MathDB

11

Part of 2012 AIME Problems

Problems(2)

Froggy

Source: 2012 AIME I Problem 11

3/16/2012
A frog begins at P0=(0,0)P_0 = (0,0) and makes a sequence of jumps according to the following rule: from Pn=(xn,yn)P_n=(x_n,y_n), the frog jumps to Pn+1P_{n+1}, which may be any of the points (xn+7,yn+2)(x_n+7, y_n+2), (xn+2,yn+7)(x_n+2,y_n+7), (xn−5,yn−10)(x_n-5, y_n-10), or (xn−10,yn−5)(x_n-10,y_n-5). There are MM points (x,y)(x,y) with ∣x∣+∣y∣≤100|x|+|y| \le 100 that can be reached by a sequence of such jumps. Find the remainder when MM is divided by 10001000.
invariantmodular arithmeticanalytic geometrygraphing linessloperotationgeometry
Recursively Defined Functions

Source: 2012 AIME II Problem 11

3/29/2012
Let f1(x)=23−33x+1f_1(x) = \frac{2}{3}-\frac{3}{3x+1}, and for n≥2n \ge 2, define fn(x)=f1(fn−1(x))f_n(x) = f_1(f_{n-1} (x)). The value of x that satisfies f1001(x)=x−3f_{1001}(x) = x - 3 can be expressed in the form mn\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm + n.
functionAMCAIMEnumber theoryrelatively primeAIME IInice words