
  • Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on o ...and what coins are silver, so the solution is <math>\boxed{9\cdot \binom 8 4=630}</math>.
    5 KB (830 words) - 01:51, 1 March 2023
  • Let <math> P </math> be the product of the nonreal roots of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math> The left-hand side of that [[equation]] is nearly equal to <math>(x - 1)^4</math>. Thus, we add 1 to each side in order to complete the fourth power
    4 KB (686 words) - 01:55, 5 December 2022
  • label("$10$",(2.5,4.5),W); label("$10$",(18.37,4.5),E);
    4 KB (567 words) - 20:20, 3 March 2020
  • ...h> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math> ...hen the previous statement says that <math>2^{111\cdot(x_1 + x_2 + x_3)} = 4</math> so taking a [[logarithm]] of that gives <math>111(x_1 + x_2 + x_3) =
    1 KB (161 words) - 19:50, 2 January 2022
  • ...cube, we need that they show an orange face. This happens in <math>\frac{4}{6} = \frac{2}{3}</math> of all orientations, so from these cubes we gain a ...the corner cubes together contribute a probability of <math>\left(\frac{1}{4}\right)^8 = \frac{1}{2^{16}}</math>
    4 KB (600 words) - 21:44, 20 November 2023
  • == Solution 4 == {{AIME box|year=2005|n=I|num-b=9|num-a=11}}
    5 KB (852 words) - 21:23, 4 October 2023
  • ...mum value of <math> d </math> is <math> m - \sqrt{n},</math> find <math> m+n. </math> === Solution 4 ===
    4 KB (707 words) - 11:11, 16 September 2021
  • ...e the number of positive integers <math> n \leq 2005 </math> with <math> S(n) </math> [[even integer | even]]. Find <math> |a-b|. </math> .... So <math>S(1), S(2)</math> and <math>S(3)</math> are odd, while <math>S(4), S(5), \ldots, S(8)</math> are even, and <math>S(9), \ldots, S(15)</math>
    4 KB (647 words) - 02:29, 4 May 2021
  • ..., and <math>3</math> <math>D</math>'s, so the string is divided into <math>4</math> partitions (<math>-D-D-D-</math>). ...<math>R</math>'s and <math>U</math>'s stay together, then there are <math>4 \cdot 3 = 12</math> places to put them.
    5 KB (897 words) - 00:21, 29 July 2022
  • Consider the [[point]]s <math> A(0,12), B(10,9), C(8,0),</math> and <math> D(-4,7). </math> There is a unique [[square]] <math> S </math> such that each of ...th>AE = BD</math>, we have <math>9 - 7 = x_E - 0</math> and <math>10 - ( - 4) = 12 - y_E</math>
    3 KB (561 words) - 14:11, 18 February 2018
  • ...| divisible]] by the [[perfect square | square]] of a prime, find <math> m+n. </math> D(MP("A",A,s)--MP("B",B,s)--MP("C",C,N,s)--cycle); D(cir);
    5 KB (906 words) - 23:15, 6 January 2024
  • r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\ D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype("2 2") + d + red)
    12 KB (2,000 words) - 13:17, 28 December 2020
  • dotfactor = 4; label("$A$",A,N);
    13 KB (2,129 words) - 18:56, 1 January 2024
  • ...itive integers <math>n</math>. Let <math>d(x)</math> be the smallest <math>n</math> such that <math>x_n=1</math>. (For example, <math>d(100)=3</math> an ...icting our assumption that <math>20</math> was the smallest value of <math>n</math>. Using [[complementary counting]], we see that there are only <math>
    9 KB (1,491 words) - 01:23, 26 December 2022
  • ...> feet. The unicorn has pulled the rope taut, the end of the rope is <math>4</math> feet from the nearest point on the tower, and the length of the rope real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;
    4 KB (729 words) - 01:00, 27 November 2022
  • ...and <math> n </math> are relatively prime positive integers, find <math> m+n. </math> ...he form <math>\frac m{19}</math> or <math>\frac n {17}</math> for <math>m, n > 0</math>.
