Difference between revisions of "2016 IMO Problems/Problem 5"
m |
|||
Line 1: | Line 1: | ||
+ | ==Problem== | ||
+ | |||
The equation | The equation | ||
<center><math>(x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots (x-2016)</math></center> | <center><math>(x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots (x-2016)</math></center> | ||
is written on the board, with <math>2016</math> linear factors on each side. What is the least possible value of <math>k</math> for which it is possible to erase exactly <math>k</math> of these <math>4032</math> linear factors so that at least one factor remains on each side and the resulting equation has no real solutions? | is written on the board, with <math>2016</math> linear factors on each side. What is the least possible value of <math>k</math> for which it is possible to erase exactly <math>k</math> of these <math>4032</math> linear factors so that at least one factor remains on each side and the resulting equation has no real solutions? | ||
+ | |||
+ | ==Solution== | ||
+ | {{solution}} | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2016|num-b=4|num-a=6}} |
Latest revision as of 00:37, 19 November 2023
Problem
The equation
is written on the board, with linear factors on each side. What is the least possible value of for which it is possible to erase exactly of these linear factors so that at least one factor remains on each side and the resulting equation has no real solutions?
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See Also
2016 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |