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The positive three-digit integer N has a ones digit of 0. What is the probability that N is divisible by 4? Express your answer as a common fraction.
...a <math>\boxed{1/2}</math> probability that <math>N</math> is divisible by 4.

427 bytes (71 words) - 17:35, 12 June 2022

Page text matches

Pythagorean Theorem
B = (4, 3); C = (4, 0);

5 KB (886 words) - 21:12, 22 January 2024
Physics
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
2005 Alabama ARML TST Problems/Problem 7
...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
2015 IMO Problems
...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
Wilson's Theorem
...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
Trivial Inequality
*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
Arithmetic mean
<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
Polynomial
* <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
Simon's Favorite Factoring Trick
...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
Factoring
<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
Mathematical problem solving
...um of the [[series]] <math>\frac11 + \frac14 + \frac19 + \cdots + \frac{1}{n^2} + \cdots</math> ...}+\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> 

2 KB (314 words) - 06:45, 1 May 2014
Pigeonhole Principle
...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
1965 AHSME Problems/Problem 40
...^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
Geometric mean
...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
Principle of Inclusion-Exclusion
...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
Cauchy-Schwarz Inequality
...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
Factorial
...[[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
==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
* [[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
Fermat's Little Theorem
...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
Euler's Totient Theorem
...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
Geometric inequality
...[[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 Proof of Lemma: 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
Elementary symmetric sum
...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
Number theory/Advanced
<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
Divisibility rules
...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
Induction
...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
Greatest common divisor
<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
Overcounting
"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
Constructive counting
...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
Euclidean algorithm
<math>r_{n-1} \pmod {r_n} \equiv 0</math> <math>r_{n-1} = r_n \cdot q_{n+1} +0</math> 

6 KB (924 words) - 21:50, 8 May 2022
Generating function
...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
Binomial Theorem
...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
Prime number
...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
Function
...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
Complementary counting
...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
Prime factorization
...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
Composite number
...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
Natural number
...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
Ellipse
...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
Base numbers
...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
Common factorizations
\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
Pi
...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
Fibonacci sequence
...<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
Muirhead's Inequality
...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
Cardinality
...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
Fundamental Theorem of Calculus
* 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
Polygon
...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
2011 AMC 10B Problems/Problem 22
<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
Vector
...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
Logarithm
...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
Law of Cosines
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
Angle bisector
A=(4,2); D=(3,4);

3 KB (575 words) - 15:27, 19 March 2023
Law of Sines
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
Sequence
...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
Geometric sequence
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
Arithmetic sequence
...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
Fermat's Last Theorem
...\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
2005 USAMO Problems/Problem 4
...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
Diophantine equation
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
Area
...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
Floor function
...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
Pascal's triangle
...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
Newton's Sums
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
1985 IMO Problems/Problem 5
...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
Trigonometry
...=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
Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality
...</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
Goldbach Conjecture
<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
Riemann Hypothesis
...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
Quadratic residues
...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
Power of a Point Theorem
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
* [[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
Collatz Problem
...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
* [[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
* [[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
* [[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
* [[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
Mock AMC
| n/a | n/a

51 KB (6,175 words) - 20:58, 6 December 2023
Bertrand's Postulate
...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
Riemann zeta function
<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
Modular arithmetic/Introduction
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
Perfect square
...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
Modular arithmetic/Intermediate
...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
2006 AIME I Problems
{{AIME Problems|year=2006|n=I}} == Problem 4 ==

7 KB (1,173 words) - 03:31, 4 January 2023
2006 AIME I Problems/Problem 15
=== Solution 4 === {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}

6 KB (910 words) - 19:31, 24 October 2023
2006 AIME I Problems/Problem 14
...> 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
2006 AIME I Problems/Problem 13
...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
2006 AIME I Problems/Problem 10
...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
2006 AIME I Problems/Problem 8
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
2006 AIME I Problems/Problem 7
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
2006 AIME I Problems/Problem 5
<math> b = 4 </math> <math>b=4</math>