    2 KB (298 words) - 20:02, 4 July 2013
  • ...and <math> n </math> are relatively prime positive integers, find <math> m+n. </math> The notation <math> [z] </math> denotes the [[floor function|great <cmath>x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots</cmath>
    2 KB (303 words) - 22:28, 11 September 2020
  • ...h> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math> ...lid has volume equal to <math>V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surface area]] <math>A = \pi r^2 + \pi r \ell</ma
    5 KB (839 words) - 22:12, 16 December 2015
  • ...<math> n</math> are [[relatively prime]] positive integers. Find <math> m+n. </math> ...ABC</math>. Thus <math>U_1</math>, and hence <math>U_2</math>, are <math>3-4-5\,\triangle</math>s.
    4 KB (618 words) - 20:01, 4 July 2013
  • ...from left to right. What is the sum of the possible remainders when <math> n </math> is divided by <math>37</math>? ...+ 10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>.
    2 KB (374 words) - 14:53, 27 December 2019
  • Solving for <math>f+l</math>, we find the sum of the two terms is <math>4</math>. <cmath>2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)</cmath>
    8 KB (1,437 words) - 21:53, 19 May 2023
  • {{AIME box|year=2004|n=I|num-b=2|num-a=4}}
    1 KB (156 words) - 17:56, 1 January 2016
  • ...h> and <math>(0,y)</math>. Because the segment has length 2, <math>x^2+y^2=4</math>. Using the midpoint formula, we find that the midpoint of the segmen \sqrt{\frac{1}{4}\left(x^2+y^2\right)}=\sqrt{\frac{1}{4}(4)}=1</math>. Thus the midpoints lying on the sides determined by vertex <mat
    3 KB (532 words) - 09:22, 11 July 2023
  • ...<math> n </math> are relatively prime positive integers. What is <math> m+n </math>? ...'s success ratio <math>\frac{349}{500}</math>. Thus, the answer is <math>m+n = 349 + 500 = \boxed{849}</math>.
    3 KB (436 words) - 18:31, 9 January 2024
  • ...3</math>, <math>x_3x_4x_1x_2</math>. Thus there are <math>5\cdot {9\choose 4}=630</math> snakelike numbers which do not contain the digit zero. ...ath>2\cdot{9 \choose 3}</math>. Thus our answer is <math>5\cdot{10 \choose 4} - 2\cdot{9 \choose 3} = \boxed{882}</math>
    3 KB (562 words) - 18:12, 4 March 2022
  • ...ath> <math>= (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2)</math> <math>= 1 - (1 + 4 + \ldots + 225)x^2 + R(x)</math>. Equating coefficients, we have <math>2C - 64 = -(1 + 4 + \ldots + 225) = -1240</math>, so <math>-2C = 1176</math> and <math>|C| =
    5 KB (833 words) - 19:43, 1 October 2023
  • Define a regular <math> n </math>-pointed star to be the union of <math> n </math> line segments <math> P_1P_2, P_2P_3,\ldots, P_nP_1 </math> such tha * each of the <math> n </math> line segments intersects at least one of the other line segments at
    4 KB (620 words) - 21:26, 5 June 2021
  • {{AIME Problems|year=2004|n=I}} ...from left to right. What is the sum of the possible remainders when <math> n </math> is divided by 37?
    9 KB (1,434 words) - 13:34, 29 December 2021
  • ...olution. Call <math>s_{n, k}</math> the number of squares below the <math>n</math> square after the final fold in a strip of length <math>2^{k}</math>. ...the <math>s_{n, k}</math> value of the pairs correspond with the <math>s_{n, k - 1}</math> values - specifically, double, and maybe <math>+ 1</math> (i
    6 KB (899 words) - 20:58, 12 May 2022
  • ...d from eight <math> 7 </math>'s in this way. For how many values of <math> n </math> is it possible to insert <math> + </math> signs so that the resulti ...1000</math>. Then the question is asking for the number of values of <math>n = a + 2b + 3c</math>.
