Wacky Polynomial

Algebra Level 5

Let $f(x) = x\left( 4x^2-3 \right) \left( 64x^6-96x^4+36x^2-3 \right)$ . Find the last three digits of the number of distinct positive reals $x$ satisfying $f^{(2014)}(x) = x$ .

Details and Assumptions

$f^{(2014)}(x)$ denotes the function $f$ applied $2014$ times successively on $x$ , i.e. $f^{(2014)}(x) = \underbrace{f ( f ( \cdots ( f(x) )\cdots ))}_{2014 \text{ times}}$ .
Any non-real / repeating solutions of the equation should be ignored. For example, the roots of the equation (counted with multiplicity) $x^5-14 x^4 + 86 x^3 - 298 x^2 + 573 x - 468=0$ are $\left \{ 3, 3, 4, 2-3i, 2+3i \right \}$ , but its number of distinct real solutions is $2$ .

This problem appeared in the Proofathon Algebra contest, and was posed by me.

The answer is 961.

1 solution

Omkar Singh
Feb 28, 2014

First of all see how many solutions are available for f(x)=x, it comes out to be 9.Because for any polynomial with degree even and having alternate powers of x such as - 2k, 2(k-1), 2(k-2), ... 0. the no. of distinct real roots will be 2k in which half will be -ve and half +ve, so it gives 9 solutions for f(x)=x. Similarly for f(f(x))=x , the total real solutions will be 81 that is equal to 9^2. Similarly for f(f(f(x)))=x i.e., f(3)(x)=x will be 729=9^3. Similarly we can determine the no. of real distinct solutions of f(2014)(x)=x will be 9^2014. Which leaves us to find the last three digits in the no. 9^2014. Which can be found by considering the final three terms of the binomial expansion of (10 - 1)^2014, because taking the modulo of the whole terms of the expansion we find that the last three terms gives the proper fraction term that is why the last three terms consideration and there proper algebraic addition we get the answer as follows- (2014 * 2013 * 10^2 / 2) -(2014*10) + (1) = 961

No, your claim is not true. The roots have to be real and distinct. The equation $\displaystyle \sum \limits_{k=0}^{9} (-1)^k \dbinom{9}{k} x^k = 0$ has only one distinct real solution $\left( x=1 \right)$ , which is a counterexample to your claim.

Sreejato Bhattacharya - 7 years, 2 months ago

Can you add a solution to this problem? Thanks!

Calvin Lin Staff - 6 years, 10 months ago

Wacky Polynomial

This problem appeared in the Proofathon Algebra contest, and was posed by me.

The answer is 961.

1 solution

0 pending reports