Maths Olympiad Prep

Library / /23 of 156

Algebra Difficulty 3.8 AMC 10/12 Find the answer China

Given quintic polynomial f(x)f(x) with its leading coefficient being 1, it satisfies f(n)=8nf(n) = 8n, n=1,2,,5n = 1, 2, \dots, 5. Then the coefficient of the term of degree 1 of f(x)f(x) is ______.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let f(x)=g(x)8xf(x) = g(x) - 8x, and then g(x)g(x) is also a quintic polynomial with its leading coefficient being 1. And there is
g(n)=f(n)8n=0,n=1,2,,5. g(n) = f(n) - 8n = 0, \quad n = 1, 2, \dots, 5.
Hence, g(x)g(x) has 5 real roots, namely, 1,2,,51, 2, \dots, 5. Therefore,
g(x)=(x1)(x2)(x5), g(x) = (x-1)(x-2)\cdots(x-5),
and thus,
f(x)=(x1)(x2)(x5)+8x. f(x) = (x-1)(x-2)\cdots(x-5) + 8x.
Consequently, the coefficient of the term of degree 1 of f(x)f(x) is
(1+12+13+14+15)5!+8=282. \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}\right) \cdot 5! + 8 = 282.

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 reproduced verbatim; metadata (topic, difficulty) added by this project.