AMC 10A,2023, Problem,21 Question and Solution

[Deepan Dutta]

Problem 21

Let  \(P(x)\) be the unique polynomial of minimal degree with the following properties:

• \(P(x)\) has a leading coefficient 1,
• \(1\) is the root of \(P(x)-1\)
• \(2\) is the root of \(P(x-2)\).
• \(3\) is the root of \(P(3x)\).
• \(4\) is the root of \(4P(x)\).

the root of \(P(x)\) are integers, with one exception. The root that is not an integer can be written as \(\frac{m}{n}\), where \(m\) and \(n\) are relatively prime integers. What is \(m+n\) ?

Solution :

As, \(2\) is the root of \(P(x-2)\) so, \(P(2-2)=0 \Rightarrow P(0)=0\), that means \(0\) is one of the integer roots of \(P(x)\).

Again, \(3\) is a root of \(P(3x)\) so, \(P(3\times3)=0\Rightarrow P(9)=0\), that means \(9\) is one of the integer roots of \(P(x)\).

Also, \(4\) is a root of \(P(3x)\) so, \(4(P(4)=0\Rightarrow P(4)=0\), that means \(4\) is one of the integer roots of \(P(x)\).

This implies that the polynomial should be degree \(4\), for \(P(x)\) being the unique polynomial of minimal degree.

Let, \(P(x)=x(x-2)(x-4)(x-9)(x-t)\), assuming \(t\) the only non-integer root of \(P(x)\).

According to the problem, \(1\) is a root of \(P(x)-1\).

\(\therefore\) \(P(1)-1=0\)

\(\Rightarrow 1\cdot (1-4)(1-9)(1-t)=1\)

\(\Rightarrow (-3)(-8)(1-t)=1\)

\(\Rightarrow 24-24t=1\)

\(\Rightarrow -24t=-23\)

\(\Rightarrow t=\frac{23}{24}\)

So, the only non-integer root of \(P(x)\) is \(\boxed{\frac{23}{24}}\)

• This topic was modified 1 year, 4 months ago by Cheenta Support.

[Author]

Posts