    11 KB (1,857 words) - 21:55, 19 June 2023
  • ...> k </math> and <math> p </math> are [[relatively prime]]. Find <math> k+m+n+p. </math> ...lack; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pen f = fontsize(8);
    3 KB (431 words) - 23:21, 4 July 2013
  • ...ctor, which gives a distance <math>\sqrt{(-375-125)^2+(-375-0)^2}=125\sqrt{4^2+3^2}=\boxed{625}</math>. {{AIME box|year=2004|n=II|num-b=10|num-a=12}}
    2 KB (268 words) - 22:20, 23 March 2023
  • ...and <math>y-x \equiv 3 \pmod 6 = 6n +3</math> for some whole number <math>n</math>. ...+31+25+\ldots+1 = 7 + 36 + 30 + 24 + \ldots + 6 + 0 = 7 + 6 \cdot (6 + 5 + 4\ldots + 1) </math>
    7 KB (1,091 words) - 18:41, 4 January 2024
  • ...est term in this sequence that is less than <math>1000</math>. Find <math> n+a_n. </math> ...(n-1)f(n)</math>, <math>a_{2n+1} = f(n)^2</math>, where <math>f(n) = nx - (n-1)</math>.{{ref|1}}
    3 KB (538 words) - 21:33, 30 December 2023
  • ...t to partitioning two items in three containers. We can do this in <math>{4 \choose 2} = 6</math> ways. We can partition the 3 in three ways and likew 4, 3, 167
    2 KB (353 words) - 18:08, 25 November 2023
  • ...and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math> draw(E--B--C--F, linetype("4 4"));
    9 KB (1,501 words) - 05:34, 30 October 2023
  • ...c{11}{24}b_3</math>, the second monkey got <math>\frac{1}{8}b_1 + \frac{1}{4}b_2 + \frac{11}{24}b_3</math>, and the third monkey got <math>\frac{1}{8}b_ <math>x = \frac{1}{4}b_1 + \frac{1}{8}b_2 + \frac{11}{72}b_3 = \frac{1}{16}b_1 + \frac{1}{8}b_2
    6 KB (950 words) - 14:18, 15 January 2024
  • ...e initial problem statement, we have <math>1000w\cdot\frac{1}{4}t=\frac{1}{4}</math>. ...lete the same amount of work, which is <math>\frac{1000}{900}\cdot\frac{1}{4}\cdot t=\frac{5}{18}t</math>.
    4 KB (592 words) - 19:02, 26 September 2020
  • ...>-digit number, for a total of <math>(2^1 - 2) + (2^2 - 2) + (2^3 -2) + (2^4 - 2) = 22</math> such numbers (or we can list them: <math>AB, BA, AAB, ABA, ...he other digit. For each choice, we have <math>2^{n - 1} - 1</math> <math>n</math>-digit numbers we can form, for a total of <math>(2^0 - 1) + (2^1 - 1
    3 KB (508 words) - 01:16, 19 January 2024
  • ...the 1-cm cubes cannot be seen. Find the smallest possible value of <math> N. </math> ...xtra layer makes the entire block <math>4\times8\times12</math>, and <math>N= \boxed{384}</math>.
    2 KB (377 words) - 11:53, 10 March 2014
  • ...rac{1}{2} \cdot \frac{r}{2} \cdot \frac{r\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}</math>. The [[central angle]] which contains the entire chord is <math>60 ...{4}}{\frac{1}{3}r^2\pi - \frac{r^2\sqrt{3}}{4}} = \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}}</math>
    2 KB (329 words) - 23:20, 4 July 2013
  • {{AIME Problems|year=2004|n=II}} ...and <math> n </math> are relatively prime positive integers, find <math> m+n. </math>
    9 KB (1,410 words) - 05:05, 20 February 2019
  • * [[2000 AIME II Problems/Problem 4|Problem 4]] {{AIME box|year=2000|n=II|before=[[2000 AIME I]]|after=[[2001 AIME I]], [[2001 AIME II | II]]}}
    1 KB (139 words) - 08:41, 7 September 2011
  • * [[2001 AIME I Problems/Problem 4|Problem 4]] {{AIME box|year=2001|n=I|before=[[2000 AIME I]], [[2000 AIME II|II]]|after=[[2001 AIME II]]}}
    1 KB (139 words) - 08:41, 7 September 2011
  • * [[2000 AIME I Problems/Problem 4|Problem 4]] {{AIME box|year=2000|n=I|before=[[1999 AIME]]|after=[[2000 AIME II]]}}
    1 KB (135 words) - 18:05, 30 May 2015
  • == Problem 4 == [[1983 AIME Problems/Problem 4|Solution]]
    7 KB (1,104 words) - 12:53, 6 July 2022
  • ...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. ...2</math> elements are adjacent. Using the well-known formula <math>\dbinom{n-k+1}{k}</math>, there are <math>\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171</
