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2001 AIME II Problems
...the largest number of students who could study both languages. Find <math>M-m</math>. x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*}

8 KB (1,282 words) - 21:12, 19 February 2019
2002 AIME II Problems
...the number formed by reversing the digits of <math>x</math>; and <math>z=|x-y|</math>. How many distinct values of <math>z</math> are possible? It is given that <math>\log_{6}a + \log_{6}b + \log_{6}c = 6</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are [[posi

7 KB (1,177 words) - 15:42, 11 August 2023
2003 AIME II Problems
The product <math>N</math> of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. Find ...th>C</math> is never immediately followed by <math>A</math>. How many seven-letter good words are there?

7 KB (1,127 words) - 09:02, 11 July 2023
1983 AIME Problems/Problem 1
== Solution 6 == {{AIME box|year=1983|before=First Question|num-a=2}}

4 KB (642 words) - 03:14, 17 August 2022
1983 AIME Problems/Problem 3
...non-negative (and 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 o ...= \sqrt{x^2+18x+45}</math>, we have <math>n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0</math>. Because the square root of a real number can't be negati

3 KB (532 words) - 05:18, 21 July 2022
1983 AIME Problems/Problem 4
...ircle is <math>\sqrt{50}</math> cm, the length of <math>AB</math> is <math>6</math> cm and that of <math>BC</math> is <math>2</math> cm. The angle <math defaultpen(linewidth(0.6)+fontsize(11));

11 KB (1,741 words) - 22:40, 23 November 2023
1983 AIME Problems/Problem 5
<math>x^2+y^2=(x+y)^2-2xy=7</math> and <math>x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10</math>

4 KB (672 words) - 10:17, 17 March 2023
1983 AIME Problems/Problem 6
Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math ...th and <math>49</math>. We notice that <math>49=7^2</math>, and both <math>6</math> and <math>8</math> are greater or less than <math>7</math> by <math>

3 KB (361 words) - 20:20, 14 January 2023
1983 AIME Problems/Problem 7
There are <math>(25-1)! = 24!</math> configurations for the knights about the table (since confi ...ts must be together, we let them be a single entity, so there are <math>(24-1)! = 23!</math> configurations for the entities.

9 KB (1,392 words) - 20:37, 19 January 2024
1983 AIME Problems/Problem 10
<div style="text-align:center;"><math>11xy,\qquad 1x1y,\qquad1xy1</math></div> <div style="text-align:center;"><math>1xxy,\qquad1xyx,\qquad1yxx.</math></div>

5 KB (855 words) - 20:26, 14 January 2023
1983 AIME Problems/Problem 11
...>2s</math>. All other edges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid? real s = 6 * 2^.5;

6 KB (971 words) - 15:35, 27 May 2024
1983 AIME Problems/Problem 12
draw((-2.6,0)--(-2.6,0.6)); draw((-2,0.6)--(-2.6,0.6));

2 KB (412 words) - 18:23, 1 January 2024
1983 AIME Problems/Problem 13
...ample, the alternating sum for <math>\{1, 2, 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 th ...th>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
In the adjoining figure, two circles with radii <math>8</math> and <math>6</math> are drawn with their centers <math>12</math> units apart. At <math>P ...dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=

13 KB (2,151 words) - 17:48, 27 May 2024
1983 AIME Problems/Problem 15
...h>. Suppose that the radius of the circle is <math>5</math>, that <math>BC=6</math>, and that <math>AD</math> is bisected by <math>BC</math>. Suppose fu 

20 KB (3,497 words) - 15:37, 27 May 2024
1984 AIME Problems/Problem 3
...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); MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N);

4 KB (726 words) - 13:39, 13 August 2023
1984 AIME Problems/Problem 5
...{\ln ab}{\ln 2} = 9</math>. The left-hand side can be interpreted as a base-2 logarithm, giving us <math>ab = 2^9 = \boxed{512}</math>. This solution is very similar to the above two, but it utilizes the well-known fact that <math> \log_{m^k}{n^k}= \log_m{n}. </math> Thus, <math> \log

6 KB (863 words) - 16:10, 16 May 2024
1984 AIME Problems/Problem 6
[[Image:1984_AIME-6.png]] ...center as that given point will always cut the circle in half, so we can re-phrase the problem:

6 KB (1,022 words) - 19:29, 22 January 2024
1984 AIME Problems/Problem 7
n-3&\mbox{if}\ n\ge 1000\\ ...))</math> is performed <math>\frac{n}{2}</math> times and then <math>f(x)=x-3</math> is performed back <math>\frac{n}{2}</math> times, the result is gre

4 KB (617 words) - 22:09, 15 May 2024
1984 AIME Problems/Problem 8
The equation <math>z^6+z^3+1=0</math> has complex roots with argument <math>\theta</math> between ...root within 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

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