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2005 AIME II Problems/Problem 4
<math>15^7 = 3^7\cdot5^7</math> so <math>15^7</math> has <math>8\cdot8 = 64</math> divisor <math>\gcd(15^7, 18^{11}) = 3^7 </math> which has 8 divisors.

3 KB (377 words) - 18:36, 1 January 2024
2005 AIME II Problems/Problem 13
...th> P(17)=10 </math> and <math> P(24)=17. </math> Given that <math> P(n)=n+3 </math> has two distinct integer solutions <math> n_1 </math> and <math> n_ ...h>(x-17)(x-24)</math> to be a factor of <math>10</math>. Hence the answer is <math>19\cdot 22=\boxed{418}</math>.

4 KB (642 words) - 14:55, 12 August 2019
2005 AIME II Problems/Problem 5
...og b=3\log a </math> or <math>\log b=2\log a </math>, so either <math> b=a^3 </math> or <math> b=a^2 </math>. ...e <math> b=a^3 </math>, note that <math> 12^3=1728 </math> while <math> 13^3=2197 </math>. Therefore, for this case, all values of <math>a</math> from <

3 KB (547 words) - 19:15, 4 April 2024
2005 AIME II Problems/Problem 6
...agical. For example, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in t ...s suggests that <math>n = 131 + 65 = 196</math>; the total number of cards is <math>196 \cdot 2 = \boxed{392}</math>.

2 KB (384 words) - 00:31, 26 July 2018
2005 AIME II Problems
...that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math> n. </math> ...he guests. Given that the probability each guest got one roll of each type is <math> \frac mn, </math> where <math> m </math> and <math> n </math> are re

7 KB (1,119 words) - 21:12, 28 February 2020
2005 AIME II Problems/Problem 7
It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}</math>. ...+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}</cmath>

2 KB (279 words) - 12:33, 27 October 2019
2005 AIME II Problems/Problem 8
.../math> and <math> p </math> are [[relatively prime]], and <math> n </math> is not divisible by the square of any [[prime]], find <math> m+n+p. </math> ...= (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7));

4 KB (693 words) - 13:03, 28 December 2021
2005 AIME II Problems/Problem 9
...positive integers <math> n </math> less than or equal to <math>1000</math> is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all rea ...t certainly hold for <math>t = \frac{\pi}2 - u</math>. Thus, the question is equivalent to asking for how many [[positive integer]]s <math>n \leq 1000</

6 KB (1,154 words) - 03:30, 11 January 2024
2005 AIME II Problems/Problem 12
...h> and <math> r </math> are [[positive]] [[integer]]s and <math> r </math> is not divisible by the [[square]] of any [[prime]], find <math> p+q+r. </math ...- y</math> again, we know have <math>xy = (400 - y)y = 150^2</math>. This is a quadratic with roots <math>200 \pm 50\sqrt{7}</math>. Since <math>y < x</

13 KB (2,080 words) - 21:20, 11 December 2022
2005 AIME II Problems/Problem 10
...at the ratio of the volume of <math> O </math> to that of <math> C </math> is <math> \frac mn, </math> where <math> m </math> and <math> n </math> are re ...,0,-3)--(0,-3,0)--(3,0,0)--(0,0,-3)--(0,3,0)--(0,0,3)--(3,0,0)--(0,3,0)--(-3,0,0));

3 KB (436 words) - 03:10, 23 September 2020
2005 AIME II Problems/Problem 11
...>a_0 = 37, a_1 = 72, a_m = 0, </math> and <math> a_{k+1} = a_{k-1} - \frac 3{a_k} </math> for <math> k = 1,2,\ldots, m-1. </math> Find <math>m. </math> <math>a_{k}a_{k+1} = a_{k-1}a_{k} - 3 </math>.

3 KB (499 words) - 18:52, 21 November 2022
Euler's identity
...er's Formula''' is <math>e^{i\theta}=\cos \theta+ i\sin\theta</math>. It is named after the 18th-century mathematician [[Leonhard Euler]]. ...umbers]] and/or [[trigonometry]]. Euler's formula replaces "[[cis]]", and is a superior notation, as it encapsulates several nice properties:

3 KB (452 words) - 23:17, 4 January 2021
2006 AMC 12A Problems/Problem 4
A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display? 1+2&9&6&3\\

2 KB (257 words) - 11:20, 2 January 2022
2005 AIME I Problems
...ose common difference is <math> k. </math> For example, <math> S_3 </math> is the sequence <math> 1,4,7,10,\ldots. </math> For how many values of <math> == Problem 3 ==

6 KB (983 words) - 05:06, 20 February 2019
2005 AIME I Problems/Problem 2
...ose common difference is <math> k</math>. For example, <math> S_3 </math> is the [[sequence]] <math> 1,4,7,10,\ldots. </math> For how many values of <ma ...h>. Thus the requested number of values is <math>12</math>, and the answer is <math>\boxed{012}</math>.

2 KB (303 words) - 01:31, 5 December 2022
2005 AIME I Problems/Problem 3
...ivisor]]s (positive integral [[divisor]]s excluding itself), each of which is less than 50? ...so <math>n</math> must be in the form <math>n=p\cdot q</math> or <math>n=p^3</math> for distinct [[prime number]]s <math>p</math> and <math>q</math>.

2 KB (249 words) - 09:37, 23 January 2024
2005 AIME I Problems/Problem 4
...5</math>, so this number works and no larger number can. Thus, the answer is <math>\boxed{294}</math>. ...factors of <math>69</math> are <math>(1,69), (3,23)</math>; <math>x</math> is maximized for the first case. Thus, <math>x = \frac{69 + 1}{2} = 35</math>,

8 KB (1,248 words) - 11:43, 16 August 2022
2005 AIME I Problems/Problem 5
...te parts to this problem: one is the color (gold vs silver), and the other is the orientation. ...t occur at all, for <math>9</math> total configurations. Thus, the answer is <math>70 \cdot 9 = \boxed{630}</math>.

5 KB (830 words) - 01:51, 1 March 2023
2005 AIME I Problems/Problem 6
Let <math> P </math> be the product of the nonreal roots of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math> The left-hand side of that [[equation]] is nearly equal to <math>(x - 1)^4</math>. Thus, we add 1 to each side in ord

4 KB (686 words) - 01:55, 5 December 2022
2005 AIME I Problems/Problem 7
...DE</math> is concurrent with line <math>BC</math>. Then, <math>ABED</math> is an isosceles trapezoid so <math>AD=BE=10</math>, and <math>BC=8</math> and ...</math>. The [[Pythagorean Theorem]] yields that <math>GC^2 = 12^2 - \sqrt{3}^2 = 141</math>, so <math>EF = GC = \sqrt{141}</math>. Therefore, <math>AB

4 KB (567 words) - 20:20, 3 March 2020
2005 AIME I Problems/Problem 8
...2^{222x+1} + 1 </math> has three [[real]] [[root]]s. Given that their sum is <math>m/n</math> where <math> m </math> and <math> n </math> are [[relative ...</math> and <math>x_1 + x_2 + x_3 = \frac{2}{111}</math>. Thus the answer is <math>111 + 2 = \boxed{113}</math>.

1 KB (161 words) - 19:50, 2 January 2022
2005 AIME I Problems/Problem 9
...[probability]] of the entire [[surface area]] of the larger cube is orange is <math> \frac{p^a}{q^br^c}, </math> where <math> p,q, </math> and <math> r < ...orientations, so from these cubes we gain a factor of <math>\left(\frac{2}{3}\right)^6</math>.

4 KB (600 words) - 21:44, 20 November 2023
2005 AIME I Problems/Problem 10
...[[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is <math>\left(\frac{35}{2}, \frac{39}{2}\right)</math>. The equation of the m ...tion for the triangle will give a smaller value of <math>p+q</math>, which is provable by following these steps over again) (alternatively, we could use

5 KB (852 words) - 21:23, 4 October 2023
2005 AIME I Problems/Problem 11
...e]] whose sides have length 8. Given the maximum value of <math> d </math> is <math> m - \sqrt{n},</math> find <math> m+n. </math> ...n it touches both other sides of the square. This can happen only when it is arranged so that the center of the semicircle lies on one diagonal of the s

4 KB (707 words) - 11:11, 16 September 2021
2005 AIME I Problems/Problem 12
...squares less than <math>n</math>. So <math>S(1), S(2)</math> and <math>S(3)</math> are odd, while <math>S(4), S(5), \ldots, S(8)</math> are even, and ...t the numbers between <math>1^2</math> and <math>2^2</math>, between <math>3^2</math> and <math>4^2</math>, and so on, all the way up to the numbers bet

4 KB (647 words) - 02:29, 4 May 2021
2005 AIME I Problems/Problem 13
...th>U</math> represent a move upwards, and <math>D</math> to be a move that is diagonal. [[Casework]] upon the number of diagonal moves: *'''Case ''' <math>d = 1</math>: It is easy to see only <math>2</math> cases.

5 KB (897 words) - 00:21, 29 July 2022
2005 AIME I Problems/Problem 14
...e the area of <math> S. </math> Find the remainder when <math> 10K </math> is divided by <math>1000</math>. Consider a point <math>E</math> such that <math>AE</math> is [[perpendicular]] to <math>BD</math>, <math>AE</math> intersects <math>BD</

3 KB (561 words) - 14:11, 18 February 2018
2005 AIME I Problems/Problem 15
...re <math> m </math> and <math> n </math> are integers and <math> n </math> is not [[divisor | divisible]] by the [[perfect square | square]] of a prime, ...thout loss of generality, let <math>AC < AB</math>, so that <math>E</math> is between <math>D</math> and <math>C</math>. Let the length of the median be

5 KB (906 words) - 23:15, 6 January 2024
2005 AIME II Problems/Problem 15
...or which the line <math> y=ax </math> contains the center of a circle that is externally [[tangent (geometry)|tangent]] to <math> w_2 </math> and interna ...centers is <math>r_1 + r_2</math>, and if they are internally tangent, it is <math>|r_1 - r_2|</math>. So we have

12 KB (2,000 words) - 13:17, 28 December 2020
2005 AIME II Problems/Problem 14
...th> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such that <math> \angle BAE\cong \angle CAD ...{BE} - 1 \Longrightarrow BE = \frac{13^2 \cdot 15}{463}</math>. The answer is <math>q = \boxed{463}</math>.