3 KB (439 words) - 18:24, 10 March 2015
2006 AIME I Problems/Problem 4
...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
2006 AIME I Problems/Problem 3
...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
2022 AIME I Problems/Problem 1
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
2021 JMC 10
...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
2006 AMC 12B Problems
...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
2006 AMC 12A Problems
== Problem 4 == [[2006 AMC 12A Problems/Problem 4|Solution]]

15 KB (2,223 words) - 13:43, 28 December 2020
2005 AMC 12A Problems
...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
2004 AMC 12A Problems
== Problem 4 == [[2004 AMC 12A Problems/Problem 4|Solution]]

13 KB (1,953 words) - 00:31, 26 January 2023
2003 AMC 12A Problems
Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $4 per pair and each T-shirt costs $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
2002 AMC 12A Problems
<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
2000 AMC 12 Problems
...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
2001 AMC 12 Problems
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
2002 AMC 12B Problems
\qquad\mathrm{(C)}\ 4 when <math>x=4</math>?

10 KB (1,547 words) - 04:20, 9 October 2022
2003 AMC 12B Problems
<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
2004 AMC 12B Problems
== Problem 4 == [[2004 AMC 12B Problems/Problem 4|Solution]]

13 KB (2,049 words) - 13:03, 19 February 2020
2005 AMC 12B Problems
...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
2006 AMC 12B Problems/Problem 14
...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
2006 AMC 12B Problems/Problem 18
...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
2006 AMC 12B Problems/Problem 20
...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
2006 AMC 12B Problems/Problem 22
...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
2006 AMC 12B Problems/Problem 23
MP("B",D(B),plain.N,f); MP("B",D(B),plain.N,f);

7 KB (1,169 words) - 14:04, 10 June 2022
2006 AMC 12A Problems/Problem 17
pair f = (4.34, 74.58); label("F", f, N);

6 KB (958 words) - 23:29, 28 September 2023
2006 AMC 12A Problems/Problem 23
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
2006 AMC 12A Problems/Problem 24
<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
2006 AMC 12A Problems/Problem 25
...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
2005 AMC 12A Problems/Problem 20
...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
2005 AMC 12B Problems/Problem 12
...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
2005 AMC 12B Problems/Problem 21
...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
2005 AMC 12B Problems/Problem 22
<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
2005 AMC 12B Problems/Problem 24
...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
2005 AMC 12B Problems/Problem 25
...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
Power of a Point Theorem/Introductory Problem 4
...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
Root (polynomial)
...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
2006 AMC 10A Problems
== Problem 4 == [[2006 AMC 10A Problems/Problem 4|Solution]]

13 KB (2,028 words) - 16:32, 22 March 2022
2006 AMC 10A Problems/Problem 12
MP('8', (16,-4), W); MP('8', (20,-8), N);

3 KB (424 words) - 10:14, 17 December 2021
2006 AMC 10A Problems/Problem 16
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
2006 AMC 10A Problems/Problem 17
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
Set
...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
Newman's Tauberian Theorem
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
2006 USAMO Problems
...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
1991 AJHSME Problems
<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
Olimpiada de Mayo/1997
* [[1997 I Problems/Problem 4|Problem 4]] * [[1997 II Problems/Problem 4|Problem 4]]

856 bytes (98 words) - 14:53, 3 July 2009
2022 USAJMO Problems/Problem 1
...<math>a_n-g_n</math> is divisible by <math>m</math> for all integers <math>n>1</math>; ...\nmid d</math> and <math>m|a+(n-1)d-gr^{n-1}</math> for all integers <math>n>1</math>.

4 KB (792 words) - 00:29, 13 April 2024
2006 AIME II Problems/Problem 2
.../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
2005 AIME II Problems/Problem 1
...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
2005 AIME II Problems/Problem 2
...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>

4 KB (628 words) - 11:28, 14 April 2024
2005 AIME II Problems/Problem 3
...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
2005 AIME II Problems/Problem 5
...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}}

