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- <cmath>|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|.</cmath> ...obtained from these paired sequences are also obtained in another <math>2^5-1</math> ways by permuting the adjacent terms <math>\{a_1,a_2\},\{a_3,a_4\},5 KB (879 words) - 11:23, 5 September 2021
- ...l be <br /> three other equivalent boards.</font></td><td><font style="font-size:85%">For those symmetric about the center, <br /> there is only one oth ...re rotationally symmetric about the center square; there are <math>\frac{49-1}{2}=24</math> such pairs. There are then <math>{49 \choose 2}-24</math> pa4 KB (551 words) - 11:44, 26 June 2020
- ...itive [[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. ...3)^n</math>. Both [[binomial expansion]]s will contain <math>n+1</math> non-like terms; their product will contain <math>(n+1)^2</math> terms, as each t3 KB (515 words) - 04:29, 27 November 2023
- ...s a path that retraces no segment. Each time that such a path reaches a non-terminal vertex, it must leave it. ...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 fro9 KB (1,671 words) - 22:10, 15 March 2024
- ...math> take on the values <math>0, 1, \ldots, 9</math>. At step i of a 1000-step process, the <math>i</math>-th switch is advanced one step, and so are ...}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}</math>. In general, the divisor-count of <math>\frac{N}{d}</math> must be a multiple of 4 to ensure that a s3 KB (475 words) - 13:33, 4 July 2016
- .../math>, which can be done in <math>4! = 24</math> ways. Then choose a three-edge path along tetrahedron <math>DBEG</math> which, because it must start a ...faces, and one face adjacent to the three B-faces, which we will call the C-face.11 KB (1,837 words) - 18:53, 22 January 2024
- ...ween a plane and a point <math>I</math> can be calculated as <math>\frac{(I-G) \cdot P}{|P|}</math>, where G is any point on the plane, and P is a vecto ...rpendicular to plane <math>ABC</math> can be found as <math>V=(A-C)\times(B-C)=\langle 8, 12, 24 \rangle</math>6 KB (1,050 words) - 18:44, 27 September 2023
- ...1, 13, 34, 3, 21, 2\}. </math> Susan makes a list as follows: for each two-element subset of <math> \mathcal{S}, </math> she writes on her list the gre Each [[element]] of the [[set]] will appear in <math>7</math> two-element [[subset]]s, once with each other number.2 KB (317 words) - 00:09, 9 January 2024
- ...that <math> \left( x^{23} + x^{22} + \cdots + x^2 + x + 1 \right) \cdot (x-1) = x^{24} - 1 </math>. The five-element sum is just <math>\sin 30^\circ + \sin 60^\circ + \sin 90^\circ + \s4 KB (675 words) - 17:23, 30 July 2022
- ...ing the starting vertex in the next move. Thus <math>P_n=\frac{1}{2}(1-P_{n-1})</math>. ...-1</math> steps plus the number of ways to get to <math>C</math> in <math>n-1</math> steps.15 KB (2,406 words) - 23:56, 23 November 2023
- {{AIME box|year=2001|n=II|num-b=14|after=Last Question}}4 KB (518 words) - 15:01, 31 December 2021
- ...plies that <math>10^{6} - 1 | 10^{6k} - 1</math>, and so any <math>\boxed{j-i \equiv 0 \pmod{6}}</math> will work. ...le 5</math>, and let <math>j-i-a = 6k</math>. Then we can write <math>10^{j-i} - 1 = 10^{a} (10^{6k} - 1) + (10^{a} - 1)</math>, and we can easily verif4 KB (549 words) - 23:16, 19 January 2024
- Let point <math>A</math> be the top-left corner of square <math>ABCD</math> and the rest of the vertices be arra ...factor. We get <math>b = 2/5a</math>, meaning the ratio of areas <math>((a-2b)/a)^2</math> = <math>(1/5)^2</math> = <math>1/25</math> = <math>m/n.</mat4 KB (772 words) - 19:31, 6 December 2023
- ...8!-64!+\cdots+1968!-1984!+2000!</math>, find the value of <math>f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j</math>. ...{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \cdots + (n! - (n-1)!)) = n!</math>.7 KB (1,131 words) - 14:49, 6 April 2023
- if and only if <math>s </math> is not a divisor of <math>p-1 </math>. ...>s|(p-1)</math>. Then for some positive integer <math>k</math>, <math>sk=p-1</math>. The conditions given are equivalent to stating that <math>sm \bmo3 KB (506 words) - 17:54, 22 June 2023
- Let <math>f(x)</math> be a non-constant polynomial in <math>x</math> of degree <math>d</math> with any non-negative integer roots, so <math>a_i > 1</math> and thus <math>b_i > 0</math9 KB (1,699 words) - 13:48, 11 April 2020
- ...is from <math>2^m(2^{i_0-m} - 1) = 2^{i_0} - 2^m</math> to <math>2^m(2^{i_0-m}+1) = 2^{i_0} + 2^m</math>. ...- 2^{i_0} = 2^p(2t - 2^{i_0-p} + 1)</math> and the number <math>2t + 2^{i_0-p} + 1</math> is odd, a jump of size <math>2^{p+1}</math> can be made from <7 KB (1,280 words) - 17:23, 26 March 2016
- For a positive integer <math>k</math> and a non-negative integer <math>n</math>, ..._1 + b_2 + b_3 + ... + b_{k-1} + b_k= n}{\binom{n}{b_1, b_2, b_3, ..., b_{k-1}, b_k} \prod_{j=1}^{k}{x_j^{b_j}}}</cmath>3 KB (476 words) - 19:37, 4 January 2023
- ...\cdots P_{n-1})\not\subseteq (a)</math>. Take any <math>b\in P_1\cdots P_{n-1}\setminus (a)</math> let <math>\gamma = b/a\in K</math>. We claim that thi ...ot\in R</math>. Now for any <math>x\in J</math>, <math>bx\in P_1\cdots P_{n-1}J\subseteq P_1\cdots P_n\subseteq (a)</math>, and so <math>bx = ar</math>9 KB (1,648 words) - 16:36, 14 October 2017
- ...</math> and <math>y</math>, define <math> x \mathop{\spadesuit} y = (x+y)(x-y) </math>. What is <math> 3 \mathop{\spadesuit} (4 \mathop{\spadesuit} 5) < ...rcle is painted blue. What is the ratio of the blue-painted area to the red-painted area?14 KB (2,059 words) - 01:17, 30 January 2024
