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- ...^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|</math>. Applying [[difference of squares]], we see that ...)^2 - (4i+3)^2}{2}\right| &= \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right|\\ &= \left|\sum_{i=0}^9 -(8i+4) \right|.2 KB (241 words) - 11:56, 13 March 2015
- The thousands digit is <math>\in \{4,5,6\}</math>. Case <math>1</math>: Thousands digit is even3 KB (440 words) - 21:20, 22 July 2021
- ...rs and <math>s\,</math> is not divisible by the square of any prime. What is <math>q+r+s\,</math>? ...th>; the [[locus]] of each of the respective conditions for <math>P</math> is the region inside the (semi)circles with diameters <math>\overline{AB}, \ov4 KB (717 words) - 22:20, 3 June 2021
- ...\text{cis}\left(\frac {(2k + 1)\pi}{10}\right)</math> where <math>k</math> is an integer between <math>0</math> and <math>9</math>. The expression to find is <math>\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}</3 KB (375 words) - 23:46, 6 August 2021
- ...the sides of the squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can be partiti ...ac {13}6n</math> squares in every row. Then <math>6|n</math>, and our goal is to maximize the value of <math>n</math>.3 KB (473 words) - 17:06, 1 January 2024
- ...multiples of <math>3</math>, the change will always be a multiple of <math>3</math>, so we just need to find the number of changes we can get from <math ...r <math>m,n</math> being [[positive integer]]s is <math>5 \times 2 - 5 - 2=3</math>.4 KB (645 words) - 15:12, 15 July 2019
- ...ths of the sides of <math>\triangle ABC\,</math> are integers, <math>BD=29^3,\,</math> and <math>\cos B=m/n\,</math>, where <math>m\,</math> and <math>n ...he form <math>29^2 x</math> and <math>29 x^2</math>, respectively, where x is an integer.3 KB (534 words) - 16:23, 26 August 2018
- ...nues until the bag is empty. The probability that the bag will be emptied is <math>p/q,\,</math> where <math>p\,</math> and <math>q\,</math> are relativ *Case 1. We draw a pair on the first two cards. The second card is the same as the first with probability <math>\frac {1}{2k - 1}</math>, then3 KB (589 words) - 14:18, 21 July 2019
- Consider the points on the [[complex plane]]. The point <math>b+37i</math> is then a rotation of <math>60</math> degrees of <math>a+11i</math> about the ...)\left(\mathrm{cis}\,60^{\circ}\right) = (a+11i)\left(\frac 12+\frac{\sqrt{3}i}2\right)=b+37i.</cmath>5 KB (788 words) - 13:53, 8 July 2023
- has at least one solution, and each solution is an ordered pair <math>(x,y)\,</math> of integers. How many such ordered pa ...h> where the signs are all independent of each other, for a total of <math>3\cdot 2\cdot 2=12</math> lattice points. They are indicated by the blue dots3 KB (442 words) - 19:51, 8 January 2024
- <center><math>y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,</math></center> ...the plane into equilateral triangles of side length <math>\tfrac{2}{\sqrt{3}}.\,</math> How many such triangles are formed?4 KB (721 words) - 16:14, 8 March 2021
- ...math>. (If <math>n\,</math> has only one digits, then <math>p(n)\,</math> is equal to that digit.) Let <center><math>S=p(1)+p(2)+p(3)+\cdots+p(999)</math></center>.2 KB (275 words) - 19:27, 4 July 2013
- \lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994 (For real <math>x\,</math>, <math>\lfloor x\rfloor\,</math> is the greatest integer <math>\le x.\,</math>)2 KB (264 words) - 13:33, 11 August 2018
- ...ngent at <math>P</math> to a circle of radius 20. Square <math>ABCD</math> is constructed with <math>A</math> and <math>B</math> on the larger circle, <m ...that <math>OE = AB - OQ = AB - 10</math>. The other leg, <math>AE</math>, is just <math>\frac 12 AB</math>.2 KB (272 words) - 03:53, 23 January 2023
- ...square]]. What is the [[remainder]] when the 1994th term of the sequence is divided by 1000? ...= 2992</math>. The value of <math>n^2 - 1 = 2992^2 - 1 \pmod{1000}</math> is <math>\boxed{063}</math>.946 bytes (139 words) - 21:05, 1 September 2023
- ...ld start with a block of <tt>H</tt>'s, the total probability is that <math>3/2</math> of it has to start with an <tt>H</tt>. ...ose sum is <math>\frac{3/64}{1-(15/32)}=\frac{3}{34}</math>, so the answer is <math>\boxed{037}</math>.6 KB (979 words) - 13:20, 11 April 2022
- ...,</math> and <math>d_{}</math> are positive integers and <math>d_{}</math> is not divisible by the square of any prime number. Find <math>m+n+d.</math> ...math>, and <math>EF = \sqrt{OE^2 - OF^2} = 9</math>. Then <math>OEF</math> is a <math>30-60-90</math> [[right triangle]], so <math>\angle OEB = \angle OE3 KB (484 words) - 13:11, 14 January 2023
- ...- \left(k^4 - 2k^3 + \frac 32k^2 - \frac 12k + \frac 1{16}\right)\\ &= 4k^3 + k. \end{align*}</cmath> ...^3 + k</math> times, and the sum for each <math>k</math> is then <math>(4k^3 + k) \cdot \frac{1}{k} = 4k^2 + 1</math>. From <math>k = 1</math> to <math>2 KB (287 words) - 01:25, 12 December 2019
- ...ity]], let <math>AP = 1.</math> It follows that <math>\triangle OPA</math> is a <math>45-45-90</math> [[right triangle]], so <math>OP = AP = 1,</math> <m ...= 1 + 1 - 2\cos \theta \Longrightarrow \cos \theta = - 3 + 2\sqrt {2} = - 3 + \sqrt{8}.</cmath>8 KB (1,172 words) - 21:57, 22 September 2022
- ...of <math>P_{}</math> cuts <math>P_{}</math> into two prisms, one of which is [[similar]] to <math>P_{},</math> and both of which have nonzero volume. G ...nce <math>x < a, y < b, z < c</math>, it follows that the only possibility is <math>y=a,z=b=1995</math>. Then,2 KB (292 words) - 19:30, 4 July 2013
