2023 AMC 12B Problem 6

Below is the professionally curated solution for Problem 6 of the 2023 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12B solutions, or check the answer key.

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Concepts:polynomial parity

Difficulty rating: 1380

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 number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?

$3$

$7$

$6$

$4$

$5$

Solution:

The exponent of the factor $(x-k)$ is $k,$ so the sign of $P$ changes at $x=k$ only when $k$ is odd, i.e. at $1,3,5,7,9.$ For $x\gt 10$ every factor is positive, so $P\gt 0.$ Sweeping left and flipping at each odd root, the positive intervals are $(10,\infty),$ $(9,10),$ $(6,7),$ $(5,6),$ $(2,3),$ and $(1,2)$ — six intervals in all.

Thus, the correct answer is C.

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