3 KB (547 words) - 19:15, 4 April 2024
2005 AIME II Problems
{{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
2005 AIME II Problems/Problem 7
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
2005 AIME II Problems/Problem 8
...> 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
2005 AIME II Problems/Problem 9
...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
2005 AIME II Problems/Problem 12
..., 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
2005 AIME II Problems/Problem 10
...</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
Euler's identity
...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
2005 AIME I Problems
{{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
2005 AIME I Problems/Problem 2
...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
2005 AIME I Problems/Problem 3
...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
2005 AIME I Problems/Problem 4
...<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
2005 AIME I Problems/Problem 5
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
2005 AIME I Problems/Problem 6
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
2005 AIME I Problems/Problem 7
label("$10$",(2.5,4.5),W); label("$10$",(18.37,4.5),E);

4 KB (567 words) - 20:20, 3 March 2020
2005 AIME I Problems/Problem 8
...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
2005 AIME I Problems/Problem 9
...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
2005 AIME I Problems/Problem 10
== Solution 4 == {{AIME box|year=2005|n=I|num-b=9|num-a=11}}

5 KB (852 words) - 21:23, 4 October 2023
2005 AIME I Problems/Problem 11
...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
2005 AIME I Problems/Problem 12
...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
2005 AIME I Problems/Problem 13
..., 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
2005 AIME I Problems/Problem 14
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
2005 AIME I Problems/Problem 15
...| 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
2005 AIME II Problems/Problem 15
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
2005 AIME II Problems/Problem 14
dotfactor = 4; label("$A$",A,N);

13 KB (2,129 words) - 18:56, 1 January 2024
2004 AIME I Problems/Problem 15
...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
2004 AIME I Problems/Problem 14
...> 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
2004 AIME I Problems/Problem 13
...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
2004 AIME I Problems/Problem 12
...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
2004 AIME I Problems/Problem 11
...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
2004 AIME I Problems/Problem 9
...<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
2004 AIME I Problems/Problem 1
...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
2004 AIME I Problems/Problem 2
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
2004 AIME I Problems/Problem 3
{{AIME box|year=2004|n=I|num-b=2|num-a=4}}

1 KB (156 words) - 17:56, 1 January 2016
2004 AIME I Problems/Problem 4
...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
2004 AIME I Problems/Problem 5
...<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
2004 AIME I Problems/Problem 6
...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
2004 AIME I Problems/Problem 7
...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
2004 AIME I Problems/Problem 8
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
2004 AIME I Problems
{{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
2004 AIME II Problems/Problem 15
...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
2004 AIME II Problems/Problem 14
...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
2004 AIME II Problems/Problem 12
...> 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
2004 AIME II Problems/Problem 11
...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
2004 AIME II Problems/Problem 10
...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
2004 AIME II Problems/Problem 9
...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
2004 AIME II Problems/Problem 8
...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
2004 AIME II Problems/Problem 7
...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
2004 AIME II Problems/Problem 6
...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
2004 AIME II Problems/Problem 5
...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
2004 AIME II Problems/Problem 4
...>-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
2004 AIME II Problems/Problem 3
...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
2004 AIME II Problems/Problem 1
...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
2004 AIME II Problems
{{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
* [[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
* [[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
* [[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
1983 AIME Problems
== Problem 4 == [[1983 AIME Problems/Problem 4|Solution]]

7 KB (1,104 words) - 12:53, 6 July 2022
2019 AIME I Problems/Problem 2
...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
1984 AIME Problems
...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
1986 AIME Problems
...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
1987 AIME Problems
.../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
1988 AIME Problems
...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
1989 AIME Problems
<center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center> == Problem 4 ==

7 KB (1,045 words) - 20:47, 14 December 2023
1990 AIME Problems
== Problem 4 == [[1990 AIME Problems/Problem 4|Solution]]

6 KB (870 words) - 10:14, 19 June 2021
1991 AIME Problems
...le <math>ABCD_{}^{}</math> has sides <math>\overline {AB}</math> of length 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</m == Problem 4 ==

7 KB (1,106 words) - 22:05, 7 June 2021
1992 AIME Problems
== Problem 4 == \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt}

8 KB (1,117 words) - 05:32, 11 November 2023
1993 AIME Problems
...etc. If the candidate went <math>\frac{n^{2}}{2}</math> miles on the <math>n^{\mbox{th}}_{}</math> day of this tour, how many miles was he from his star ...many contestants caught <math>n\,</math> fish for various values of <math>n\,</math>.

