# Search results

• B = (4, 3); C = (4, 0);
5 KB (842 words) - 19:38, 26 January 2020
• 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 $n$, then $n$ will put an equivalent force in the opposite direction.
9 KB (1,355 words) - 23:53, 2 August 2020
• ...e series: <center>$3+\frac{11}4+\frac 94 + \cdots + \frac{n^2+2n+3}{2^n}+\cdots$.</center> ...^{\infty} \left(\frac{2n}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)[/itex]
1 KB (193 words) - 21:13, 18 May 2021
• ...gers $n\geq 3$, there exists a balanced set consisting of $n$ points. </li> ...ath> for which there exists a balanced centre-free set consisting of $n$ points. </li>
4 KB (692 words) - 22:33, 15 February 2021
• ...emainder 1. Show that there is an integer ${n}$ such that $n^2 + 1$ is divisible by ${p}$.
4 KB (639 words) - 22:48, 9 January 2020
• *Show that $\sum_{k=1}^{n}a_k^2 \geq a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1$. [[Inequality_Introductory_Problem_2|Solution]] *Show that $x^2+y^4\geq 2x+4y^2-4$ for all real $x$.
2 KB (399 words) - 22:10, 29 May 2021
• <center>$AM=\frac{x_1+x_2+\cdots+x_n}{n}$</center> is the arithmetic mean of the ${n}$ numbers $x_1,x_2,\ldots,x_n$.
699 bytes (110 words) - 12:44, 20 September 2015
• * $5x^4 - 2x^2 + 9$, in the variable $x$ A polynomial in one variable is a function $P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0$. Here, $a_i$ is the <m
6 KB (1,046 words) - 13:07, 14 July 2021
• *Two different [[prime number]]s between $4$ and $18$ are chosen. When their sum is subtracted from th ...n divided by $5$, find the value of the remainder of when $n$ is divided by $5$.
4 KB (631 words) - 13:42, 14 July 2021
• <cmath>a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})</cmath> If $n=2$, this creates the difference of squares factorization, <cmath>a^2-
3 KB (530 words) - 13:34, 14 July 2021
• ...um of the [[series]] $\frac11 + \frac14 + \frac19 + \cdots + \frac{1}{n^2} + \cdots$<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[/itex]<br>
2 KB (314 words) - 06:45, 1 May 2014
• ...mong any $n$ integers, there are two with the same modulo-$n-1$ residue. ...boxes then at least one box must hold at least $\left\lceil \frac{k}{n} \right\rceil$ objects. Here $\lceil \cdot \rceil$ denote
4 KB (691 words) - 00:54, 2 April 2021
• ...^4 + 6x^3 + 11x^2 + 3x + 31[/itex] is the square of an integer. Then $n$ is: $\textbf{(A)}\ 4 \qquad 3 KB (571 words) - 17:00, 9 July 2018 • ...numbers. Note that if [itex]n$ is even, we take the positive $n$th root. It is analogous to the [[arithmetic mean]] (with addition r ...1 and 2 is $\sqrt[4]{6\cdot 4\cdot 1 \cdot 2} = \sqrt[4]{48} = 2\sqrt[4]{3}$.
2 KB (282 words) - 22:04, 11 July 2008
• ...four guys in order. By the same logic as above, this is $2!\binom{6}{4}=30$. Again, $|A\cap C|$ would be putting five guys in ord If $(A_i)_{1\leq i\leq n}$ are finite sets, then:
9 KB (1,733 words) - 20:46, 26 October 2020
• <cmath> \frac{a_1+a_2+\ldots+a_n}{n}\geq\sqrt[n]{a_1a_2\cdots a_n} </cmath> <cmath> \sum_{i=1}^{n}\frac{a_i}{n} \geq \prod\limits_{i=1}^{n}a_i^{\frac{1}{n}} . </cmath>
4 KB (562 words) - 01:30, 6 March 2021
• ...[[recursion|recursive definition]] for the factorial is $n!=n \cdot (n-1)!$. * $4! = 24$
10 KB (799 words) - 10:45, 12 May 2021
• ==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) - 12:06, 15 July 2021
• * [[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 $a^{\varphi(n)} \equiv 1 \pmod{n}$, where $\varphi(n)$ denotes [[Euler's totient function]]. In particular, $\varphi( ...let [itex]S = \{\text{natural numbers relatively prime to and less than}\ n\}$.
11 KB (1,827 words) - 02:02, 15 October 2020

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