13 KB (2,129 words) - 18:56, 1 January 2024
2004 AIME I Problems/Problem 15
f(x)=\begin{cases}1 & \text{if }x = 1\\ \frac x{10} & \text{if }x\text{ is divisible by 10}\\ x+1 & \text{otherwise}\end{cases} ...st <math>n</math> such that <math>x_n=1</math>. (For example, <math>d(100)=3</math> and <math>d(87)=7</math>.) Let <math>m</math> be the number of posit

9 KB (1,491 words) - 01:23, 26 December 2022
2004 AIME I Problems/Problem 14
...> and <math> c </math> are [[positive]] [[integer]]s, and <math> c </math> is prime. Find <math> a+b+c. </math> real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;

4 KB (729 words) - 01:00, 27 November 2022
2004 AIME I Problems/Problem 13
...oots of the form <math> z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34, </math> with <math> 0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{3 ...nomial]] <math>P</math> is very difficult to work with directly, but there is one obvious transformation to make: sum the [[geometric series]]:

2 KB (298 words) - 20:02, 4 July 2013
2004 AIME I Problems/Problem 12
...on <math> [z] </math> denotes the [[floor function|greatest integer]] that is less than or equal to <math> z. </math> <math>\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor</math> is even when

2 KB (303 words) - 22:28, 11 September 2020
2004 AIME I Problems/Problem 11
...s a 3-inch radius. The entire [[surface]] of the cone, including its base, is painted. A [[plane]] [[parallel]] to the base of the cone divides the cone ...face area]] <math>A = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]], we ge

5 KB (839 words) - 22:12, 16 December 2015
2004 AIME I Problems/Problem 10
...[[probability]] that the circle will not touch diagonal <math> AC </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relativ ...nter of the circle must be in the <math>34 \times 13</math> rectangle that is one unit away from the sides of rectangle <math>ABCD</math>. We want to fin

5 KB (836 words) - 07:53, 15 October 2023
2004 AIME I Problems/Problem 9
...<math> U_1 </math> is similar to <math> U_2 </math> and <math> V_1 </math> is similar to <math> V_2. </math> The minimum value of the area of <math> U_1 ...h>ABC</math>. Thus <math>U_1</math>, and hence <math>U_2</math>, are <math>3-4-5\,\triangle</math>s.

4 KB (618 words) - 20:01, 4 July 2013
2004 AIME I Problems/Problem 1
...to right. What is the sum of the possible remainders when <math> n </math> is divided by <math>37</math>? ...2) + 10(n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>.

2 KB (374 words) - 14:53, 27 December 2019
2004 AIME I Problems/Problem 2
...atest element of <math>A</math> and the greatest element of <math>B</math> is <math>99</math>. Find <math>m.</math> ...must be <math>2</math>. Therefore, the largest element in <math>A</math> is <math>2 + \frac{m-1}{2}</math>.

8 KB (1,437 words) - 21:53, 19 May 2023
2004 AIME I Problems/Problem 4
...<math> S </math> enclose a region whose [[area]] to the nearest hundredth is <math>k</math>. Find <math> 100k</math>. ...e at each corner of the square. The area enclosed by all of the midpoints is <math>4-4\cdot \left(\frac{\pi}{4}\right)=4-\pi \approx .86</math> to the n

3 KB (532 words) - 09:22, 11 July 2023
2004 AIME I Problems/Problem 5
...</math> and <math> n </math> are relatively prime positive integers. What is <math> m+n </math>? From here, we see the largest possible value of <math>a+b</math> is <math>349</math>.

3 KB (436 words) - 18:31, 9 January 2024
2004 AIME I Problems/Problem 6
...s [[odd integer | odd]] and <math> a_i>a_{i+1} </math> if <math> i </math> is [[even integer | even]]. How many snakelike integers between 1000 and 9999 ...into two cases: one in which zero is one of the digits and one in which it is not. In the latter case, suppose we pick digits

3 KB (562 words) - 18:12, 4 March 2022
2004 AIME I Problems/Problem 7
...ath>Q(x)</math> is some polynomial [[divisibility | divisible]] by <math>x^3</math>. ...x)</math>, where <math>R(x)</math> is some polynomial divisible by <math>x^3</math>.

5 KB (833 words) - 19:43, 1 October 2023
2004 AIME I Problems/Problem 8
There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are sim ...of this <math>n</math>-gon in a counterclockwise direction: <math>0, 1, 2, 3, \ldots, n-1.</math>

4 KB (620 words) - 21:26, 5 June 2021
2004 AIME I Problems
...to right. What is the sum of the possible remainders when <math> n </math> is divided by 37? ...t element of <math> A </math> and the greatest element of <math> B </math> is 99. Find <math> m. </math>

9 KB (1,434 words) - 13:34, 29 December 2021
2004 AIME II Problems/Problem 15
...th>256</math> by <math>1</math> strip of quadruple thickness. This process is repeated <math>8</math> more times. After the last fold, the strip has beco Number the squares <math>0, 1, 2, 3, ... 2^{k} - 1</math>. In this case <math>k = 10</math>, but we will consi

6 KB (899 words) - 20:58, 12 May 2022
2004 AIME II Problems/Problem 14
...ht <math> 7 </math>'s in this way. For how many values of <math> n </math> is it possible to insert <math> + </math> signs so that the resulting expressi ...g by <math>7</math>, <math>a + 11b + 111c = 1000</math>. Then the question is asking for the number of values of <math>n = a + 2b + 3c</math>.

11 KB (1,857 words) - 21:55, 19 June 2023
2004 AIME II Problems/Problem 13
...of triangle <math> ABC </math> and the area of triangle <math> EBD </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relativ ...B \parallel CE, BC \parallel AD, </math> it follows that <math>ABCF</math> is a [[parallelogram]], and so <math>\triangle ABC \cong \triangle CFA</math>.

3 KB (486 words) - 22:15, 7 April 2023
2004 AIME II Problems/Problem 12
..., </math> and <math> p </math> are [[positive integer]]s, <math> n </math> is not [[divisibility | divisible]] by the [[perfect square | square]] of any real r = (-60 + 48 * 3^.5)/23;

3 KB (431 words) - 23:21, 4 July 2013
2004 AIME II Problems/Problem 10
...ath> S, </math> the [[probability]] that it is divisible by <math>9</math> is <math> p/q, </math> where <math> p </math> and <math> q </math> are relativ ...{40}{2}</math> because we’re choosing 2 1s to go in 40 digit slots. This is equal to 780; we have found <math>q</math>, our denominator.

7 KB (1,091 words) - 18:41, 4 January 2024
2004 AIME II Problems/Problem 9
...ression. Let <math> a_n </math> be the greatest term in this sequence that is less than <math>1000</math>. Find <math> n+a_n. </math> ...th>. This happens with <math>f(7)f(8) = 29 \cdot 33 = 957</math>, and this is the <math>2(8) = 16</math>th term of the sequence.

3 KB (538 words) - 21:33, 30 December 2023
2004 AIME II Problems/Problem 8
...the prime factorization of <math>2004^{2004}</math> is <math>2^{4008}\cdot 3^{2004}\cdot 167^{2004}</math>. ...ample, the number of divisors of <math>2004=2^2\cdot 3^1\cdot 167^1</math> is <math>(2+1)(1+1)(1+1)=12</math>.

2 KB (353 words) - 18:08, 25 November 2023
2004 AIME II Problems/Problem 7
...th> CF=3 </math> are given. The perimeter of rectangle <math> ABCD </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relativ pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C;

9 KB (1,501 words) - 05:34, 30 October 2023
2004 AIME II Problems/Problem 6
...t the end of the process are in the [[ratio]] <math> 3: 2: 1, </math>what is the least possible total for the number of bananas? ...c{11}{24}b_3</math>, and the third monkey got <math>\frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3</math>.

6 KB (950 words) - 14:18, 15 January 2024
2004 AIME II Problems/Problem 5
...urther behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the <math>800</math> worke ...0}{800}(60)=\frac{150}{8}</math>. The train then has <math>60-15-\frac{50}{3}-\frac{150}{8}=230/24</math> minutes left to travel 250 miles, and doing th

4 KB (592 words) - 19:02, 26 September 2020
2004 AIME II Problems/Problem 4
...ath>n</math>-digit number, for a total of <math>(2^1 - 2) + (2^2 - 2) + (2^3 -2) + (2^4 - 2) = 22</math> such numbers (or we can list them: <math>AB, BA ...s we can form, for a total of <math>(2^0 - 1) + (2^1 - 1) + (2^2 - 1) + (2^3 - 1) = 11</math> such numbers (or we can list them: <math>A0, A00, A0A, AA0

3 KB (508 words) - 01:16, 19 January 2024
2004 AIME II Problems/Problem 3
...gruent]] 1-cm [[cube (geometry) | cube]]s [[face]] to face. When the block is viewed so that three of its faces are visible, exactly <math>231</math> of ...s close together as possible, which occurs when the smaller block is <math>3 \times 7 \times 11</math>. Then the extra layer makes the entire block <ma

2 KB (377 words) - 11:53, 10 March 2014
2004 AIME II Problems/Problem 2
...ability]] that they get the same color combination, irrespective of order, is <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relat ...c{28}{153}</math>. So the probability that they both pick two red candies is <math>\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323}</math>. The same

2 KB (330 words) - 13:42, 1 January 2015
2004 AIME II Problems/Problem 1
...y [[prime]]. Find the [[remainder]] when the product <math> abcdef </math> is divided by 1000. .../math>; the rest of the area of the circle is then equal to <math>\frac{2}{3}r^2\pi</math>.

