Difference between revisions of "2019 AIME I Problems/Problem 10"
m (AIME needs to be capitilized) |
Itsameyushi (talk | contribs) (→Problem 10) |
||
Line 2: | Line 2: | ||
==Problem 10== | ==Problem 10== | ||
+ | For distinct complex numbers <math>z_1,z_2,\dots,z_{673}</math>, the polynomial | ||
+ | <cmath> (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 </cmath>can be expressed as <math>x^{2019} + 20x^{2018} + 19x^{2017}+g(x)</math>, where <math>g(x)</math> is a polynomial with complex coefficients and with degree at most <math>2016</math>. The value of | ||
+ | <cmath> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| </cmath>can be expressed in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
+ | |||
==Solution== | ==Solution== | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=9|num-a=11}} | {{AIME box|year=2019|n=I|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:22, 14 March 2019
The 2019 AIME I takes place on March 13, 2019.
Problem 10
For distinct complex numbers , the polynomial can be expressed as , where is a polynomial with complex coefficients and with degree at most . The value of can be expressed in the form , where and are relatively prime positive integers. Find .
Solution
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.