8 KB (1,275 words) - 06:55, 2 September 2021
1994 AIME Problems
..., where <math>m\,</math> and <math>n\,</math> are integers. Find <math>m + n\,</math>. == Problem 4 ==

7 KB (1,141 words) - 07:37, 7 September 2018
1995 AIME Problems
...> and <math>n</math> are relatively prime positive integers. Find <math>m-n.</math> ...h> and <math>n</math> are relatively prime positive integers, find <math>m+n.</math>

6 KB (1,000 words) - 00:25, 27 March 2024
1996 AIME Problems
.../math> is it true that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even integer? ...tive integer <math>n</math> for which the expansion of <math>(xy-3x+7y-21)^n</math>, after like terms have been collected, has at least 1996 terms.

6 KB (931 words) - 17:49, 21 December 2018
1997 AIME Problems
...and <math>n</math> are relatively prime positive integers. Find <math>m + n.</math> == Problem 4 ==

7 KB (1,098 words) - 17:08, 25 June 2020
1998 AIME Problems
== Problem 4 == ...ath>n</math> are [[relatively prime]] [[positive integer]]s. Find <math>m+n.</math>

7 KB (1,084 words) - 02:01, 28 November 2023
1999 AIME Problems
...nd <math>n_{}</math> are relatively prime positive integers. Find <math>m+n.</math> Find the sum of all positive integers <math>n</math> for which <math>n^2-19n+99</math> is a perfect square.

7 KB (1,094 words) - 13:39, 16 August 2020
2000 AIME I Problems
{{AIME Problems|year=2000|n=I}} ...he least positive integer <math>n</math> such that no matter how <math>10^{n}</math> is expressed as the product of any two positive integers, at least

7 KB (1,204 words) - 03:40, 4 January 2023
2001 AIME I Problems
{{AIME Problems|year=2001|n=I}} == Problem 4 ==

7 KB (1,212 words) - 22:16, 17 December 2023
2002 AIME I Problems
{{AIME Problems|year=2002|n=I}} ...h> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.

8 KB (1,374 words) - 21:09, 27 July 2023
2003 AIME I Problems
{{AIME Problems|year=2003|n=I}} <center><math> \frac{((3!)!)!}{3!} = k \cdot n!, </math></center>

6 KB (965 words) - 16:36, 8 September 2019
2000 AIME II Problems
{{AIME Problems|year=2000|n=II}} ...and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.

6 KB (947 words) - 21:11, 19 February 2019
2001 AIME II Problems
{{AIME Problems|year=2001|n=II}} .../math> forms a perfect square. What are the leftmost three digits of <math>N</math>?

8 KB (1,282 words) - 21:12, 19 February 2019
2002 AIME II Problems
{{AIME Problems|year=2002|n=II}} == Problem 4 ==

7 KB (1,177 words) - 15:42, 11 August 2023
2003 AIME II Problems
{{AIME Problems|year=2003|n=II}} ...is the sum of the other two. Find the sum of all possible values of <math>N</math>.

7 KB (1,127 words) - 09:02, 11 July 2023
1983 AIME Problems/Problem 3
...he roots of this equation are real, since its discriminant is <math>18^2 - 4 \cdot 1 \cdot 20 = 244</math>, which is positive. Thus by [[Vieta's formula ...he square root of a real number can't be negative, the only possible <math>n</math> is <math>5</math>.

3 KB (532 words) - 05:18, 21 July 2022
1983 AIME Problems/Problem 4
<cmath> AC = \sqrt{AB^2 + BC^2} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}. </cmath> string n[] = {"O","$T_1$","B","C","M","A","$T_3$","M","$T_2$"};

11 KB (1,741 words) - 22:40, 23 November 2023
1983 AIME Problems/Problem 6
Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <mat ...ively, we could have noted that <math>a^b\equiv a^{b\pmod{\phi{(n)}}}\pmod n</math>. This way, we have <math>6^{83}\equiv 6^{83\pmod {42}}\equiv 6^{-1}\

3 KB (361 words) - 20:20, 14 January 2023
1983 AIME Problems/Problem 11
label("E",E,N); label("F",F,N);

5 KB (865 words) - 21:11, 6 February 2023
1983 AIME Problems/Problem 13
...simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<!-- don't remove the following tag, for PoTW on the Wiki front pa ...<math>S</math> be a non-[[empty set | empty]] [[subset]] of <math>\{1,2,3,4,5,6\}</math>.