2 KB (329 words) - 23:20, 4 July 2013
2004 AIME II Problems
...re of any prime. Find the remainder when the product <math> abcdef </math> is divided by 1000. ...obability that they get the same color combination, irrespective of order, is <math> m/n, </math> where <math> m </math> and <math> n </math> are relativ

9 KB (1,410 words) - 05:05, 20 February 2019
1983 AIME Problems
== Problem 3 == What is the product of the real roots of the equation <math>x^2 + 18x + 30 = 2 \sqr

7 KB (1,104 words) - 12:53, 6 July 2022
2019 AIME I Problems/Problem 2
...9, 20</math> distinct from <math>J</math>. The value of <math>B - J</math> is at least <math>2</math> with a probability that can be expressed in the for ...because <math>B \ne J</math>, so the probability that <math>B-J < 0</math> is <math>\frac{1}{2}</math> by symmetry.

5 KB (830 words) - 22:15, 28 December 2023
1984 AIME Problems
..._{98}</math> if <math>a_1</math>, <math>a_2</math>, <math>a_3\ldots</math> is an arithmetic progression with common difference 1, and <math>a_1+a_2+a_3+\ ...sitive multiple of <math>15</math> such that every digit of <math>n</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>.

6 KB (933 words) - 01:15, 19 June 2022
1986 AIME Problems
What is the sum of the solutions to the equation <math>\sqrt[4]{x} = \frac{12}{7 - == Problem 3 ==

5 KB (847 words) - 15:48, 21 August 2023
1987 AIME Problems
An ordered pair <math>(m,n)</math> of non-negative integers is called "simple" if the addition <math>m+n</math> in base <math>10</math> re What is the largest possible distance between two points, one on the sphere of radi

6 KB (869 words) - 15:34, 22 August 2023
1988 AIME Problems
...rder -- the correct five buttons. The sample shown below has <math>\{1, 2, 3, 6, 9\}</math> as its combination. Suppose that these locks are redesigned == Problem 3 ==

6 KB (902 words) - 08:57, 19 June 2021
1989 AIME Problems
== Problem 3 == Suppose <math>n_{}^{}</math> is a positive integer and <math>d_{}^{}</math> is a single digit in base 10. Find <math>n_{}^{}</math> if

7 KB (1,045 words) - 20:47, 14 December 2023
1990 AIME Problems
The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are ne Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>.

6 KB (870 words) - 10:14, 19 June 2021
1991 AIME Problems
...overline {AB}</math> of length 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</math> into 168 congruent segments with points == Problem 3 ==

7 KB (1,106 words) - 22:05, 7 June 2021
1992 AIME Problems
...n its decimal representation, there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are == Problem 3 ==

8 KB (1,117 words) - 05:32, 11 November 2023
1993 AIME Problems
== Problem 3 == <center><math>\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\

8 KB (1,275 words) - 06:55, 2 September 2021
1994 AIME Problems
...erfect square. What is the remainder when the 1994th term of the sequence is divided by 1000? ...<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

7 KB (1,141 words) - 07:37, 7 September 2018
1995 AIME Problems
...</math> The total area enclosed by at least one of <math>S_{1}, S_{2}, S_{3}, S_{4}, S_{5}</math> can be written in the form <math>m/n,</math> where <m == Problem 3 ==

6 KB (1,000 words) - 00:25, 27 March 2024
1996 AIME Problems
...magic square, the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Fin ...that <math>n<1000</math> and that <math>\lfloor \log_{2} n \rfloor</math> is a positive even integer?

6 KB (931 words) - 17:49, 21 December 2018
1997 AIME Problems
== Problem 3 == ...number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?

7 KB (1,098 words) - 17:08, 25 June 2020
1998 AIME Problems
For how many values of <math>k</math> is <math>12^{12}</math> the [[least common multiple]] of the positive integers == Problem 3 ==

7 KB (1,084 words) - 02:01, 28 November 2023
1999 AIME Problems
Find the smallest prime that is the fifth term of an increasing arithmetic sequence, all four preceding ter ...rigin cuts this figure into two congruent polygons. The slope of the line is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are relativ

7 KB (1,094 words) - 13:39, 16 August 2020
2000 AIME I Problems
...ositive integer <math>n</math> such that no matter how <math>10^{n}</math> is expressed as the product of any two positive integers, at least one of thes ...<math>D</math> across the y-axis. The area of pentagon <math>ABCDE</math> is <math>451</math>. Find <math>u + v</math>.

7 KB (1,204 words) - 03:40, 4 January 2023
2001 AIME I Problems
...h>\mathcal{S}</math>, and the mean of <math>\mathcal{S}\cup\{2001\}</math> is <math>27</math> more than the mean of <math>\mathcal{S}</math>. Find the me == Problem 3 ==

7 KB (1,212 words) - 22:16, 17 December 2023
2002 AIME I Problems
...it arrangement that reads the same left-to-right as it does right-to-left) is <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively pr size(250);real x=sqrt(3);

8 KB (1,374 words) - 21:09, 27 July 2023
2003 AIME I Problems
<center><math> \frac{((3!)!)!}{3!} = k \cdot n!, </math></center> ...k </math> and <math> n </math> are positive integers and <math> n </math> is as large as possible, find <math> k + n. </math>

6 KB (965 words) - 16:36, 8 September 2019
2000 AIME II Problems
<center><math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math></center> A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola <math

6 KB (947 words) - 21:11, 19 February 2019
2001 AIME II Problems
...t and 85 percent of the school population, and the number who study French is between 30 percent and 40 percent. Let <math>m</math> be the smallest numbe == Problem 3 ==

8 KB (1,282 words) - 21:12, 19 February 2019
2002 AIME II Problems
...s between <math>100</math> and <math>999</math>, inclusive; <math>y</math> is the number formed by reversing the digits of <math>x</math>; and <math>z=|x ...7,12,10)</math>, <math>Q=(8,8,1)</math>, and <math>R=(11,3,9)</math>. What is the [[surface area]] of the cube?

7 KB (1,177 words) - 15:42, 11 August 2023
2003 AIME II Problems
...> of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of <math>N</m ...whose digits are all different. What is the remainder when <math>N</math> is divided by 1000?

7 KB (1,127 words) - 09:02, 11 July 2023
1983 AIME Problems/Problem 1
<math>x^{120}=w^5</math>, <math>y^{120}=w^3</math>, and <math>(xyz)^{120}=w^{10}</math>. ...=w</math>. It now becomes clear that one way to find <math>\log_z w</math> is to find what <math>x^{12}</math> and <math>y^{12}</math> are in terms of <m

4 KB (642 words) - 03:14, 17 August 2022
1983 AIME Problems/Problem 2
It is best to get rid of the [[absolute value]]s first. Adding these together, we find that the sum is equal to <math>30-x</math>, which attains its minimum value (on the given i

1 KB (184 words) - 20:16, 14 January 2023
1983 AIME Problems/Problem 3
What is the product of the [[real]] [[root]]s of the [[equation]] <math>x^2 + 18x + ...d moreover, plugging in <math>y=-6</math>, we get <math>-6=6</math>, which is obviously false). Hence we have <math>y=10</math> as the only solution for

3 KB (532 words) - 05:18, 21 July 2022
1983 AIME Problems/Problem 4
...d that of <math>BC</math> is <math>2</math> cm. The angle <math>ABC</math> is a right angle. Find the square of the distance (in centimeters) from <math> ...tem to get <math>x = 1</math> and <math>y = 5</math>, such that the answer is <math>1^2 + 5^2 = \boxed{026}</math>.

11 KB (1,741 words) - 22:40, 23 November 2023
1983 AIME Problems/Problem 5
.../math> is <math>7</math> and the sum of the cubes is <math>10</math>. What is the largest real value that <math>x + y</math> can have? One way to solve this problem is by [[substitution]]. We have

4 KB (672 words) - 10:17, 17 March 2023
1983 AIME Problems/Problem 6
After some quick division, our answer is <math>\boxed{035}</math>. === Solution 3 (cheap and quick) ===

3 KB (361 words) - 20:20, 14 January 2023
1983 AIME Problems/Problem 7
...ch other. If <math>P</math> is written as a fraction in lowest terms, what is the sum of the numerator and denominator? ...24}=1-\frac{420}{552}=1-\frac{35}{46}=\frac{11}{46}</math>, and the answer is <math>11+46=\boxed{057}</math>.

9 KB (1,392 words) - 20:37, 19 January 2024
1983 AIME Problems/Problem 8
What is the largest <math>2</math>-digit [[prime]] factor of the integer <math>n = ...h>3p<200</math>. The largest such prime is <math>\boxed{061}</math>, which is our answer.

2 KB (243 words) - 20:23, 14 January 2023
1983 AIME Problems/Problem 9
...e x\sin x \le \frac{\pi}{2}</math>, this value of <math>\frac{2}{3}</math> is attainable by the [[Intermediate Value Theorem]]). ...We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>.

4 KB (722 words) - 20:25, 14 January 2023
1983 AIME Problems/Problem 10
...h>, <math>1005</math> and <math>1231</math> have something in common: each is a <math>4</math>-digit number beginning with <math>1</math> that has exactl ...ath>, <math>x\neq1</math>, and <math>y\neq1</math>. Hence, there are <math>3\cdot9\cdot8=216</math> numbers of this form.

5 KB (855 words) - 20:26, 14 January 2023
1983 AIME Problems/Problem 11
...dges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid? triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6);

5 KB (865 words) - 21:11, 6 February 2023
1983 AIME Problems/Problem 12
...from their intersection point <math>H</math> to the center <math>O</math> is a positive rational number. Determine the length of <math>AB</math>. draw((-2,-2*sqrt(3))--(-2,2*sqrt(3)));

2 KB (412 words) - 18:23, 1 January 2024
1983 AIME Problems/Problem 13
...3, 6,9\}</math> is <math>9-6+3-2+1=5</math> and for <math>\{5\}</math> it is simply <math>5</math>. Find the sum of all such alternating sums for <math> Let <math>S</math> be a non-[[empty set | empty]] [[subset]] of <math>\{1,2,3,4,5,6\}</math>.

