New Year's Countdown Day 12: Twelve Days of Christmas

We know that $p(x) = 25x - 24$ when $x = 1, 2, \dots. 10.$ Thus, we can factor $p$ as

$p(x) = 25x - 24 + q(x) \displaystyle \prod_{i = 1}^{10} (x - i),$

for some polynomial $q(x)$ with real coefficients. Since we want to minimize the degree of $p,$ $q$ must be a constant function, so set $q(x) = c.$ Now, we use the relationship between $p(11)$ and $p(12)$ to find the value of $c$ :

$\begin{aligned} p(11) + 5 &= p(12) \\ 25(11) - 24 + c(10)(9)\cdots(1) + 5 &= 25(12) - 24 + c(11)(10)\cdots(2) \\ 256 + 10! \cdot c &= 276 + 11! \cdot c \\ (11! - 10!)c &= -20 \\ 10(10!)c &= -20 \\ c &= -\dfrac{2}{10!}. \end{aligned}$

Thus,

$p(x) = 25x - 24 - \dfrac{2}{10!} \displaystyle \prod_{i = 1}^{10} (x - i). \\$

Note that both $p(11) = 251 - \dfrac{2}{10!}(10!) = 249$ and $p(12) = 276 - \dfrac{2}{10!}(11!) = 254$ are positive integers, so they can represent the number of presents my true love will give to me.

Plugging in $x = 13$ yields

$p(13) = 25(13) - 24 - \dfrac{2}{10!}(12)(11)\cdots(3) = 301 - 12 \cdot 11 = \boxed{169}.$