    5 KB (830 words) - 22:15, 28 December 2023
  • ...h>n</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>. ...<math>t_{2}</math>, and <math>t_{3}</math> in the figure, have areas <math>4</math>, <math>9</math>, and <math>49</math>, respectively. Find the area of
    6 KB (933 words) - 01:15, 19 June 2022
  • ...m of the solutions to the equation <math>\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}</math>? == Problem 4 ==
    5 KB (847 words) - 15:48, 21 August 2023
  • .../math> of non-negative integers is called "simple" if the addition <math>m+n</math> in base <math>10</math> requires no carrying. Find the number of sim == Problem 4 ==
    6 KB (869 words) - 15:34, 22 August 2023
  • ...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>. == Problem 4 ==
    6 KB (902 words) - 08:57, 19 June 2021
  • <center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center> == Problem 4 ==
    7 KB (1,045 words) - 20:47, 14 December 2023
  • == Problem 4 == [[1990 AIME Problems/Problem 4|Solution]]
    6 KB (870 words) - 10:14, 19 June 2021
  • ...le <math>ABCD_{}^{}</math> has sides <math>\overline {AB}</math> of length 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</m == Problem 4 ==
    7 KB (1,106 words) - 22:05, 7 June 2021
  • == Problem 4 == \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt}
    8 KB (1,117 words) - 05:32, 11 November 2023
  • ...etc. If the candidate went <math>\frac{n^{2}}{2}</math> miles on the <math>n^{\mbox{th}}_{}</math> day of this tour, how many miles was he from his star ...many contestants caught <math>n\,</math> fish for various values of <math>n\,</math>.
    8 KB (1,275 words) - 06:55, 2 September 2021
  • ..., where <math>m\,</math> and <math>n\,</math> are integers. Find <math>m + n\,</math>. == Problem 4 ==
    7 KB (1,141 words) - 07:37, 7 September 2018
  • ...> and <math>n</math> are relatively prime positive integers. Find <math>m-n.</math> ...h> and <math>n</math> are relatively prime positive integers, find <math>m+n.</math>
    6 KB (1,000 words) - 00:25, 27 March 2024
  • .../math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even integer? ...tive integer <math>n</math> for which the expansion of <math>(xy-3x+7y-21)^n</math>, after like terms have been collected, has at least 1996 terms.
    6 KB (931 words) - 17:49, 21 December 2018
  • ...and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> == Problem 4 ==
    7 KB (1,098 words) - 17:08, 25 June 2020
  • == Problem 4 == ...ath>n</math> are [[relatively prime]] [[positive integer]]s. Find <math>m+n.</math>
    7 KB (1,084 words) - 02:01, 28 November 2023
  • ...nd <math>n_{}</math> are relatively prime positive integers. Find <math>m+n.</math> Find the sum of all positive integers <math>n</math> for which <math>n^2-19n+99</math> is a perfect square.
    7 KB (1,094 words) - 13:39, 16 August 2020
  • {{AIME Problems|year=2000|n=I}} ...he least positive integer <math>n</math> such that no matter how <math>10^{n}</math> is expressed as the product of any two positive integers, at least
    7 KB (1,204 words) - 03:40, 4 January 2023
  • {{AIME Problems|year=2001|n=I}} == Problem 4 ==
    7 KB (1,212 words) - 22:16, 17 December 2023
  • {{AIME Problems|year=2002|n=I}} ...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
    8 KB (1,374 words) - 21:09, 27 July 2023
  • {{AIME Problems|year=2003|n=I}} <center><math> \frac{((3!)!)!}{3!} = k \cdot n!, </math></center>
    6 KB (965 words) - 16:36, 8 September 2019
  • {{AIME Problems|year=2000|n=II}} ...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
    6 KB (947 words) - 21:11, 19 February 2019
  • {{AIME Problems|year=2001|n=II}} .../math> forms a perfect square. What are the leftmost three digits of <math>N</math>?