5 KB (894 words) - 22:02, 5 April 2024
1983 AIME Problems/Problem 14
...units apart. At <math>P</math>, one of the points of intersection, a line is drawn in such a way that the chords <math>QP</math> and <math>PR</math> hav <asy>size(160); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=in

13 KB (2,149 words) - 18:44, 5 February 2024
1983 AIME Problems/Problem 15
...is expressed as a fraction <math>\frac{m}{n}</math> in lowest terms, what is the product <math>mn</math>? add(pathticks(A--F,1,0.5,0,3.5));

19 KB (3,221 words) - 01:05, 7 February 2023
1984 AIME Problems/Problem 1
..._{98}</math> if <math>a_1</math>, <math>a_2</math>, <math>a_3\ldots</math> is an [[arithmetic progression]] with common difference 1, and <math>a_1+a_2+a One approach to this problem is to apply the formula for the sum of an [[arithmetic series]] in order to fi

4 KB (576 words) - 21:03, 23 December 2023
1984 AIME Problems/Problem 2
...[multiple]] of <math>15</math> such that every [[digit]] of <math>n</math> is either <math>8</math> or <math>0</math>. Compute <math>\frac{n}{15}</math>. Any multiple of 15 is a multiple of 5 and a multiple of 3.

1 KB (187 words) - 20:05, 29 May 2021
1984 AIME Problems/Problem 3
...smaller [[triangle]]s <math>t_{1}</math>, <math>t_{2}</math>, and <math>t_{3}</math> in the figure, have [[area]]s <math>4</math>, <math>9</math>, and < D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12);

4 KB (726 words) - 13:39, 13 August 2023
1984 AIME Problems/Problem 4
...moved, the average of the remaining numbers drops to <math>55</math>. What is the largest number that can appear in <math>S</math>? ...math> numbers other than the <math>68,</math> and the sum of these numbers is <math>s.</math>

2 KB (319 words) - 03:38, 16 January 2023
1984 AIME Problems/Problem 5
...b^3 = 2^{15}</math> and that <math>\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 2^{21}</math>. If we multiply the two equations together, we get that < ...2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2</math>. This means that <math>\frac{\ln ab}{\ln 2} = 9</math>.

5 KB (782 words) - 14:49, 1 August 2023
1984 AIME Problems/Problem 6
...l [[area]] of the parts of the three circles to the other side of it. What is the [[absolute value]] of the [[slope]] of this line? The line passes through the center of the bottom circle; hence it is the circle's [[diameter]] and splits the circle into two equal areas. For t

6 KB (1,022 words) - 19:29, 22 January 2024
1984 AIME Problems/Problem 7
The [[function]] f is defined on the [[set]] of [[integer]]s and satisfies <math>f(n)=\begin{case n-3&\mbox{if}\ n\ge 1000\\

4 KB (617 words) - 18:01, 9 March 2022
1984 AIME Problems/Problem 8
The equation <math>z^6+z^3+1=0</math> has complex roots with argument <math>\theta</math> between <mat ...ithin the desired range that satisfies our original equation <math>x^6 + x^3 + 1 = 0</math>.

3 KB (430 words) - 19:05, 7 February 2023
1984 AIME Problems/Problem 9
...rc</math> angle. Find the [[volume]] of the tetrahedron in <math>\mbox{cm}^3</math>. triple A=(0,0,0),B=(3,0,0),C=(1.8,10,0),D=(1.5,4,4),Da=(D.x,D.y,0),Db=(D.x,0,0);

6 KB (947 words) - 20:44, 26 November 2021
1984 AIME Problems/Problem 10
..., where <math>c</math> is the number of correct answers and <math>w</math> is the number of wrong answers. (Students are not penalized for problems left ...ightarrow 4c-w=89</math> so <math>w\equiv 3\pmod{4}</math>. But if <math>w=3</math>, then <math>c=23</math>, which was the result given; otherwise <math

7 KB (1,163 words) - 23:53, 28 March 2022
1984 AIME Problems/Problem 11
...not change the probability of the birch trees being near each other. That is, in the end, you multiply the numerator by the number of ways to arrange th ...5} = 792</math> total ways to arrange the twelve trees, so the probability is <math>\frac{56}{792} = \frac{7}{99}</math>.

7 KB (1,115 words) - 00:52, 7 September 2023
1984 AIME Problems/Problem 12
...math>x</math>. If <math>x=0</math> is a root for <math>f(x)=0</math>, what is the least number of roots <math>f(x)=0</math> must have in the interval <ma Since <math>0</math> is a root, all multiples of <math>10</math> are roots, and anything congruent

3 KB (588 words) - 14:37, 22 July 2020
1984 AIME Problems/Problem 14
What is the largest even integer that cannot be written as the sum of two odd compo ...+ 6n</math> for nonnegative <math>n</math> are odd composites. We now have 3 cases:

8 KB (1,346 words) - 01:16, 9 January 2024
1984 AIME Problems/Problem 15
...\frac{w^2}{6^2-7^2}=1 </math> <math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></div> Rewrite the system of equations as <cmath>\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.</cmath>

6 KB (1,050 words) - 18:07, 16 January 2024
1984 AIME Problems/Problem 13
Find the value of <math>10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).</math> ...= \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}</math>. Let <math>a = \cot^{-1}(3)</math>, <math>b=\cot^{-1}(7)</math>, <math>c=\cot^{-1}(13)</math>, and <ma

3 KB (473 words) - 12:06, 18 December 2018
1995 IMO Problems/Problem 2
...h> \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}. </cmath> ...c^3(a+b)} &= \frac{x^3}{xyz(1/y+1/z)} + \frac{y^3}{xyz(1/z+1/x)} + \frac{z^3}{xyz(1/x+1/z)} \\

6 KB (1,122 words) - 12:23, 6 January 2022
1985 AIME Problems/Problem 15
...o fold into a [[polyhedron]]. What is the [[volume]] (in <math>\mathrm{cm}^3</math>) of this polyhedron? ...e]], so the volume is <math>\frac12 \cdot 12^3 = 864</math>, so our answer is <math>\boxed{864}</math>.

2 KB (245 words) - 22:44, 4 March 2024
1985 AIME Problems/Problem 14
...us we must have <math>n > 10</math>, so <math>n = 15</math> and the answer is <math>15 + 10 = \boxed{25}</math>. ...e weakest <math>10</math> who gained <math>45</math> points vs them, which is a contradiction since it must be larger. Thus, <math>n=\boxed{25}</math>.

5 KB (772 words) - 22:14, 18 June 2020
1985 AIME Problems/Problem 13
...>\ldots</math> are of the form <math>a_n=100+n^2</math>, where <math>n=1,2,3,\ldots</math> For each <math>n</math>, let <math>d_n</math> be the greatest ...r | divides]] <math>100+n^2</math>, it must divide <math>2n+1</math> if it is going to divide the entire [[expression]] <math>100+n^2+2n+1</math>.

4 KB (671 words) - 20:04, 6 March 2024
1985 AIME Problems/Problem 12
...te end. Let <math>p = \frac{n}{729}</math> be the probability that the bug is at vertex <math>A</math> when it has crawled exactly <math>7</math> meters. ...gers <math>k,</math> let <math>P(k)</math> be the probability that the bug is at vertex <math>A</math> when it has crawled exactly <math>k</math> meters.

17 KB (2,837 words) - 13:34, 4 April 2024
1985 AIME Problems/Problem 11
...>-plane and is [[tangent line | tangent]] to the <math>x</math>-axis. What is the length of its [[major axis]]? ...ath>F_2 Y \leq F’_2 Y</math> with equality if and only if <math>Y</math> is on the <math>x</math>-axis. Now, we have

5 KB (932 words) - 17:00, 1 September 2020
1985 AIME Problems/Problem 10
where <math>x</math> is a [[real number]], and <math>\lfloor z \rfloor</math> denotes the greatest ...or \le 3</math>. But according to this the maximum we can get is <math>1+2+3 = 6</math>, so we only need to try the first 6 numbers.

12 KB (1,859 words) - 18:16, 28 March 2022
1985 AIME Problems/Problem 9
...[[rational number]], is expressed as a [[fraction]] in lowest terms, what is the sum of its numerator and denominator? pair O = (0,0), A = r*expi(pi/3);

5 KB (763 words) - 16:20, 28 September 2019
1985 AIME Problems/Problem 8
...difference, is as small as possible. For this minimum <math>M</math>, what is <math>100M</math>? Then <math>M</math> is the greatest of the <math>7</math> absolute values. So basically you are as

2 KB (377 words) - 02:17, 16 February 2021
1985 AIME Problems/Problem 7
.../math> are [[positive integer]]s such that <math>a^5 = b^4</math>, <math>c^3 = d^2</math>, and <math>c - a = 19</math>. Determine <math>d - b</math>. ...th>s - t^2 = 1</math>. Then <math>s = 10, t = 3</math> and so <math>d = s^3 = 1000</math>, <math>b = t^5 = 243</math> and <math>d-b=\boxed{757}</math>.