5 KB (894 words) - 22:02, 5 April 2024
1983 AIME Problems/Problem 14
...abel("$P$",P,1.5*dir(80)); label("$R$",R,NE); label("12",waypoint(O1--O2,0.4),S);</asy> ...,b=(20,0),m=(9.72456,5.31401),n=(20.58055,1.77134),p=(15.15255,3.54268),q=(4.29657,7.08535),r=(26,0);

13 KB (2,149 words) - 18:44, 5 February 2024
1983 AIME Problems/Problem 15
...rational number. If this number is expressed as a fraction <math>\frac{m}{n}</math> in lowest terms, what is the product <math>mn</math>? label("$D$",D,N);

19 KB (3,221 words) - 01:05, 7 February 2023
1984 AIME Problems/Problem 1
A somewhat quicker method is to do the following: for each <math>n \geq 1</math>, we have <math>a_{2n - 1} = a_{2n} - 1</math>. We can substi == Solution 4 ==

4 KB (576 words) - 21:03, 23 December 2023
1984 AIME Problems/Problem 3
...h>t_{2}</math>, and <math>t_{3}</math> in the figure, have [[area]]s <math>4</math>, <math>9</math>, and <math>49</math>, respectively. Find the area of pair A=(0,0),B=(12,0),C=(4,5);

4 KB (726 words) - 13:39, 13 August 2023
1984 AIME Problems/Problem 5
...</math>. If we multiply the two equations together, we get that <math>a^4b^4 = 2^{36}</math>, so taking the fourth root of that, <math>ab = 2^9 = \boxed ...ac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2</math>. This means that <math>\frac{\ln ab}{\ln 2} = 9</math

5 KB (782 words) - 14:49, 1 August 2023
1984 AIME Problems/Problem 7
...nction]] f is defined on the [[set]] of [[integer]]s and satisfies <math>f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\

4 KB (617 words) - 18:01, 9 March 2022
1984 AIME Problems/Problem 9
triple M,N,P[],Q[]; int n=s.length;

6 KB (947 words) - 20:44, 26 November 2021
1984 AIME Problems/Problem 11
...robability]] that no two birch trees are next to one another. Find <math>m+n</math>. ...e out one tree between each pair of birch trees. So you would remove <math>4</math> trees that aren't birch. What you are left with is a unique arrangem

7 KB (1,115 words) - 00:52, 7 September 2023
1984 AIME Problems/Problem 14
...>21</math>, ... , and in general <math>9 + 6n</math> for nonnegative <math>n</math> are odd composites. We now have 3 cases: ...th> can be expressed as <math>9 + (9+6n)</math> for some nonnegative <math>n</math>. Note that <math>9</math> and <math>9+6n</math> are both odd composi

8 KB (1,346 words) - 01:16, 9 January 2024
1985 AIME Problems/Problem 14
...tal, for an average of 9. Thus we must have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = \boxed{25}</math>. ...nts vs them, which is a contradiction since it must be larger. Thus, <math>n=\boxed{25}</math>.

5 KB (772 words) - 22:14, 18 June 2020
1985 AIME Problems/Problem 13
...<math>a_{n+1}</math>. Find the maximum value of <math>d_n</math> as <math>n</math> ranges through the [[positive integer]]s. ...h>2n+1</math> if it is going to divide the entire [[expression]] <math>100+n^2+2n+1</math>.

