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2008 AIME I Problems/Problem 3
...btracting the second from twice the first gives <math>4j + 5s = 57</math>. Mod 4, we need <math>s\equiv1\pmod{4}</math>. Thus, <math>(j,s) = (13,1),(8,5), ...math>y+z-2x=-17</math>. Note that <math>-17</math> is odd, so one of <math>x,y,z</math> is odd. We see from our second expression that <math>z</math> mu

3 KB (564 words) - 22:15, 28 November 2023
2008 AIME I Problems/Problem 4
...that satisfy the equation <math>x^2 + 84x + 2008 = y^2</math>. Find <math>x + y</math>. ...(x+42)^2 + 244</math>. Thus <math>244 = y^2 - (x+42)^2 = (y - x - 42)(y + x + 42)</math> by [[difference of squares]].

4 KB (732 words) - 22:17, 28 November 2023
2008 AIME I Problems/Problem 9
...> (the solution sets, for the sake of consistency, are in the form <math>z,x,y</math>). Summing the results for all the solutions gives us <math>5130</m ...eft(x^3+x^4+x^6\right)^{10}</cmath> where we want the coefficient of <math>x^{41}.</math>

6 KB (909 words) - 15:39, 8 August 2022
2005 USAMO Problems/Problem 3
...th> and <math>Q' =Q</math>. All our angles will be directed, and measured mod <math>\pi</math>. ...<math>\triangle APC.</math> Let its equation be <cmath>-a^2yz-b^2xz-c^2xy+(x+y+z)(ux+vy+wz)=0.</cmath> Substituting in <math>A,C</math> gives <math>u=w=

6 KB (1,117 words) - 01:17, 11 October 2021
2005 USAMO Problems/Problem 2
\begin{align*}x^6 + x^3 + x^3y + y &= 147^{157} \\ x^3 + x^3y + y^2 + y + z^9 &= 157^{147}\end{align*}

7 KB (1,053 words) - 10:38, 12 August 2015
2008 USAMO Problems/Problem 1
...' Let <math>g</math> be a multiplicative generator of the nonzero integers mod p. Set <math>r \equiv g^{(p-1)/3}</math>. Then <math>r-1 \not\equiv 0 \pm By [[Dirichlet's Theorem]], there are infinitely many primes congruent to 1 (mod 3). Let <math>p_0, \dotsc, p_n</math> be <math>(n+1)</math> such primes, a

11 KB (1,964 words) - 03:38, 17 August 2019
2008 USAMO Problems/Problem 6
...ative (because it's equivalent to addition of n-dimensional vectors <math>\mod{2}</math>). So it suffices to show this set of orders is closed under comp ...= BA^{ - 1} \in H_1</math>. Therefore, <math>H_1</math> and <math>\{T_2X|X \in H_1\}</math> are disjoint and of the same size. Moreover, the product

13 KB (2,414 words) - 14:37, 11 July 2016
Normal subgroup
...(Hence the notation <math>\mathbb{Z}/n\mathbb{Z}</math> for the integers mod <math>n</math>.) ...nt to its converse (by replacing <math>x</math>, <math>y</math> with <math>x^{-1}</math>, <math>y^{-1}</math>).

15 KB (2,840 words) - 12:22, 9 April 2019
Order (group theory)
...then it is also the least positive integer <math>n</math> for which <math>x^n=e</math>. ...>a</math> relatively prime to <math>n</math>, the order of <math>a</math> (mod <math>n</math>) usually means the order of <math>a</math> in the multiplica

1 KB (183 words) - 10:43, 11 April 2020
Lagrange's Theorem
...ng <math>x\mapsto ba^{-1}x</math> induces a bijection from the left cosets mod <math>K</math> contained in an arbitrary <math>H</math>-coset <math>aH</mat

2 KB (303 words) - 12:24, 9 April 2019
Sylow Theorems
...te group, the number of Sylow <math>p</math>-subgroups is equivalent to 1 (mod <math>p</math>). ...et <math>G \times T</math> by the law <math>g(\alpha, x) \mapsto (g\alpha, x)</math>; extend this action canonically to the subsets of <math>G \times S<

11 KB (2,071 words) - 12:25, 9 April 2019
2003 AMC 12B Problems/Problem 18
...uch that <math>7x^5 = 11y^{13}.</math> The minimum possible value of <math>x</math> has a prime factorization <math>a^cb^d.</math> What is <math>a + b + Substitute <math>a^cb^d</math> into <math>x</math>. We then have <math>7(a^{5c}b^{5d}) = 11y^{13}</math>. Divide both s

6 KB (914 words) - 11:07, 7 September 2023
KGS math club/solution 2 1
Lemma 2: x(n) = 2^n * (n^2 + 1) (letting x(0) = 1, x(1) = 4 and x(2) = 20) For n > 3: assume x(n) = 2^n * (n^2 + 1) and similarly for x(n-1) and x(n-2).

2 KB (393 words) - 12:54, 11 August 2023
2001 IMO Shortlist Problems/C1
...z = 2001</math> and we want to maximize xyz. By AM-GM, xyz is maximal when x = y = z = 667, so m = <math>667^3</math>. ...a_j,a_k)</math> that satisfies the condition must leave different residues mod 3, which means the indexes must be in different <math>A_i</math>. Thus, the

4 KB (703 words) - 12:45, 27 November 2017
2008 IMO Problems/Problem 3
...}</math>. Observe that the function <math>x+\sqrt{p-x^2}</math> over <math>x\in(2,\sqrt{p-4})</math> reaches its minima on the ends, so <math>a+b</math> Lemma: For every prime <math>p \equiv 1 \mod{4}</math>, there exists a positive integer <math>n \le \dfrac{p - \sqrt{p-4

5 KB (1,002 words) - 01:09, 19 November 2023
Fermat's Two Squares Theorem
* What numbers does a form <math>F(x,y) = ax^2 + bxy + y^2</math> represent? ...rs <math>\mathbb{Z}[i]</math>, resolves these questions for the form <math>x^2 + y^2</math>.

4 KB (612 words) - 12:10, 30 May 2019
2009 AMC 12A Problems/Problem 15
We know that <math>i^x</math> cycles every <math>4</math> powers so we group the sum in <math>4</m ...ne of reason, <math>i^{n}=1,i</math>. This only works for <math>n\equiv 1 \mod 4</math>. Therefore, we have:

4 KB (634 words) - 16:34, 3 December 2020
2009 AMC 10B Problems/Problem 13
...th>-axis. Which side will touch the point <math>x=2009</math> on the <math>x</math>-axis? xaxis("$x$",-8,14,EndArrow(3));

2 KB (399 words) - 18:39, 9 February 2023
2005 USAMO Problems/Problem 6
...ath>s\left(\sum_{x\in X} x\right) = k</math> for any nonempty subset <math>X\subset S</math>. Prove that there are constants <math>0 < C_1 < C_2</math> ...nonempty subset <math>X</math> of <math>S</math> with <math>\sum_{x\in X} x</math> a multiple of <math>10^q - 1</math>, and hence Lemma 2 shows that <m

4 KB (725 words) - 23:59, 29 March 2016
2009 AIME II Problems/Problem 13
|(x-\omega^1)\ldots(x-\omega^6)(x-\omega^8)\ldots(x-\omega^{13})| for <math>x=1</math>.

8 KB (1,279 words) - 20:27, 17 May 2024

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