1 KB (222 words) - 11:04, 4 November 2022
1985 AIME Problems/Problem 6
As shown in the figure, [[triangle]] <math>ABC</math> is divided into six smaller triangles by [[line]]s drawn from the [[vertex | v ...> share the same altitude from <math>C</math>, so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are

5 KB (789 words) - 03:09, 23 January 2023
1985 AIME Problems/Problem 5
...e sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492? ...-b) = a</math> and <math>a_8 = a - (a - b) = b</math>. Since the sequence is recursively defined by the first 2 terms, after this point it must continue

2 KB (410 words) - 13:37, 1 May 2022
1985 AIME Problems/Problem 4
.... Find the value of <math>n</math> if the the [[area]] of the small square is exactly <math>\frac1{1985}</math>. ...frac{1}{\sqrt{2n^2 - 2n + 1}}</math>. But the height of the parallelogram is the side of the little square, so <math>2n^2 - 2n + 1 = 1985</math>. Solvi

3 KB (484 words) - 21:40, 2 March 2020
1985 AIME Problems/Problem 3
...nd <math>c</math> are [[positive integer]]s which satisfy <math>c=(a + bi)^3 - 107i</math>, where <math>i^2 = -1</math>. ...6</math> (since we know <math>a</math> is positive). Thus <math>c = 6^3 - 3\cdot 6 = \boxed{198}</math>.

1 KB (205 words) - 18:58, 10 March 2024
1986 AIME Problems/Problem 15
...]s through <math>A</math> and <math>B</math> lie along the lines <math>y=x+3</math> and <math>y=2x+4</math> respectively, find the area of triangle <mat ...}{1 + \tan \theta_1 \tan \theta_2} = \frac{2-1}{1 + 2 \cdot 1 } = \frac{1}{3}. </cmath>

11 KB (1,722 words) - 09:49, 13 September 2023
1986 AIME Problems/Problem 14
...find the shortest distance between <math>AC</math> and that corner, which is <math>\frac {wl}{\sqrt {w^2 + l^2}}</math>. ...ctively. (This would give us the guess that the sides are of the ratio 1:2:3, but let's provide the complete solution.)

2 KB (346 words) - 13:13, 22 July 2020
1986 AIME Problems/Problem 13
...ord of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by <tt>TH</tt>, <t ...mine what happens to the last coin toss. Adding <tt>HH</tt> or <tt>TT</tt> is simply an [[identity]] for the last coin toss, so we will ignore them for n

3 KB (445 words) - 19:44, 8 January 2023
1986 AIME Problems/Problem 12
...Suppose no two disjoint subsets of <math>S</math> have the same sum. What is the largest sum a set <math>S</math> with these properties can have? ...at least <math>\dbinom{6}{0} + \dbinom{6}{1} + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57</math> of its subsets have at most four elements (the n

2 KB (364 words) - 19:41, 1 September 2020
1986 AIME Problems/Problem 11
The [[polynomial]] <math>1-x+x^2-x^3+\cdots+x^{16}-x^{17}</math> may be written in the form <math>a_0+a_1y+a_2y^ ...By the [[Binomial Theorem]], this is <math>(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}</math>.

6 KB (872 words) - 16:51, 9 June 2023
1986 AIME Problems/Problem 10
...h>m \equiv 358</math> mod <math>666</math>. We see that there are no other 3-digit integers that are <math>358</math> mod <math>666</math>, so <math>m = === Solution 3 ===

3 KB (565 words) - 16:51, 1 October 2023
1986 AIME Problems/Problem 9
...=450</math>, and <math>AC=510</math>. An interior [[point]] <math>P</math> is then drawn, and [[segment]]s are drawn through <math>P</math> [[parallel]] ...}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}=\frac{10}{\frac{10}{306}} = \boxed{306}</math></center>

11 KB (1,850 words) - 18:07, 11 October 2023
1986 AIME Problems/Problem 8
...l [[divisor]]s of a number excluding itself) of <math>1000000</math>. What is the integer nearest to <math>S</math>? ...math>48</math> are proper. The sum of multiple logarithms of the same base is equal to the logarithm of the products of the numbers.

3 KB (487 words) - 20:52, 16 September 2020
1986 AIME Problems/Problem 7
...nteger]]s which are [[exponent|powers]] of 3 or sums of distinct powers of 3. Find the <math>100^{\mbox{th}}</math> term of this sequence. ...must change it back to base 10 for the answer, which is <math>3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed {981}</math>.

5 KB (866 words) - 00:00, 22 December 2022
1986 AIME Problems/Problem 5
...largest [[positive integer]] <math>n</math> for which <math>n^3+100</math> is [[divisible]] by <math>n+10</math>? ...h>; we can double-check manually and we find that indeed <math>900\mid 890^3+100</math>.

2 KB (338 words) - 19:56, 15 October 2023
1986 AIME Problems/Problem 4
...e fifth given equation gives <math>x_5 = 65</math>, so our answer is <math>3\cdot17 + 2\cdot65 = \boxed{181}</math>. <cmath>3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}</cmath>

1 KB (212 words) - 16:25, 17 November 2019
1986 AIME Problems/Problem 3
If <math>\tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>? Since <math>\cot</math> is the reciprocal function of <math>\tan</math>:

3 KB (545 words) - 23:44, 12 October 2023
1986 AIME Problems/Problem 2
== Solution 3 (Geometry) == ...rt5,2\sqrt6,</math> and <math>2\sqrt7,</math> by Heron's Formula, the area is the square root of the original expression.

3 KB (460 words) - 00:44, 5 February 2022
1987 AIME Problems/Problem 15
...{440}{441}T_2</math>. Additionally, the area of triangle <math>ABC</math> is equal to both <math>T_1 + T_2 + 441</math> and <math>T_3 + T_4 + T_5 + 440. ...and <math>ABC</math> is <math>441</math>, and the ratio between the sides is <math>\sqrt {441} = 21</math>. As a result, <math>AB = 21\sqrt {440} = \sqr

5 KB (838 words) - 18:05, 19 February 2022
1987 AIME Problems/Problem 14
...+ 2b^2 - 2ab\right)\left(a^2 + 2b^2 + 2ab\right).</math> Each of the terms is in the form of <math>x^4 + 324.</math> Using Sophie Germain, we get that x^4 + 324 &= x^4 + 4\cdot 3^4 \\

7 KB (965 words) - 10:42, 12 April 2024
1987 AIME Problems/Problem 12
...</math>, where <math>n</math> is a [[positive integer]] and <math>r</math> is a [[positive]] [[real number]] less than <math>1/1000</math>. Find <math>n< ...<math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>. H

4 KB (673 words) - 19:48, 28 December 2023
1987 AIME Problems/Problem 11
...the largest possible value of <math>k</math> for which <math>3^{11}</math> is expressible as the sum of <math>k</math> consecutive [[positive integer]]s. <math>3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)</ma

3 KB (418 words) - 18:30, 20 January 2024
1987 AIME Problems/Problem 10
...teps are visible on the escalator at a given time? (Assume that this value is constant.) ...er the time it took Bob to climb, the [[ratio]] of their distances covered is the same as the ratio of their speeds, so <math>\frac{e}{b} = \frac{x - 75}

7 KB (1,187 words) - 16:21, 27 January 2024
1987 AIME Problems/Problem 8
What is the largest positive integer <math>n</math> for which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + .../math>. Thus, <math>48n < 56k < 49n</math>. <math>k</math> is unique if it is within a maximum [[range]] of <math>112</math>, so <math>n = 112</math>.

2 KB (393 words) - 16:59, 16 December 2020
1987 AIME Problems/Problem 6
...re <math>XY = YB + BC + CZ = ZW = WD + DA + AX</math>, and <math>PQ</math> is [[parallel]] to <math>AB</math>. Find the [[length]] of <math>AB</math> (i ...umber is also equal to one quarter the area of the entire rectangle, which is <math>\frac{19\cdot AB}{4}</math>, so we have <math>AB = XY + 87</math>.

3 KB (530 words) - 07:46, 1 June 2018
1987 AIME Problems/Problem 5
If we move the <math>x^2</math> term to the left side, it is factorable with [[SFFT|Simon's Favorite Factoring Trick]]: ...ath>y^2 - 10 = 39</math>, so <math>y^2 = 49</math>. Thus, <math>3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}</math>.

1 KB (160 words) - 04:44, 21 January 2023
1987 AIME Problems/Problem 4
...h is when <math>x = 60</math> and <math>y = \pm 15</math>. Since the graph is [[symmetry|symmetric]] about the y-axis, we just need [[casework]] upon <ma *<math>x - 60 > 0</math>. Then <math>y = -\frac{3}{4}x+60</math>.

2 KB (371 words) - 17:25, 13 February 2024
1987 AIME Problems/Problem 3
...nice'' if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers? ...of the distinct proper divisors of <math>n</math>. A number <math>n</math> is ''nice'' in one of two instances:

3 KB (511 words) - 09:29, 9 January 2023
1987 AIME Problems/Problem 2
What is the largest possible [[distance]] between two [[point]]s, one on the [[sphe {{AIME box|year=1987|num-b=1|num-a=3}}

697 bytes (99 words) - 18:46, 14 February 2014
1987 AIME Problems/Problem 1
An [[ordered pair]] <math>(m,n)</math> of [[non-negative]] [[integer]]s is called "simple" if the [[addition]] <math>m+n</math> in base <math>10</math ...pair]]s will be <math>(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}</math>.

1 KB (191 words) - 14:42, 17 September 2016
1988 AIME Problems/Problem 15
...typed during the day, and the boss delivers them in the order <math>1, 2, 3, 4, 5, 6, 7, 8, 9</math>. ...ch typing orders are possible? (That there are no letters left to be typed is one of the possibilities.)

7 KB (1,186 words) - 10:16, 4 June 2023
1988 AIME Problems/Problem 14
...icular]] to <math>y=2x</math>, so the slope of <math>\overline{PP'}</math> is <math>\frac{-1}{2}</math>. Thus <math>\frac{y' - y}{x' - x} = \frac{-1}{2} ...math>, which is unchanged by the reflection, into the expression. But this is not necessary. We see that <math>b=-7</math>, <math>c=-12</math>, so <math>

4 KB (700 words) - 17:21, 3 May 2021
1988 AIME Problems/Problem 13
...</math> and <math>b</math> are integers such that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1</math>. ...Fibonacci sequence. Carrying out this pattern, we find that the remainder is <cmath>(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0.</cmath> Since the

10 KB (1,585 words) - 03:58, 1 May 2023
1988 AIME Problems/Problem 12
...the product <math>abc</math> if <math>a + b + c = 43</math> and <math>d = 3</math>. ...]] <math>\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = 1</math> is a form of [[Ceva's Theorem]].

