I don't know how to count

Algebra Level 4

Consider a quintic function $f(x)$ such that $f(k) = k$ for $k=1,2,3,4,5$ . Given that $f(x) = g(x) + q(x)$ such that $g(x)$ has all real roots and its leading coefficient is an integer and $q(x)$ is a linear function, which of these answer choices is $(f(6) - 6)$ always divisible by?

2\cdot 3\cdot 5^2

5!

2^2\cdot 5^2

2\cdot3^2\cdot 5

2^2\cdot 3\cdot 5

6!

3\cdot 4!

1 solution

Vishwak Srinivasan
Jul 26, 2015

Since each function is given to us: let $f(x) - x = g(x)$

By observing the values given, the zeroes of $g(x)$ are $1,2,3,4,5$ .

Hence, we have a system of polynomials for $g(x)$ .

$g(x) = k(x-1)(x-2)(x-3)(x-4)(x-5)$

where $k$ is an arbitrary non zero integer constant

$f(x) - x = k(x-1)(x-2)(x-3)(x-4)(x-5)$

By shifting the $x$ to R.H.S, it can be seen that $q(x) = x$

$f(6) - 6 = k(5)(4)(3)(2)(1) = k(5!)$

Hence $f(6) - 6$ is a multiple of $5!$ .

I'm not quite sure how you got $f(x) - x = g(x)$ . I understand the rest, but could you please explain that to me?

Ethan Ooamii - 5 years, 7 months ago

$f(1) = 1, f(2) = 2, f(3) = 3, f(4) = 4, f(5) = 5$

$\implies f(1) - 1 =0 , f(2) -2 = 0, f(3) - 3 =0 , f(4) - 4 = 0 , f(5) - 5 =0$

$\implies f(x) - x = 0 , x = 1 ,2 ,3, 4, 5$

Let $g(x) = f(x) - x$ . $\implies g(x) = 0 , x = 1,2,3,4,5$

Vishwak Srinivasan - 5 years, 7 months ago

I don't know how to count

1 solution

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