2023 AMC 10B Problem 12

The roots are $1, 2, \ldots, 10$, and they cut the number line into $11$ pieces. Between two roots the sign of $P$ cannot change, so we just need the sign on each piece. Start where the sign is obvious and walk left.

1. Far right. For $x > 10$, for example $x = 11$, every factor $x - k$ is positive, so $P(x)$ is positive.
2. What happens when we pass a root. Passing $x = k$ changes the sign of only one factor, $(x - k)$, from positive to negative. That factor appears $k$ times in $P$.
- If $k$ is odd, an odd number of signs flip, so the sign of $P$ flips.
- If $k$ is even, an even number of signs flip, and they cancel out, so the sign of $P$ stays the same.

(For example, $(-1)^3 = -1$ but $(-1)^2 = +1$.)
3. Walk left from $+$.

1. Count the $+$ signs. The piece to the right of $10$ is positive, and the table has $5$ more: $(9,10)$, $(6,7)$, $(5,6)$, $(2,3)$, $(1,2)$. That is $6$ intervals.

The answer is $\boxed{\textbf{(C)}\ 6}$.

A polynomial made of factors like $(x - k)$ can only change sign at a root, and it changes sign there only when that factor appears an odd number of times. Starting on the far right, where every factor is positive, and walking left with the rule "odd power flips, even power does not" gives the whole sign chart without plugging in any numbers.

Group the roots in pairs from the right: $10$ and $9$, $8$ and $7$, and so on. Passing an even root changes nothing, so the two pieces on either side of an even root always have the same sign, and only the odd roots switch between blocks. From the right, the blocks are: right of $10$ together with $(9,10)$ positive; $(7,8)$ and $(8,9)$ negative; $(5,6)$ and $(6,7)$ positive; $(3,4)$ and $(4,5)$ negative; $(1,2)$ and $(2,3)$ positive; left of $1$ negative. Three positive blocks of two pieces each make $6$.