4 KB (671 words) - 20:04, 6 March 2024
1985 AIME Problems/Problem 12
...when it has crawled exactly <math>7</math> meters. Find the value of <math>n</math>. P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\

17 KB (2,837 words) - 13:34, 4 April 2024
1985 AIME Problems/Problem 10
...o <math>\frac n2</math> to <math>\frac {n+1}2</math> for any integer <math>n</math> (same reasoning as above). So now we only need to test every 10 numb *<math>4</math>: We can partition as <math>1+1+2</math>, and from the previous case

12 KB (1,859 words) - 18:16, 28 March 2022
1985 AIME Problems/Problem 5
...is chosen so that <math>a_n = a_{n - 1} - a_{n - 2}</math> for each <math>n \ge 3</math>. What is the sum of the first 2001 terms of this sequence if t ...h>n</math> times, <math>a_{j + 6n} = a_j</math> for all [[integer]]s <math>n</math> and <math>j</math>.

2 KB (410 words) - 13:37, 1 May 2022
1985 AIME Problems/Problem 4
...division points closest to the opposite vertices. Find the value of <math>n</math> if the the [[area]] of the small square is exactly <math>\frac1{1985 ...2 - 2n + 1 = 1985</math>. Solving this [[quadratic equation]] gives <math>n = \boxed{32}</math>.

3 KB (484 words) - 21:40, 2 March 2020
1986 AIME Problems/Problem 15
...> and <math>B</math> lie along the lines <math>y=x+3</math> and <math>y=2x+4</math> respectively, find the area of triangle <math>ABC</math>. label("$B$", B, N);

11 KB (1,722 words) - 09:49, 13 September 2023
1986 AIME Problems/Problem 13
...s this just becomes the ball-and-urn argument. We want to add 5 balls into 4 urns, which is the same as 3 dividers; hence this gives <math>{{5+3}\choose .../math> and multiplication, the answer is <math>{{2+4-1}\choose2} \cdot{{5+4-1}\choose5}=560</math> ~Slight edits in LaTeX by EthanSpoon

3 KB (445 words) - 19:44, 8 January 2023
1986 AIME Problems/Problem 11
=== Solution 4(calculus) === ...ix} ,</math> where the term <math>\dbinom{n}{k}</math> is negated if <math>n+k</math> is odd.

6 KB (872 words) - 16:51, 9 June 2023
1986 AIME Problems/Problem 10
...>. Play the role of the magician and determine <math>(abc)</math> if <math>N= 3194</math>. Let <math>n=abc</math> then

3 KB (565 words) - 16:51, 1 October 2023
1986 AIME Problems/Problem 9
D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle); pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N));

11 KB (1,850 words) - 18:07, 11 October 2023
1986 AIME Problems/Problem 8
...So the overall power of <math>2</math> and <math>5</math> is <math>7(1+2+3+4+5+6) = 7 \cdot 21 = 147</math>. However, since the question asks for proper ...d(n))/2}</math>, where <math>d(n)</math> is the number of divisor of <math>n</math>.

3 KB (487 words) - 20:52, 16 September 2020
1986 AIME Problems/Problem 7
The increasing [[sequence]] <math>1,3,4,9,10,12,13\cdots</math> consists of all those positive [[integer]]s which a ...o determine the 100th number. <math>100</math> is equal to <math>64 + 32 + 4</math>, so in binary form we get <math>1100100</math>. However, we must cha

5 KB (866 words) - 00:00, 22 December 2022
1986 AIME Problems/Problem 5
...]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>? ...<math>900</math>. The greatest [[integer]] <math>n</math> for which <math>n+10</math> divides <math>900</math> is <math>\boxed{890}</math>; we can doub

2 KB (338 words) - 19:56, 15 October 2023
1987 AIME Problems/Problem 15
== Solution 4 (Algebra) == label("$A$",A,N);

5 KB (838 words) - 18:05, 19 February 2022
1987 AIME Problems/Problem 14
...(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.</cmath> ...^2 + 2b^2 + 2ab\right).</math> Each of the terms is in the form of <math>x^4 + 324.</math> Using Sophie Germain, we get that

7 KB (965 words) - 10:42, 12 April 2024
1987 AIME Problems/Problem 12
...s a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n</math>. In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.

4 KB (673 words) - 19:48, 28 December 2023

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