2023 AMC 10A Problem 21

Question

There is a unique polynomial

P(x)

of least degree with leading coefficient

1

satisfying all of the following:

1

is a root of

P(x) - 1,

2

is a root of

P(x - 2),

3

is a root of

P(3x),

and

4

is a root of

4P(x).

All the roots of

P(x)

except one are integers. If the one non-integer root can be written as

\frac{m}{n},

where

m

and

n

are relatively prime positive integers, what is

m + n?

41

43

45

47

49

Translate each condition into a value:

P(1) = 1,

P(0) = 0,

P(9) = 0,

P(4) = 0

The integer roots

0, 4, 9

force

P(x) = x(x - 4)(x - 9)(x - c);

use

P(1) = 1

to find

c

Translate each condition into a value:

P(1) = 1,

P(0) = 0,

P(9) = 0,

and

P(4) = 0.

So

0, 4, 9

are roots. Could a cubic do it? A monic cubic with those roots has

P(1) = (1)(-3)(-8) = 24 \ne 1,

so no. The least-degree monic polynomial is degree

4:

P(x) = x(x - 4)(x - 9)(x - c).

Now

P(1) = (1)(-3)(-8)(1 - c)

= 24(1 - c)

= 1,

so

1 - c = \frac{1}{24}

and

c = \frac{23}{24}.

That’s the lone non-integer root, so

m + n = 23 + 24 = 47.

Thus, D is the correct answer.

Problem 21 in other years