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2023 AIME I Problems/Problem 11
...h (1,2), because 1 is still unusable as it is consecutive with 2. The only difference is we now have only 4-10 to work with. Using the same pattern as before, we ==Solution 3 (Double Recursive Equations)==

15 KB (2,414 words) - 06:57, 26 November 2023
2023 AIME II Problems
...th>x,y,</math> and <math>z</math> be real numbers satisfying the system of equations ...<math>2 \times 6</math> grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by <math>3.</math> One

8 KB (1,370 words) - 07:14, 13 June 2024
2021 Fall AMC 10A Problems/Problem 17
...the heights at pillar <math>C</math> and pillar <math>D</math> is half the difference between the heights at <math>B</math> and <math>E,</math> so Solving these equations, we get <math>h_E = \boxed{\textbf{(D) } 17}</math>.

10 KB (1,705 words) - 00:13, 17 July 2023
1972 AHSME Problems/Problem 16
Let <math>a</math> be first and <math>b</math> be second. We can then get equations based on our knowledge: <math>b-a = 9-b</math> and <math>b/a = a/3</math>. for some common difference <math>d</math> and common ratio <math>r</math>. We can use these to obtain

2 KB (370 words) - 13:44, 4 April 2024
2020 AMC 8 Problems/Problem 25
...he side lengths have a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, the answer must be <math>\dfrac{3322 - 2020}{2} = \dfr ...h>r+s</math>. Then, <math>r+s+s+r+s=3322</math>. Now, we have 2 systems of equations.

4 KB (606 words) - 13:22, 1 January 2024
Quartic Equation
These types of equations are extremely hard to solve; however, there are very clever methods for sol ==Solving Quartic Equations==

13 KB (2,376 words) - 16:56, 19 February 2024
2021 AIME II Problems/Problem 15
...root is equal to the perfect square, <math>m^2</math>. Thus, after using a difference of squares, we have ..., which yields <math>n=\boxed{258}</math>, which, if plugged in to for our equations of <math>f(n)</math> and <math>g(n)</math>, will yield the desired ratio, a

14 KB (2,569 words) - 09:28, 28 March 2024
Jadhav Theorem
...icable for any 3 terms of an Arithmetic Progression with a constant common difference between them. This theorem is derived by Jyotiraditya Jadhav. ...the first and the third term (ac) will always be the square of the common difference (d).

2 KB (346 words) - 05:52, 1 April 2021
2022 AMC 10B Problems
...ence between a pair of primes is equal to <math>2</math>, and the positive difference between the cubes of the two primes is <math>31106</math>. What is the sum ...the first <math>n</math> terms of an arithmetic sequence that has a common difference of <math>2</math>. The quotient <math>\frac{S_{3n}}{S_n}</math> does not de

15 KB (2,224 words) - 13:10, 20 February 2024
2021 Fall AMC 12A Problems/Problem 24
Let <math>k</math> be the common difference of the arithmetic progression of the side-lengths. It follows that <math>b, ...f these sides form an arithmetic sequence, we have the following system of equations:

9 KB (1,500 words) - 01:18, 29 August 2022
2023 AIME I Problems/Problem 5
...{2}</math>, and therefore <math>b + d = a\sqrt{2}</math>. By squaring both equations, we obtain Thus, the equations <math>PA \cdot PC = 56</math> and <math>PB \cdot PD = 90</math> can be writ

19 KB (3,107 words) - 23:31, 17 January 2024
Chebyshev polynomials of the first kind
...) = \cos ny \cos y + \sin ny \sin y. \end{align*}</cmath> The sum of these equations is <cmath> \cos ((n+1)y) + \cos ((n-1)y) = 2 \cos ny \cos y;</cmath> rearra

10 KB (1,919 words) - 15:24, 26 June 2023
Chebyshev polynomials of the second kind
...- y) = \sin((n+1)y)\cos y - \cos((n+1)y)\sin y.</cmath> The sum of these equations is <cmath>\sin((n+2)y) + \sin ny = 2\sin((n+1)y)\cos y;</cmath> rearranging

2 KB (392 words) - 22:12, 11 March 2022
2022 AMC 10A Problems/Problem 3
...h>40</math> less than the second number. What is the absolute value of the difference between the first and second numbers? ...ath> and the second number is <math>47.</math> Their absolute value of the difference is <math>|42-47|=\boxed{\textbf{(E) } 5}.</math>

2 KB (333 words) - 11:35, 24 March 2024
2023 AMC 8 Problems/Problem 25
...h>241-20=221</math>, and the maximum–<math>250-13=237</math>. There is a difference of <math>13</math> between them, so only <math>17</math> and <math>18</math ...ge to satisfy this inequality. On the other hand, we can now find that the difference will be <math>17</math>, which satisfies this inequality.

4 KB (580 words) - 17:58, 17 May 2024
2023 AMC 10A Problems/Problem 23
...is another pair of two integers that multiply to <math>n</math> but have a difference of 23, one integer must be greater than <math>a</math>, and one must be sma ...h>(2y+2x+43)</math> and <math>(2y-2x+3)</math> must be integer, we get two equations.

8 KB (1,343 words) - 16:19, 7 June 2024
2023 AMC 12A Problems/Problem 6
...of <math>\overline{AB}</math> is <math>(6, 2)</math>. What is the positive difference between the <math>x</math>-coordinates of <math>A</math> and <math>B</math> ..._{2}(6+m)=2+n</math> and <math>\log_{2}(6-m)=2-n</math>. Now add these two equations to obtain <math>\log_{2}(6+m)+\log_{2}(6-m)=4</math>. By logarithm rules, w

3 KB (547 words) - 20:33, 10 June 2024
2024 AIME I Problems/Problem 15
...this antioptimal box. (If the height and width weren't the same, the extra difference between them could be used to make the length longer.) Thus, let the width ...idden inside this equation and call this <math>m</math>. Now we have three equations:

12 KB (2,049 words) - 16:13, 7 June 2024
2002 AMC 12P Problems/Problem 13
Essentially, this boils down to writing <math>217</math> as a difference of squares. We know <math>217 = (7)(31)</math>, so we assume there exist po Solving this system of equations gives <math>a = 109</math> and <math>b = 108</math>. However, <math>108 > 1

2 KB (305 words) - 20:35, 10 March 2024

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