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- ...rom the given conditions, <math>\max(j, m) = 3</math>, <math>\max(m, p) = \max(p, j) = 4</math>. Thus we must have <math>p = 4</math> and at least one of Now, for the powers of 5: we have <math>\max(k, n) = \max(n, q) = \max(q, k) = 3</math>. Thus, at least two of <math>k, n, q</math> must be equal3 KB (547 words) - 09:00, 18 June 2024
- ...ers <math>(2^{x})(3^{y})(5^{z})</math>, where <math>x, y</math>, and <math>z</math> take on the values <math>0, 1, \ldots, 9</math>. At step i of a 100 ...{j}}3^{y_{j}}5^{z_{j}}</math>. Let <math>N = 2^{9}3^{9}5^{9}</math> be the max switch label. To find the divisor multiples in the range of <math>d_{i}</ma3 KB (475 words) - 12:33, 4 July 2016
- ...efficients such that the sequence <math>\{ p(f(n^2))-2n) \}_{n \in \mathbb{Z} \ge 0}</math> is bounded above. (In particular, this requires <math>f(n^2) ...p|(2n + a_j)</math> for some <math>j\leq k</math>. Hence <math>p - 2n\leq \max\{a_1, a_2, \ldots, a_k\}</math>. The prime divisors of <math>c</math> form9 KB (1,699 words) - 12:48, 11 April 2020
- | <math>\textstyle\max</math>||\max||<math>\textstyle\min</math>||\min||<math>\Pr</math>||\Pr .../math>\mathcal{Z}<math>||\mathcal{Z}||</math>\mathfrak{Z}<math>||\mathfrak{Z}12 KB (1,798 words) - 20:55, 22 September 2024
- pair z = expi(3pi/10); int max = 7;4 KB (641 words) - 20:24, 21 April 2014
- ...th>x\le p</math>, <math>x\in \mathbb{Z}^+</math>, <math>\exists Y\subseteq Z</math>, <math>\sigma(Y)=x</math>, <math>\nexists \sigma(Y)=p+1</math>. <b>Lemma</b>) If <math>Z</math> is a <math>n</math>-<math>p</math> set, <math>p\leq 2^n-1</math>.5 KB (858 words) - 06:52, 19 July 2016
- ...the region outside the hexagon, and let <math>S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace</math>. Then the area of <math>S</math> has the form <ma ...xt{cis}\,\theta</math> is in <math>R</math>, then the point <math>\frac{1}{z} = \frac{1}{r} \text{cis}\, \left(-\theta\right)</math> is in <math>S</math6 KB (894 words) - 17:56, 25 December 2022
- ...<cmath>[a,b,c]=2^{\max(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3)} \cdots</cmath> ...in(x_1,y_1,z_1)} \cdot 3^{\max(x_2,y_2,z_2) - \min(x_2,y_2,z_2)} \cdot 5^{\max(x_3,y_3,z_3) - \min(x_3,y_3,z_3)} \cdots.</math> Squaring the LHS will just5 KB (1,018 words) - 10:14, 6 October 2023
- <cmath>d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\}</cmath> <cmath>d=\max\{d_i:1\le i\le n\}.</cmath>3 KB (505 words) - 08:24, 10 September 2020
- ...)</math>, <math>f_{n+2}\le\frac{f_{n+1}+f_{n}}{2}< \max (f_{n},f_{n+1})\le\max (f_{n},f_{n-1})</math> ...(f_{n},f_{n-1})> \max (f_{n+1},f_{n+2})</math> and that means that <math>\max (f_{2n},f_{2n+1})</math> is a strictly decreasing function and it must reac3 KB (593 words) - 06:56, 19 July 2016
- ...at <math>(7,3)</math> with radius <math>8</math>, and we want to find the max of <math>3x + 4y</math>. Let <math>z = 3x + 4y</math>. Solving for <math>y</math>, we get <math>y = (z - 3x)/4</math>. Substituting into the given equation, we get9 KB (1,441 words) - 16:51, 22 October 2023
- <cmath>\max(a_1, a_2, \dots, a_n) \le n \cdot \min(a_1, a_2, \dots, a_n),</cmath> ...ferred from the second, since <math>x+y>z \longrightarrow x^2 + y^2 +2xy > z^2</math> and <math>2xy</math> is trivially greater than <math>0.</math> So5 KB (930 words) - 01:17, 7 March 2018
- <cmath>\max(a_1, a_2, \dots, a_n) \le n \cdot \min(a_1, a_2, \dots, a_n),</cmath> ...ctions <math>f : \mathbb{Z}^+ \to \mathbb{Z}^+</math> (where <math>\mathbb{Z}^+</math> is the set of positive integers) such that <math>f(n!) = f(n)!</m3 KB (507 words) - 02:37, 7 June 2020
- Notice that for any <math>\theta</math>, <math>\max{\cos \theta} = 1</math>. This is achieved when <math>\theta = 0</math>, or ...ath>\text{LHS} = \text{RHS} = 2</math>, and we get (<math>m, n \in \mathbb{Z}</math>),5 KB (761 words) - 09:08, 17 August 2023
- ...math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> c ...<math>z = 10\operatorname{cis}{(\beta)}</math>. Then, <math>\dfrac{w - z}{z} = \dfrac{\operatorname{cis}{(\alpha)} - 10\operatorname{cis}{(\beta)}}{10\5 KB (782 words) - 19:25, 10 October 2023
- <math> ny+z = 1 </math> <cmath> |x|+|y|\leq\sqrt{2(x^{2}+y^{2})}\leq 2\mbox{Max}(|x|, |y|) </cmath>18 KB (2,788 words) - 18:25, 22 August 2024
- ...atisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600 \text{ and lcm}(y,z)=900</math>? ...<math>(b,h)=(1,0),(0,1)</math> or <math>(1,1)</math>. Similarly if <math>\max(d,g)=2</math> then <math>(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)</math>. Thus o7 KB (1,135 words) - 17:47, 16 October 2024
- ...}) = 31 \cdot 20 \cdot \frac{4}{5} = 31 \cdot 16</math> (Note that at this max value, since <math>\beta</math> and <math>\gamma</math> are both acute, <ma ...ath>z</math> be a complex number with <math>|z|=1</math>, <math>\text{Arg}(z)\ge 0^\circ</math> and <math>\text{Arg}(zw)\le90^\circ</math>. Then we rep9 KB (1,526 words) - 01:31, 29 December 2021
- ...that all of <math>x,y,z</math> are nonnegative integers. (*Note that <math>MAX(xyz)</math> is the largest possible product of <math>xyz</math>, and <math>12 KB (1,915 words) - 16:38, 29 April 2021
- ...= \frac{3}{4}x</math>. Now extend line <math>CD</math> to the point <math>Z</math> on <math>AE</math>, forming parallelogram <math>ZABC</math>. As <mat ...sider reading the barycentric coordinates article written by Evan Chen and Max Schindler here: https://artofproblemsolving.com/wiki/index.php/Resources_fo7 KB (1,186 words) - 19:40, 21 October 2024
