Competition problem — Algebra — Maths Olympiad Prep

Library / /14 of 100

Algebra Difficulty 3.8 AMC 10/12 Find the answer China

Let $f(x) = ax + b$ , with $a, b$ real numbers; $f_1(x) = f(x)$ , $f_{n+1}(x) = f(f_n(x))$ , $n = 1, 2, \dots$ . If $f_7(x) = 128x + 381$ , then $a + b = \underline{\hspace{2cm}}$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

$\begin{aligned} f_n(x) &= a^n x + (a^{n-1} + a^{n-2} + \dots + a + 1)b \\ &= a^n x + \frac{a^n - 1}{a - 1} \times b. \end{aligned}$
As $f_7(x) = 128x + 381$ , we have $a^7 = 128$ and $\frac{a^7 - 1}{a - 1} \times b = 381$ . Then $a = 2$ , $b = 3$ . The answer is $a + b = 5$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.