2023 AMC 10B Problem 12

Question

When the roots of the polynomial

P(x) = (x-1)^1(x-2)^2(x-3)^3 \cdots (x-10)^{10}

are removed from the real number line, what remains is the union of

11

disjoint open intervals. On how many of those intervals is

P(x)

positive?

3

7

6

4

5

For

x \gt 10

every factor is positive; the sign of

P

can only change at a root

Crossing the root

x = i

flips the sign exactly when the exponent

i

is odd

For

x \gt 10,

every factor

(x - i)^i

is positive, so

P(x) \gt 0.

Now move left. Crossing

x = i

flips the sign only when

i

is odd, that is at

i = 9, 7, 5, 3, 1.

So the eleven intervals, right to left, carry signs

+, +, -, -, +, +, -, -, +, +, -.

Six are positive. Therefore, the answer is C.

Problem 12 in other years