f(x) is a polynomial with integer coefficients and degree at most 10. There are N distinct integer values for which f(n)=2, and M distinct integer values for which f(m)=−2. What is the maximum possible value of NM?
maths - Steve, Friday, May 3, 2013 at 10:39am
Consider a function of degree 10 which has 9 turning points. If the local max/min values are such that |f(x)| > 2, then there are two values of x near each extremum where f(x) is 2 or -2.
So, counting the two values of x outside the interval containing all the roots, there will be 10 places where f(x)=2 and 10 more where f(x) = -2.
10*10 = 100
maths - pria, Friday, May 3, 2013 at 11:16am
maths - Steve, Friday, May 3, 2013 at 12:22pm
Hmmm. Got any suggestions, or just an answer key? See any problem with my logic?
maths - 12k, Friday, May 3, 2013 at 12:56pm
maths - Steve, Friday, May 3, 2013 at 5:03pm
Ahh. I see where I went wrong. You want integer values. I'll have to think on it a bit more.
maths - Steve, Friday, May 3, 2013 at 5:46pm
y = ax^10 + bx^9 + ... + k
We can only have 11 solutions to finding values for a,b,...k given
y(x)=±2 for 11 different values of x
Max MN for M+N=11 would be 6*5 = 30
How to get 75? That's strange, since 75=3*25 or 5*15
That's an odd distribution of values for x
Here's another idea. If y=±2 for some set of x values,
y^2 = 4 for those same values. If y is degree 10, y^2 is degree 20. So, if we set up a system of equations with 21 coefficients, we can get 21 points where y^2 = 4. That would allow us to have 11*10=110 for MN. Still don't get 75.
maths - 12k, Saturday, May 4, 2013 at 5:24am
Sry made a mid calculation,like the way u did it,is 110 correct