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2011 AIME I Problems/Problem 15
Factoring the perfect square, we get: <math>ab-(b+a)^2=-2011</math> or <math>(b+a)^2- {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}}

8 KB (1,302 words) - 04:07, 24 July 2023
2011 AIME I Problems/Problem 7
...an integer is quite obviously the number of factors of <math>2010</math>. Factoring <math>2010</math>, we obtain <math>2010 = 2 \times 3 \times 5 \times 67</ma {{AIME box|year=2011|n=I|num-b=6|num-a=8}}

10 KB (1,677 words) - 19:59, 29 May 2024
2011 AIME II Problems/Problem 9
...volves products <math>x_i x_j</math> with <math>i \equiv j \pmod 3</math>. Factoring out say <math>x_1</math> and <math>x_4</math> we see that the constraint is {{AIME box|year=2011|n=II|num-b=8|num-a=10}}

3 KB (466 words) - 15:06, 16 January 2023
2012 AIME I Problems/Problem 10
...condition implies <math>x^2\equiv 256 \pmod{1000}</math>. Rearranging and factoring, <cmath>(x-16)(x+16)\equiv 0\pmod {1000}.</cmath> {{AIME box|year=2012|n=I|num-b=9|num-a=11}}

8 KB (1,338 words) - 02:03, 25 November 2023
2013 AIME I Problems/Problem 14
...{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}.</cmath> Factoring the denominator, we have <cmath>10-4\sin^2{\theta}-3\sin{\theta}=(5-4\sin\t {{AIME box|year=2013|n=I|num-b=13|num-a=15}}

10 KB (1,641 words) - 20:03, 3 January 2024
2014 AIME I Problems/Problem 9
...quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer the ...imate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting a

10 KB (1,653 words) - 00:30, 27 January 2024
2015 AIME I Problems/Problem 3
Rearrange this expression and factor the left side (this factoring can be done using <math>(a^3-b^3) = (a-b)(a^2+a b+b^2)</math> or synthetic Recognizing that AIME answers are <math>0</math> through <math>999</math>, the numbers whose cube

6 KB (1,031 words) - 23:19, 23 January 2024
2015 AIME I Problems/Problem 11
...ath>. Simplifying, we have <math>y^2+y^2\frac{x^2-y^2}{(x+y)^2}=64</math>. Factoring out the <math>y^2</math>, we have <math>y^2\left(1+\frac{x^2-y^2}{(x+y)^2}\ [[File:2015 AIME I 11.png|270px|right]]

5 KB (906 words) - 17:43, 27 September 2023
2015 AIME II Problems/Problem 14
...^2)}{(x - y)^2} = 13</math>. Now, we clean this up to obtain the following factoring of <math>0 = 2 \cdot (2x - 3y) \cdot (3x - 2y)</math>. This implies that <m {{AIME box|year=2015|n=II|num-b=13|num-a=15}}

10 KB (1,751 words) - 22:21, 26 November 2023
2016 AIME II Problems/Problem 1
...have <math>a(k^2+k+1)=444</math>. Adding <math>k</math> to both sides and factoring, ...5)=ak^2-25-(ak-9)</math>. Simplifying, <math>k^2-2k+1=\frac{12}{a}</math>. Factoring,

5 KB (788 words) - 02:50, 1 March 2024
2017 AIME I Problems/Problem 6
We actually do not need to spend time factoring <math>x^2 - 120x + 3024</math>. Since the problem asks for <math>|x_1 - x_2 ...math> which using trivial algebra gives you <math>x^2-120x+3024</math> and factoring gives you <math>(x-84)(x-36)</math> and so your answer is <math>\boxed{048}

4 KB (726 words) - 16:18, 5 January 2024
2017 AIME II Problems/Problem 6
==Solution 7 (alternative factoring)== {{AIME box|year=2017|n=II|num-b=5|num-a=7}}

7 KB (1,076 words) - 14:13, 12 June 2024
2018 AIME I Problems/Problem 1
The factoring condition is equivalent to the discriminant <math>a^2-4b</math> being equal {{AIME box|year=2018|n=I|before=First Problem|num-a=2}}

10 KB (1,670 words) - 16:38, 15 January 2024
2018 AIME I Problems/Problem 5
...>x^2+xy-2y^2=0</cmath> upon simplification and division by <math>3</math>. Factoring <math>x^2+xy-2y^2=0</math> gives <cmath>(x+2y)(x-y)=0</cmath> Then, <cmath> {{AIME box|year=2018|n=I|num-b=4|num-a=6}}

3 KB (543 words) - 12:52, 30 July 2023
2018 AIME I Problems/Problem 6
...f the roots of such a polynomial are distinct? One can proceed as follows. Factoring gives us that <math>(z^{840}+1)(z^{600}-1)=0,</math> so this implies that < {{AIME box|year=2018|n=I|num-b=5|num-a=7}}

7 KB (1,211 words) - 00:23, 20 January 2024
2019 AIME II Problems/Problem 11
...frac{7 \cdot 189 \cdot 5}{8 \sqrt{5} \cdot 8\sqrt{5}}\right)}</cmath> Then factoring yields <cmath>\sqrt{\frac{21^2(8^2-15)}{(8\sqrt{5})^2}} =\frac{147}{8\sqrt{ [[File:AIME-II-2019-11.png|500px|right]]

12 KB (1,985 words) - 19:52, 28 January 2024
2019 AIME I Problems/Problem 8
Now factoring <math>m^5+n^5</math> as solution 4 yields <math>m^5+n^5=(m+n)(m^4-m^3n+m^2n {{AIME box|year=2019|n=I|num-b=7|num-a=9}}

10 KB (1,878 words) - 13:19, 1 February 2024
2019 AIME I Problems/Problem 12
...c{(z^2-19z)-19(z^2-19z)-(z^2-19z)}{z^2-19z-z}</math> being pure imaginary. Factoring and simplifying, we find that this is simply equivalent to <math>(z-19)(z+1 {{AIME box|year=2019|n=I|num-b=11|num-a=13}}

8 KB (1,534 words) - 22:17, 28 December 2023
2023 AIME I Problems/Problem 10
Factoring our approximation gives <math>U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6 {{AIME box|year=2023|n=I|num-b=9|num-a=11}}

10 KB (1,578 words) - 09:48, 24 April 2024
2021 AIME I Problems/Problem 2
...ractions by multiplying both sides by <math>11^2=121,</math> then solve by factoring: {{AIME box|year=2021|n=I|num-b=1|num-a=3}}

8 KB (1,294 words) - 00:59, 23 August 2022

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