4 KB (727 words) - 23:37, 7 March 2024
1988 AIME Problems/Problem 11
...h> be [[complex number]]s. A line <math>L</math> in the [[complex plane]] is called a mean [[line]] for the [[point]]s <math>w_1, w_2, \dots, w_n</math> ...14 + 43i</math>, there is a unique mean line with <math>y</math>-intercept 3. Find the [[slope]] of this mean line.

2 KB (422 words) - 00:22, 6 September 2020
1988 AIME Problems/Problem 10
The number of segments joining the vertices of the polyhedron is <math>{48\choose2} = 1128</math>. We must now subtract out those segments t ...t each of its endpoints, the number of edges <math>E</math> is <math>\frac{3}{2}V = 72</math>.

5 KB (811 words) - 19:10, 25 January 2021
1988 AIME Problems/Problem 9
...<math>20k + 8 \equiv 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework: ...le value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution.

6 KB (893 words) - 08:15, 2 February 2023
1988 AIME Problems/Problem 8
& = \frac{91}{3}\cdot f(4,6) \\ ==Solution 3 (Number Theory)==

4 KB (538 words) - 13:24, 12 October 2021
1988 AIME Problems/Problem 7
.../math> divides <math>BC</math> into [[segment]]s of length 3 and 17. What is the area of triangle <math>ABC</math>? ...So, <math>\tan \alpha = \frac {17}{h}</math> and <math>\tan \beta = \frac {3}{h}</math>. Using the tangent addition formula <math>\tan (\alpha + \beta)

1 KB (178 words) - 23:25, 20 November 2023
1988 AIME Problems/Problem 6
It is possible to place positive integers into the vacant twenty-one squares of t ...ce is <math>103 - 2b</math>, so that square also has a value of <math>2b + 3(103 - 2b) = 309 - 4b</math>. Equating, we get <math>148 - 3a = 309 - 4b \Lo

5 KB (878 words) - 23:06, 20 November 2023
1988 AIME Problems/Problem 4
What is the smallest possible value of <math>n</math>? ...,1</math> and find that the <math>LHS</math> is <math>3</math> and the RHS is <math>1.</math> Similarly testing <math>1,-1,-1,1</math> yields <math>4</ma

2 KB (394 words) - 10:21, 27 January 2024
1988 AIME Problems/Problem 3
2^{3 \log_2(\log_8x)} &= \log_2x\\ (\log_8x)^3 &= \log_2x\\

3 KB (481 words) - 21:52, 18 November 2020
1988 AIME Problems/Problem 2
Note that this revolves between the two numbers. Since <math>1988</math> is even, we thus have <math>f_{1988}(11) = f_{4}(11) = \boxed{169}</math>. {{AIME box|year=1988|num-b=1|num-a=3}}

696 bytes (103 words) - 19:16, 27 February 2018
1988 AIME Problems/Problem 1
...order -- the correct five buttons. The sample shown below has <math>\{1,2,3,6,9\}</math> as its [[combination]]. Suppose that these locks are redesigne ...ath>9</math>, the number of ways to choose a set of <math>x</math> buttons is <math>\sum^{9}_{k=1}{10 \choose k}</math>.

1 KB (181 words) - 18:23, 26 August 2019
1989 AIME Problems/Problem 15
...ven that <math>AP=6</math>, <math>BP=9</math>, <math>PD=6</math>, <math>PE=3</math>, and <math>CF=20</math>, find the area of <math>\triangle ABC</math> ...<math>RST</math>. We'll make use of the following fact: if <math>P</math> is a point in the interior of triangle <math>XYZ</math>, and line <math>XP</ma

13 KB (2,091 words) - 00:20, 26 October 2023
1989 AIME Problems/Problem 14
...+i</math> using the integers <math>0,1,2,\ldots,n^2</math> as digits. That is, the equation is true for a unique choice of non-negative integer <math>m</math> and digits

2 KB (408 words) - 17:28, 16 September 2023
1989 AIME Problems/Problem 13
...members of <math>S</math> differ by <math>4</math> or <math>7</math>. What is the largest number of [[element]]s <math>S</math> can have? ...we can take at most one from each of the pairs: <math>[2,9]</math>, <math>[3,7]</math>, <math>[4,11]</math>, <math>[6,10]</math>. Now, <math>1989 = 180\

2 KB (274 words) - 04:07, 17 December 2023
1989 AIME Problems/Problem 11
...r D\rfloor</math>? (For real <math>x</math>, <math>\lfloor x\rfloor</math> is the [[floor function|greatest integer]] less than or equal to <math>x</math ...> (in other words, <math>x \in [1,1000]</math>). Indeed, <math>D(x)</math> is symmetric about <math>x = 500.5</math>; consider replacing all of numbers <

5 KB (851 words) - 18:01, 28 December 2022
1989 AIME Problems/Problem 10
pair A = (0,0), B = (3, 0), C = (1, 4); ...wo angles in the triangle. So, the cotangent of any angle in the triangle is directly proportional to the sum of the squares of the two adjacent sides,

8 KB (1,401 words) - 21:41, 20 January 2024
1989 AIME Problems/Problem 9
Taking the given equation modulo <math>2,3,</math> and <math>5,</math> respectively, we have n^5&\equiv0\pmod{3}, \\

6 KB (874 words) - 15:50, 20 January 2024
1989 AIME Problems/Problem 8
...+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7</cmath> for some <math>k\in\{1,2,3\}.</math> f(3)&=9a+3b+c&&=123,

8 KB (1,146 words) - 04:15, 20 November 2023
1989 AIME Problems/Problem 6
...t, frozen lake. The [[distance]] between <math>A</math> and <math>B</math> is <math>100</math> meters. Allie leaves <math>A</math> and skates at a [[spee pair A=(0,0),B=(10,0),C=6*expi(pi/3);

5 KB (864 words) - 19:55, 2 July 2023
1989 AIME Problems/Problem 5
...st terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>. ...<math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>.

2 KB (258 words) - 00:07, 25 June 2023
1989 AIME Problems/Problem 4
...[[perfect square]] and <math>a+b+c+d+e</math> is a [[perfect cube]], what is the smallest possible value of <math>c</math>? ...dot y^3</math>, <math>3^3</math> must be a factor of <math>c</math>. <math>3^35^2 = \boxed{675}</math>, which works as the solution.

3 KB (552 words) - 12:41, 3 March 2024
1989 AIME Problems/Problem 3
Suppose <math>n</math> is a [[positive integer]] and <math>d</math> is a single [[digit]] in [[base 10]]. Find <math>n</math> if ...must be [[divisible]] by 37, and the only digit for which this is possible is <math>d = 9</math>. Thus <math>4d + 1 = 37</math> and <math>n = \boxed{750

3 KB (499 words) - 22:17, 29 March 2024
1989 AIME Problems/Problem 2
...mber and <math>{10 \choose 2} = 45</math> have 2 members. Thus the answer is <math>1024 - 1 - 10 - 45 = \boxed{968}</math>. ...{n \choose 0}+{n \choose 1} + {n \choose 2} + \dots + {n \choose n}</math> is equivalent to <math>2^n</math>

911 bytes (135 words) - 08:30, 27 October 2018
1989 AIME Problems/Problem 1
...+1} = \sqrt{(870)(868) +1} = \sqrt{(868 +1)^2} = \boxed{869}</math>. This is because we have that <math>a=868</math> as per the equation <math>(a+1)^2 = == Solution 3 (Symmetry with Generalization) ==

4 KB (523 words) - 00:12, 8 October 2021
1990 AIME Problems/Problem 14
...rline{AP}</math> and <math>\overline{BP}</math> are joined, and the figure is then creased along segments <math>\overline{CP}</math> and <math>\overline{ label("$13\sqrt{3}$", A--D, S);

7 KB (1,086 words) - 08:16, 29 July 2023
1990 AIME Problems/Problem 13
...9^{4000}_{}</math> has 3817 [[digit]]s and that its first (leftmost) digit is 9, how many [[element]]s of <math>T_{}^{}</math> have 9 as their leftmost d If there is exactly 1 n-digit power of 9, then such a number <math>m</math> cannot begi

5 KB (762 words) - 01:18, 10 February 2023
1990 AIME Problems/Problem 12
...]s of the 12-gon can be written in the form <math>a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},</math> where <math>a^{}_{}</math>, <math>b^{}_{}</math>, <ma *The length of each of the 12 sides is <math>2 \cdot 12\sin 15</math>. <math>24\sin 15 = 24\sin (45 - 30) = 24\fra

6 KB (906 words) - 13:23, 5 September 2021
1990 AIME Problems/Problem 11
...h>, which decreases as <math>a</math> increases. Thus, <math>n = 23</math> is the greatest possible value to satisfy the given conditions. ...annot be less than or equal to <math>n</math>, else the product of <math>n-3</math> consecutive positive integers will be less than <math>n!</math>.

3 KB (519 words) - 09:28, 28 June 2022
1990 AIME Problems/Problem 10
...unity]]. The set <math>C = \{zw : z \in A ~ \mbox{and} ~ w \in B\}</math> is also a set of complex roots of unity. How many distinct elements are in <m ...^8, n^{16}, \ldots n^{144}\}</math> and of set <math>B</math> as <math>\{n^3, n^6, \ldots n^{144}\}</math>. <math>n^x</math> can yield at most <math>144

3 KB (564 words) - 04:47, 4 August 2023
1990 AIME Problems/Problem 9
A [[fair]] coin is to be tossed <math>10_{}^{}</math> times. Let <math>\frac{i}{j}^{}_{}</math ...obability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9 + 64 = \boxed{073}</math>.