    8 KB (1,282 words) - 21:12, 19 February 2019
  • {{AIME Problems|year=2002|n=II}} == Problem 4 ==
    7 KB (1,177 words) - 15:42, 11 August 2023
  • {{AIME Problems|year=2003|n=II}} ...is the sum of the other two. Find the sum of all possible values of <math>N</math>.
    7 KB (1,127 words) - 09:02, 11 July 2023
  • ...he roots of this equation are real, since its discriminant is <math>18^2 - 4 \cdot 1 \cdot 20 = 244</math>, which is positive. Thus by [[Vieta's formula ...he square root of a real number can't be negative, the only possible <math>n</math> is <math>5</math>.
    3 KB (532 words) - 05:18, 21 July 2022
  • <cmath> AC = \sqrt{AB^2 + BC^2} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}. </cmath> string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"};
    11 KB (1,741 words) - 22:40, 23 November 2023
  • Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <mat ...ively, we could have noted that <math>a^b\equiv a^{b\pmod{\phi{(n)}}}\pmod n</math>. This way, we have <math>6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\
    3 KB (361 words) - 20:20, 14 January 2023
  • label("E",E,N); label("F",F,N);
    5 KB (865 words) - 21:11, 6 February 2023
  • ...simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<!-- don't remove the following tag, for PoTW on the Wiki front pa ...<math>S</math> be a non-[[empty set | empty]] [[subset]] of <math>\{1,2,3,4,5,6\}</math>.
    5 KB (894 words) - 14:29, 22 August 2020
  • ...abel("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);</asy> ...,b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0);
    13 KB (2,149 words) - 18:44, 5 February 2024
  • ...rational number. If this number is expressed as a fraction <math>\frac{m}{n}</math> in lowest terms, what is the product <math>mn</math>? label("$D$",D,N);
    19 KB (3,221 words) - 01:05, 7 February 2023
  • A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>. We can substi == Solution 4 ==
    4 KB (576 words) - 21:03, 23 December 2023
  • ...h>t_{2}</math>, and <math>t_{3}</math> in the figure, have [[area]]s <math>4</math>, <math>9</math>, and <math>49</math>, respectively. Find the area of pair A=(0,0),B=(12,0),C=(4,5);
    4 KB (726 words) - 13:39, 13 August 2023
  • ...</math>. If we multiply the two equations together, we get that <math>a^4b^4 = 2^{36}</math>, so taking the fourth root of that, <math>ab = 2^9 = \boxed ...ac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2</math>. This means that <math>\frac{\ln ab}{\ln 2} = 9</math
    5 KB (782 words) - 14:49, 1 August 2023
  • ...nction]] f is defined on the [[set]] of [[integer]]s and satisfies <math>f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\
    4 KB (617 words) - 18:01, 9 March 2022
  • triple M,N,P[],Q[]; int n=s.length;
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  • ...robability]] that no two birch trees are next to one another. Find <math>m+n</math>. ...e out one tree between each pair of birch trees. So you would remove <math>4</math> trees that aren't birch. What you are left with is a unique arrangem
    7 KB (1,115 words) - 00:52, 7 September 2023
  • ...>21</math>, ... , and in general <math>9 + 6n</math> for nonnegative <math>n</math> are odd composites. We now have 3 cases: ...th> can be expressed as <math>9 + (9+6n)</math> for some nonnegative <math>n</math>. Note that <math>9</math> and <math>9+6n</math> are both odd composi
    8 KB (1,346 words) - 01:16, 9 January 2024
  • ...tal, for an average of 9. Thus we must have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = \boxed{25}</math>. ...nts vs them, which is a contradiction since it must be larger. Thus, <math>n=\boxed{25}</math>.
    5 KB (772 words) - 22:14, 18 June 2020
  • ...<math>a_{n+1}</math>. Find the maximum value of <math>d_n</math> as <math>n</math> ranges through the [[positive integer]]s. ...h>2n+1</math> if it is going to divide the entire [[expression]] <math>100+n^2+2n+1</math>.
