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- ...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>? ...the second pair gives <math>98</math>. We now check that <math>130</math> is optimal, setting <math>a=m-2</math>, <math>b=n-2</math> in order to simplif2 KB (390 words) - 21:05, 29 May 2023
- ...n,</math> its complex power sum is defined to be <math>a_1i + a_2i^2+ a_3i^3 + \cdots + a_ni^n,</math> where <math>i^2 = - 1.</math> Let <math>S_n</mat ...we will just define to have a power sum of zero) with <math>9</math> in it is equal to the number of subsets without a <math>9</math>. To easily see this2 KB (384 words) - 14:47, 14 June 2024
- ...and <math>c</math> is not divisible by the square of any [[prime]]. What is <math>a^{2} + b^{2} + c^{2}</math>? ...math> gives <math>x = \frac {\sqrt {5} - 1}{2}</math> and <math>y = \frac {3 - \sqrt {5}}{2}</math>.5 KB (884 words) - 14:33, 18 June 2024
- ...math> and <math>c</math> are [[positive]] [[integer]]s, and <math>c</math> is not divisible by the square of any [[prime]]. Find <math>a + b + c.</math> ...e the area of the unshaded region over the area of the total region, which is the probability that the mathematicians do not meet:4 KB (624 words) - 19:00, 19 June 2024
- ...he preceding term from the one before that. The last term of the sequence is the first [[negative]] term encounted. What positive integer <math>x</math The best way to start is to just write out some terms.2 KB (354 words) - 19:37, 24 September 2023
- Note that this is an algebraic bijection, we have simplified the problem and essentially remo ...hrough 50; thus the answer is <math>n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600</math>, and <math>\frac n{100} = \boxed{196}</math>.5 KB (684 words) - 18:52, 19 June 2024
- ...values. The [[probability]] that all three players obtain an [[odd]] sum is <math>m/n,</math> where <math>m</math> and <math>n</math> are [[relatively ...that it matters in what order the people pick the tiles; the final answer is the same if we assume the opposite, that order doesn't matter.)5 KB (917 words) - 02:37, 12 December 2022
- ...ac 12</math> of the 30 by 30 [[square]] it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of th ...positive integers, so <math>(0,0)</math> doesn't count; hence, the answer is <math>480</math>.6 KB (913 words) - 16:34, 6 August 2020
- For how many values of <math>k</math> is <math>12^{12}</math> the [[least common multiple]] of the positive integers It is evident that <math>k</math> has only 2s and 3s in its prime factorization,2 KB (289 words) - 22:50, 23 April 2024
- ...ed by folding the triangle along the sides of its midpoint triangle. What is the volume of this pyramid? ...om <math>P</math> to <math>\triangle ABC</math>. The crux of this problem is the following lemma.7 KB (1,169 words) - 15:28, 13 May 2024
- ...ath> and <math>CA=15,</math> and the [[tangent]] of angle <math>PAB</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relativ /* constructing P, C is there as check */7 KB (1,184 words) - 13:25, 22 December 2022
- ...]] to <math>\overline{AB}</math> at <math>P_{},</math> and its [[radius]] is <math>21</math>. Given that <math>AP=23</math> and <math>PB=27,</math> fin We want the perimeter, which is <math>2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}</math>.3 KB (472 words) - 19:03, 21 June 2024
- ...s equal to the second half), but it quickly becomes apparent that this way is difficult to pull off. Instead, we look to [[telescope]] the sum. Using the ...}{\sin 175} \Longrightarrow s = \tan \frac{175}{2},</cmath> and our answer is <math>\boxed{177}</math>.4 KB (614 words) - 04:38, 8 December 2023
- ...segments form a [[triangle]] whose vertices are among the ten given points is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relat ...e picked. Since the triangle accounts for 3 segments, there are <math>45 - 3 = 42</math> segments remaining.3 KB (524 words) - 17:25, 17 July 2023
- ...ction]] has the property that the image of each point in the complex plane is [[equidistant]] from that point and the [[origin]]. Given that <math>|a+bi ...h passes through <math>\left(\frac 12, \frac12\right)</math>, so its graph is <math>x + y = 1</math>. Substituting <math>x = (a-b)</math> and <math>y = (6 KB (1,010 words) - 19:01, 24 May 2023
- ...9</math>. At step i of a 1000-step process, the <math>i</math>-th switch is advanced one step, and so are all the other switches whose labels divide th ...f <math>\frac{N}{d}</math> must be a multiple of 4 to ensure that a switch is in position A:3 KB (475 words) - 13:33, 4 July 2016
- ...er manipulation <math>y = \frac {x}{\sqrt {3}}</math> and <math>y = \sqrt {3}x</math>, respectively, which are still linear functions. Basically the squ ...2\pi) = \frac {1}{12}(2400\pi - 1200\pi) = 100\pi</math>. Hence the answer is <math>\boxed{314}</math>.2 KB (354 words) - 16:42, 20 July 2021
- ...h> is <math>43/99</math> and the [[area]] of octagon <math>ABCDEFGH</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relat ...of all <math>8</math> of them is <math>\frac{86}{99}</math> and the answer is <math>\boxed{185}</math>.3 KB (398 words) - 13:27, 12 December 2020
- ...all [[positive integer]]s <math>n</math> for which <math>n^2-19n+99</math> is a [[perfect square]]. ...math>. This gives <math>n=1, 9, 10,</math> or <math>18</math>, and the sum is <math>1+9+10+18=\boxed{38}</math>.2 KB (296 words) - 01:18, 29 January 2021
- ...his figure into two [[congruent]] [[polygon]]s. The [[slope]] of the line is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relat ...th>\frac{45 + \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