3 KB (425 words) - 19:31, 30 July 2021
1990 AIME Problems/Problem 8
...ng columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules: 1) The marksman first chooses a column from which a target is to be broken.

3 KB (491 words) - 04:24, 4 November 2022
1990 AIME Problems/Problem 7
...s lengths of side <math>15,\ 20,\ 25</math>, indicating that it is a <math>3-4-5</math> [[right triangle]]. At this point, we just need to find another ...2},\ \frac{15}{2}</math>. It follows that <math>\frac{QP'}{RP'} = \frac{5}{3}</math>, and so <math>P' = \left(\frac{5x_R + 3x_Q}{8},\frac{5y_R + 3y_Q}{8

8 KB (1,319 words) - 11:34, 22 November 2023
1990 AIME Problems/Problem 6
...them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, s ...ember is proportional to the percentage of tagged fish in May, <math>\frac{3}{42} = \frac{60}{x} \Longrightarrow \boxed{x = 840}</math>.

2 KB (325 words) - 13:16, 26 June 2022
1990 AIME Problems/Problem 5
Let <math>n^{}_{}</math> be the smallest positive [[integer]] that is a multiple of <math>75_{}^{}</math> and has exactly <math>75_{}^{}</math> p ...fore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}</math>.

1 KB (175 words) - 03:45, 21 January 2023
1990 AIME Problems/Problem 4
...- 29 \Longleftrightarrow 0 = (x - 13)(x + 3)</math>. The positive [[root]] is <math>\boxed{013}</math>. {{AIME box|year=1990|num-b=3|num-a=5}}

1 KB (156 words) - 07:35, 4 November 2022
1990 AIME Problems/Problem 3
...q s\geq 3)</math> such that each [[interior angle]] of <math>P_1^{}</math> is <math>\frac{59}{58}</math> as large as each interior angle of <math>P_2^{}< The formula for the interior angle of a regular sided [[polygon]] is <math>\frac{(n-2)180}{n}</math>.

3 KB (516 words) - 19:18, 16 April 2024
1990 AIME Problems/Problem 2
Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>. ...>52-6\sqrt{43}</math>, the only feasible possibility is <math>(\sqrt{43} - 3)^2</math>.

5 KB (765 words) - 23:00, 26 August 2023
1990 AIME Problems/Problem 1
The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are ne ...r is the biggest non-square and non-cube less than <math>529</math>, which is <math>\boxed{528}</math>.

2 KB (283 words) - 23:11, 25 June 2023
2006 AMC 10B Problems/Problem 2
...</math>, define <math> x \spadesuit y = (x+y)(x-y) </math>. What is <math> 3 \spadesuit (4 \spadesuit 5) </math>? ...4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) = 3 \spadesuit (-9) = (3+(-9))(3-(-9)) = \boxed{\textbf{(A)}-72}</math>

633 bytes (85 words) - 10:33, 19 December 2021
2006 AMC 10B Problems/Problem 4
...side the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area? <math> \textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 6\qquad \textbf{(D) } 8\qquad \textbf{(E) } 9 </math>

1 KB (172 words) - 10:47, 19 December 2021
2006 AMC 10B Problems/Problem 5
...of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square? ...tely enclosed in a square with a side length of 5. The area of this square is <math>5^2 = 25</math>.

1 KB (242 words) - 18:35, 15 August 2023
2006 AMC 10B Problems/Problem 7
Which of the following is equivalent to <math> \sqrt{\frac{x}{1-\frac{x-1}{x}}} </math> when <math> x ...math>. As no other option choice fits, <math>\boxed{\textbf{(A)}-x}</math> is the correct solution.

1 KB (179 words) - 10:33, 19 August 2022
2006 AMC 10B Problems/Problem 11
What is the tens digit in the sum <math> 7!+8!+9!+...+2006!</math> <math> \textbf{(A) } 1\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 6\qquad \textbf{(E) } 9 </math>

1 KB (170 words) - 14:00, 26 January 2022
2006 AMC 10B Problems/Problem 12
...y=\frac{1}{4}x+b </math> intersect at the point <math> (1,2) </math>. What is <math> a+b </math>? <math> \textbf{(A) } 0\qquad \textbf{(B) } \frac{3}{4}\qquad \textbf{(C) } 1\qquad \textbf{(D) } 2\qquad \textbf{(E) } \frac{9

1 KB (220 words) - 20:07, 27 November 2023
2006 AMC 10B Problems/Problem 14
...ac1a </math> are the roots of the equation <math> x^2-px+q=0 </math>. What is <math>q</math>? ...of the form <math> x^2 + bx + c = 0 </math>, the product of the [[root]]s is <math>c</math> ([[Vieta's Formulas]]).

2 KB (264 words) - 21:10, 19 September 2023
1991 AIME Problems/Problem 15
...unique positive integer <math>n^{}_{}</math> for which <math>S_n^{}</math> is also an integer. Find this <math>n^{}_{}</math>. ...he value of <math>S_n</math>. The minimum value of <math>S_n</math>, then, is the length of the straight line connecting the bottom vertex of the first r

4 KB (658 words) - 16:58, 10 November 2023
1991 AIME Problems/Problem 14
A [[hexagon]] is inscribed in a [[circle]]. Five of the sides have length <math>81</math> an ...18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18));

2 KB (284 words) - 03:56, 23 January 2023
1991 AIME Problems/Problem 13
...blue. What is the largest possible number of red socks in the drawer that is consistent with this data? ...cks are drawn randomly, without replacement, both are red or both are blue is given by

7 KB (1,328 words) - 20:24, 5 February 2024
2006 AMC 10B Problems/Problem 15
...D</math> is <math>24</math> and <math> \angle BAD = 60^\circ </math>. What is the area of rhombus <math>BFDE</math>? ...n, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3));

3 KB (445 words) - 22:01, 20 August 2022
1991 AIME Problems/Problem 12
...ath>\overline{CD}</math>, and <math>\overline{DA}</math>, respectively. It is given that <math>PB^{}_{}=15</math>, <math>BQ^{}_{}=20</math>, <math>PR^{}_ <center><asy>defaultpen(fontsize(12)+linewidth(1.3)); pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P=(23.4,28.8), Q

8 KB (1,270 words) - 23:36, 27 August 2023
1991 AIME Problems/Problem 11
...re <math>a,b,c^{}_{}</math> are positive integers and <math>c^{}_{}</math> is not divisible by the square of any prime. Find <math>a+b+c^{}_{}</math>. _Diagram by 1-1 is 3_

4 KB (740 words) - 19:33, 28 December 2022
1991 AIME Problems/Problem 10
...s written as a [[fraction]] in [[irreducible fraction|lowest terms]], what is its [[numerator]]? aab & 4 & 2 & 3 \\

5 KB (813 words) - 06:10, 25 February 2024
2006 AMC 10B Problems/Problem 17
...randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are th ...} \frac{1}{6}\qquad \textbf{(C) } \frac{1}{5}\qquad \textbf{(D) } \frac{1}{3}\qquad \textbf{(E) } \frac{1}{2} </math>

1 KB (211 words) - 04:32, 4 November 2022
1991 AIME Problems/Problem 9
...and that <math>\csc x+\cot x=\frac mn,</math> where <math>\frac mn</math> is in lowest terms. Find <math>m+n^{}_{}.</math> ...problem is much easier computed if we consider what <math>\sec (x)</math> is, then find the relationship between <math>\sin( x)</math> and <math>cos (x)

10 KB (1,590 words) - 14:04, 20 January 2023
2006 AMC 10B Problems/Problem 18
...}}{a_{n-2}} </math> for each positive integer <math> n \ge 3 </math>. What is <math> a_{2006} </math>? ...\mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3 </math>

1 KB (158 words) - 01:33, 29 May 2023
1991 AIME Problems/Problem 8
Since <math>-a</math> is an integer, we need <math>\sqrt{a^2-24a}</math> to be an integer (let this Which implies that <math>b^2 + 144</math> is a [[perfect square]] also (let this be <math>c^2</math>). Then

2 KB (310 words) - 11:25, 13 June 2023
1991 AIME Problems/Problem 5
...will be between <math>0 < \frac{a}{b} < 1</math>. Therefore, the solution is <math>\frac{2^8}{2} = \boxed{128}</math>.

919 bytes (141 words) - 20:00, 4 July 2022
1991 AIME Problems/Problem 4
The [[range]] of the [[sine]] function is <math>-1 \le y \le 1</math>. It is [[periodic function|periodic]] (in this problem) with a period of <math>\fr ...ath>x < 1</math>, we can count <math>4</math> more solutions. The solution is <math>154 + 1 + 4 = \boxed{159}</math>.

2 KB (300 words) - 16:01, 26 November 2019
1991 AIME Problems/Problem 3
...th> for <math>k = 0,1,2,\ldots,1000</math>. For which <math>k_{}^{}</math> is <math>A_k^{}</math> the largest? ...lues of <math>k>165.8</math>, the largest possible value of <math>k</math> is <math>\boxed{166}</math>.

5 KB (865 words) - 12:13, 21 May 2020
1991 AIME Problems/Problem 2
...line {AB}</math> of [[length]] 4 and <math>\overline {CB}</math> of length 3. Divide <math>\overline {AB}</math> into 168 [[congruent]] [[segment]]s wit pair A=(0,0),B=(4,0),C=(4,3),D=(0,3);

4 KB (595 words) - 12:51, 17 June 2021
1991 AIME Problems/Problem 1
...nd so there are no integral solutions for <math>(x,y)</math>. The solution is <math>5^2 + 11^2 = \boxed{146}</math>. ...ctor of <math>11</math> is <math>(5,11)</math>, and checking shows that it is correct.

4 KB (628 words) - 22:05, 7 June 2021
1992 AIME Problems/Problem 15
...ne a positive integer <math>n^{}_{}</math> to be a factorial tail if there is some positive integer <math>m^{}_{}</math> such that the decimal representa .../math> is a multiple of <math>5</math>, <math>f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)</math>.