    4 KB (671 words) - 20:04, 6 March 2024
  • ...when it has crawled exactly <math>7</math> meters. Find the value of <math>n</math>. P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\
    17 KB (2,829 words) - 21:21, 21 February 2024
  • ...o <math>\frac n2</math> to <math>\frac {n+1}2</math> for any integer <math>n</math> (same reasoning as above). So now we only need to test every 10 numb *<math>4</math>: We can partition as <math>1+1+2</math>, and from the previous case
    12 KB (1,859 words) - 18:16, 28 March 2022
  • ...is chosen so that <math>a_n = a_{n - 1} - a_{n - 2}</math> for each <math>n \ge 3</math>. What is the sum of the first 2001 terms of this sequence if t ...h>n</math> times, <math>a_{j + 6n} = a_j</math> for all [[integer]]s <math>n</math> and <math>j</math>.
    2 KB (410 words) - 13:37, 1 May 2022
  • ...division points closest to the opposite vertices. Find the value of <math>n</math> if the the [[area]] of the small square is exactly <math>\frac1{1985 ...2 - 2n + 1 = 1985</math>. Solving this [[quadratic equation]] gives <math>n = \boxed{32}</math>.
    3 KB (484 words) - 21:40, 2 March 2020
  • ...> and <math>B</math> lie along the lines <math>y=x+3</math> and <math>y=2x+4</math> respectively, find the area of triangle <math>ABC</math>. label("$B$", B, N);
    11 KB (1,722 words) - 09:49, 13 September 2023
  • ...s this just becomes the ball-and-urn argument. We want to add 5 balls into 4 urns, which is the same as 3 dividers; hence this gives <math>{{5+3}\choose .../math> and multiplication, the answer is <math>{{2+4-1}\choose2} \cdot{{5+4-1}\choose5}=560</math> ~Slight edits in LaTeX by EthanSpoon
    3 KB (445 words) - 19:44, 8 January 2023
  • === Solution 4(calculus) === ...ix} ,</math> where the term <math>\dbinom{n}{k}</math> is negated if <math>n+k</math> is odd.
    6 KB (872 words) - 16:51, 9 June 2023
  • ...>. Play the role of the magician and determine <math>(abc)</math> if <math>N= 3194</math>. Let <math>n=abc</math> then
    3 KB (565 words) - 16:51, 1 October 2023
  • D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle); pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N));
    11 KB (1,850 words) - 18:07, 11 October 2023
  • ...So the overall power of <math>2</math> and <math>5</math> is <math>7(1+2+3+4+5+6) = 7 \cdot 21 = 147</math>. However, since the question asks for proper ...d(n))/2}</math>, where <math>d(n)</math> is the number of divisor of <math>n</math>.
    3 KB (487 words) - 20:52, 16 September 2020
  • The increasing [[sequence]] <math>1,3,4,9,10,12,13\cdots</math> consists of all those positive [[integer]]s which a ...o determine the 100th number. <math>100</math> is equal to <math>64 + 32 + 4</math>, so in binary form we get <math>1100100</math>. However, we must cha
    5 KB (866 words) - 00:00, 22 December 2022
  • ...]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>? ...<math>900</math>. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides <math>900</math> is <math>\boxed{890}</math>; we can doub
    2 KB (338 words) - 19:56, 15 October 2023
  • == Solution 4 (Algebra) == label("$A$",A,N);
    5 KB (838 words) - 18:05, 19 February 2022
  • ...(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.</cmath> ...^2 + 2b^2 + 2ab\right).</math> Each of the terms is in the form of <math>x^4 + 324.</math> Using Sophie Germain, we get that
    6 KB (822 words) - 04:38, 21 January 2023
  • ...s a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>. In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.