2 KB (358 words) - 01:54, 2 October 2020
1992 AIME Problems/Problem 14
== Solution 3 == A consequence of Ceva's theorem sometimes attributed to Gergonne is that <math>\frac{AO}{OA'}=\frac{AC'}{C'B}+\frac{AB'}{B'C}</math>, and simil

4 KB (667 words) - 01:26, 16 August 2023
1992 AIME Problems/Problem 13
...ath>. So if we can write: <math>b^2=-(a+n)^2+m</math>, then <math>m</math> is the maximum value of <math>b^2</math> (this follows directly from the [[tri Then the area is <math>9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}</math>.

4 KB (703 words) - 02:40, 29 December 2023
1992 AIME Problems/Problem 9
.../math>. A circle with center <math>P^{}_{}</math> on <math>AB^{}_{}</math> is drawn tangent to <math>BC^{}_{}</math> and <math>AD^{}_{}</math>. Given tha ...<math>AP=x=\frac{161}{3}</math>. This gives us a final answer of <math>161+3=\boxed{164}</math>

5 KB (874 words) - 10:27, 22 August 2021
1992 AIME Problems/Problem 8
...1,a_3-a_2,a_4-a_3,\ldots)</math>, whose <math>n^{\mbox{th}}_{}</math> term is <math>a_{n+1}-a_n^{}</math>. Suppose that all of the terms of the sequence ...binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...</math>

5 KB (778 words) - 21:36, 3 December 2022
1992 AIME Problems/Problem 7
...</math> is <math>120^{}_{}</math>, the area of face <math>BCD^{}_{}</math> is <math>80^{}_{}</math>, and <math>BC=10^{}_{}</math>. Find the volume of the ...ot \sin 30^\circ=8</math>. Therefore, the volume is <math>\frac{8\cdot120}{3}=\boxed{320}</math>.

800 bytes (114 words) - 17:40, 14 March 2017
1992 AIME Problems/Problem 6
...irs of consecutive integers in <math>\{1000,1001,1002,\ldots,2000\}</math> is no carrying required when the two integers are added? 0\leq C\leq 8 & 1 & A & B & C+1 & 0\leq A,B,C\leq 4 & 5^3 \\

3 KB (455 words) - 02:03, 10 July 2021
1992 AIME Problems/Problem 4
In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\\vspace{4pt}

3 KB (476 words) - 14:13, 20 April 2024
1992 AIME Problems/Problem 3
...es, winning three and losing one. At the end of the weekend, her win ratio is greater than <math>.503</math>. What's the largest number of matches she co ...ches won, so that <math>\frac{n}{2n}=\frac{1}{2}</math>, and <math>\frac{n+3}{2n+4}>\frac{503}{1000}</math>.

2 KB (251 words) - 08:05, 2 January 2024
1992 AIME Problems/Problem 2
...s [[decimal representation]], there are at least two digits and each digit is less than any digit to its right. How many ascending positive integers are ...ition for 0 is at the leftmost end of the number, i.e. a leading 0), there is exactly one ascending number with those digits.

2 KB (336 words) - 05:18, 4 November 2022
1992 AIME Problems/Problem 1
...ac{49}{30}.</math> Following this pattern, our answer is <math>4(10)+8(1+2+3+\cdots+9)=\boxed{400}.</math> ...us the sum of the smallest <math>8</math> rational numbers satisfying this is <math>\frac12\cdot8\cdot1=4</math>. Now refer to solution 1.

1 KB (190 words) - 20:02, 23 February 2022
2006 AMC 10B Problems/Problem 19
..., respectively. What is the area of the shaded region in the figure, which is bounded by <math>BD</math>, <math>BE</math>, and the minor arc connecting < pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B;

5 KB (873 words) - 15:39, 29 May 2023
2006 AMC 10B Problems/Problem 20
...6,178)</math>, <math>D=(8,y)</math>, for some integer <math>y</math>. What is the area of rectangle <math>ABCD</math>? Therefore the area of rectangle <math>ABCD</math> is <math> 200\sqrt{101}\cdot2\sqrt{101} = 40,400 \Rightarrow E </math>

4 KB (594 words) - 15:45, 30 July 2023
2006 AMC 10B Problems/Problem 21
...math>6</math>, on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the two dice? <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> are <math>2x</ma

3 KB (484 words) - 19:09, 15 October 2023
2006 AMC 10B Problems/Problem 22
...ath>, and <math>N</math> are positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? ...nly possible positive integer pairs <math>(N , 4B+5J)</math> whose product is <math>253</math> are: <math> (1,253) ; (11,23) ; (23,11) ; (253,1) </math>

2 KB (394 words) - 00:51, 25 November 2023
2006 AMC 10B Problems/Problem 23
...te sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral? pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);

5 KB (861 words) - 00:53, 25 November 2023
2006 AMC 10B Problems/Problem 24
...ath> and <math>BC</math> are common external tangents to the circles. What is the area of the [[concave]] [[hexagon]] <math>AOBCPD</math>? fill((-3,7)--(-3,-7)--(-7,-7)--(-7,7)--cycle, white);</asy>

4 KB (558 words) - 14:38, 6 April 2024
2006 AMC 10B Problems/Problem 25
..."and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children? ...the license plate. Since at least one of <math>4</math> or <math>8</math> is contained in <math>S</math>, we have <math>4 | m</math>.

5 KB (878 words) - 14:39, 3 December 2023
1993 AIME Problems/Problem 15
A=(8,0); B=origin; C=(3,4); H=(3,0); draw(A--B--C--cycle); draw(C--H); ...s of the two right triangles, the distance between the two tangency points is simply <math>\frac{n-2}{2n+2}=\frac{n-2}{2(n+1)}</math>.

3 KB (449 words) - 21:39, 21 September 2023
1993 AIME Problems/Problem 14
...a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its cente ...mbinations of positive and negative. Then by symmetry, the other rectangle is also centered at the origin, <math>O</math>.

3 KB (601 words) - 09:25, 19 November 2023
1993 AIME Problems/Problem 13
...again. If <math>t\,</math> is written as a fraction in lowest terms, what is the sum of the numerator and denominator? ...3}.</cmath> Finally, the sum of the numerator and denominator is <math>160+3=\boxed{163}.</math>

8 KB (1,231 words) - 20:06, 26 November 2023
1993 AIME Problems/Problem 12
...th>\overline{P_n L}</math>. Given that <math>P_7 = (14,92)\,</math>, what is <math>k + m\,</math>? ...>(r,s)</math> and we want to find <math>(u,v)</math> so <math>(r,s)</math> is the midpoint of <math>(u,v)</math> and <math>(p,q)</math>, then <math>u=2r-

4 KB (611 words) - 13:59, 15 July 2023
1993 AIME Problems/Problem 11
...st in the first game, and that the probability that he wins the sixth game is <math>m/n\,</math>, where <math>m\,</math> and <math>n\,</math> are relativ .../math>, and the probability of the second person winning is <math>\frac{1}{3}</math>.

7 KB (1,058 words) - 20:57, 22 December 2020
1993 AIME Problems/Problem 10
...h>T</math> triangular faces and <math>P</math> pentagonal faces meet. What is the value of <math>100P+10T+V</math>? ...=\frac{3t+5p}{2}</math>, (the factor <math>2</math> in the [[denominator]] is because we are counting twice each edge, since two adjacent faces share one

4 KB (716 words) - 20:50, 17 April 2022
1993 AIME Problems/Problem 9
...will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as <math>1993</math>? ...}</math>. Therefore, one of <math>1993 - n</math> or <math>1994 + n</math> is odd, and each of them must be a multiple of <math>125</math> or <math>16</m

3 KB (488 words) - 02:06, 22 September 2023
1993 AIME Problems/Problem 8
...distinct subsets of <math>S\,</math> so that the union of the two subsets is <math>S\,</math>? The order of selection does not matter; for example, the ...math> elements of <math>S.</math> So our final answer is then <math>\frac {3^6 - 1}{2} + 1 = \boxed{365}.</math>

9 KB (1,400 words) - 14:09, 12 January 2024
1993 AIME Problems/Problem 7
...the box. If <math>p</math> is written as a fraction in lowest terms, what is the sum of the numerator and denominator? There is a total of <math>P(1000,6)</math> possible ordered <math>6</math>-tuples <m

5 KB (772 words) - 09:04, 7 January 2022
1993 AIME Problems/Problem 6
What is the smallest [[positive]] [[integer]] that can be expressed as the sum of n ...e find that the least possible value of <math>b = 45</math>, so the answer is <math>10(45) + 45 = 495</math>.

3 KB (524 words) - 18:06, 9 December 2023
1993 AIME Problems/Problem 5
...h>n \ge 1\,</math>, define <math>P_n(x) = P_{n - 1}(x - n)\,</math>. What is the [[coefficient]] of <math>x\,</math> in <math>P_{20}(x)\,</math>?<!-- do ...0</math> into the function definition, we get <math>P_0(x-210) = (x - 210)^3 + 313(x - 210)^2 - 77(x - 210) - 8</math>. We only need the coefficients of

2 KB (355 words) - 13:25, 31 December 2018
1993 AIME Problems/Problem 4
...k) = (c-a)(d-c) = 93</math>. Hence <math>(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)</math>. ...1,c - 28,c,c + 3),</math> <math>(c - 1,c + 92,c,c + 93),</math> <math>(c - 3,c + 28,c,c + 31)</math>. The last two solutions don't follow <math>a < b <

8 KB (1,343 words) - 16:27, 19 December 2023
1993 AIME Problems/Problem 3
<center><math>\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline n & 0 & 1 & 2 & 3 & \dots & 13 & 14 & 15 \\ :(b) those who caught <math>3</math> or more fish averaged <math>6</math> fish each;

2 KB (364 words) - 00:05, 9 July 2022
1993 AIME Problems/Problem 2
...^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

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