    4 KB (673 words) - 19:48, 28 December 2023
  • ...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
  • ...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
  • ...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
  • ...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
  • 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
  • ...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
  • ...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
  • ...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
  • 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
  • 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
  • ...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
  • 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
  • ...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
  • 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
  • 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
  • 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
  • ...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
  • ...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
  • 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
  • ...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
  • ...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) - 03:33, 16 January 2023
  • == 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
  • 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
  • label("$P$", P, N); label("P", P, N);
    7 KB (1,086 words) - 08:16, 29 July 2023
  • ...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
  • 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
  • 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
  • ...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
  • ...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
  • ...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) - 13:17, 26 June 2022
  • <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
  • 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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...}^{}</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
  • ...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
  • ...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
  • ...<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
  • ...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
  • \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) - 16:14, 21 January 2024
  • ...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
  • ...> 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
  • ...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
  • ...</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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...), 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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...<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
  • <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
  • ...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
  • ...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
  • ...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
  • ...}</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
  • 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
  • ...> 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
  • 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
  • ...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
  • ..._{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
  • 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
  • ...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
  • <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
  • ...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
  • ...> 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
  • 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
  • ...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
  • ...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
  • 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
  • ...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
  • 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
  • ...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
  • ...[[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
  • .../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
  • ...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
  • ...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
    8 KB (1,160 words) - 10:11, 22 March 2024
  • ...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
  • ...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
  • ...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
  • ..., 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
  • 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
  • | 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
  • ...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
  • ...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
  • ...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
  • ===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
  • ...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
  • ...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
  • ...}</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
  • ...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
  • ...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
  • ...}</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
  • 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
  • ...}</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
  • ...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
  • ...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
  • ...=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
  • ...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
  • ...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
  • <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
  • ...>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
  • ...<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) - 10:41, 10 September 2023
  • x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\
    6 KB (966 words) - 21:48, 29 January 2024
  • ...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
  • ...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
  • {{AIME box|year=2000|n=I|num-b=2|num-a=4}}
    679 bytes (98 words) - 00:51, 2 November 2023
  • ...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
  • ...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
  • 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
  • ...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
  • ...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
  • ===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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...</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
  • ...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
  • <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
  • ...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
  • *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
  • ...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
  • ...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
  • == 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
  • ...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 (750 words) - 21:09, 23 December 2021
  • <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
  • {{AIME box|year=2002|n=I|num-b=4|num-a=6}}
    1 KB (220 words) - 20:50, 12 November 2022
  • ...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
  • ...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
  • ...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
  • ...</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
  • ...<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
  • 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
  • ...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
  • <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
  • 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
  • ...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
  • ...> 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
  • *<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
  • ...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
  • <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
  • ...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
  • ...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
  • 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
  • ...<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
  • ...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 (330 words) - 14:14, 13 January 2024
  • ...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
  • ...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
  • D=(7.6667,-4); G=(6.3333,4);
    5 KB (787 words) - 17:38, 30 July 2022
  • ...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
  • ...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
  • ...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
  • ...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
  • r[1] = 4.096; label("$y = mx$", (8,12*sqrt(221)/49*8), N);
    7 KB (1,182 words) - 09:56, 7 February 2022
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • 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
  • 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
  • ...math>, from which we can deduce that <math>c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48</math>. {{AIME box|year=2002|n=II|num-b=2|num-a=4}}
    1,016 bytes (161 words) - 03:31, 6 December 2019
  • ...<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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...</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
  • ...<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
  • ...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
  • ...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
  • ...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
  • ...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));
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  • 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
  • ...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
  • ...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\\
    5 KB (904 words) - 20:38, 4 March 2024
  • ...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
  • ...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);
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  • ...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
    3 KB (483 words) - 11:19, 9 August 2023
  • <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
  • ...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 (192 words) - 18:08, 11 August 2022
  • <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
  • ...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
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  • 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
  • ...<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
  • ...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 (284 words) - 15:56, 11 February 2024
  • ...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
  • ...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
  • ...+(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
  • ...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
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  • ...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}}
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  • ...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
  • dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);
    3 KB (528 words) - 18:14, 16 December 2021
  • ...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
  • ...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>.
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  • 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) - 16:57, 19 February 2024
  • <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
  • ...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
  • ...{\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
  • ...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
  • ...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
  • draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);
    7 KB (988 words) - 16:35, 16 January 2024
  • ...[[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
  • 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
  • ...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
  • ...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
  • ....</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
  • <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
  • ...}</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
  • 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
  • ...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
  • ...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
  • ...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
  • 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
  • ...<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
  • ...}\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
  • ...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
  • ...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>.
    870 bytes (136 words) - 10:49, 4 April 2012
  • ...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
  • ...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
  • ...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
  • label("$A$",A,N); draw(D--P--E,linetype("4 4"));
    2 KB (358 words) - 23:22, 3 May 2014
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • ...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
  • <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
  • ...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
  • * [[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
  • ...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
  • ...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
  • ...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
  • ....</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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