Question:
Let $p(x)$ be a polynomial of degree 100 such that
\begin{align*}
p(1) &= 1, \\
p(2) &= 2, \\
p(3) &= 3, \\
&\dots, \\
p(99) &= 99, \\
p(100) &= 100, \\
p(101) &= 102.
\end{align*}Find $p(102).$

Answer:
Let $q(x) = p(x) - x.$  Then $q(x)$ has degree 100, and $q(1) = q(2) = \dots = q(100) = 0,$ so
\[q(x) = c(x - 1)(x - 2) \dotsm (x - 100)\]for some constant $c.$  Since $p(101) = 102,$ $q(101) = 1.$  Setting $x = 101$ in the equation above, we get
\[q(101) = 100! \cdot c,\]so $c = \frac{1}{100!}.$  Then
\[q(x) = \frac{(x - 1)(x - 2) \dotsm (x - 100)}{100!}.\]In particular,
\[q(102) = \frac{101 \cdot 100 \dotsm 2}{100!} = 101,\]so $p(102) = q(102) + 102 = 101 + 102 = \boxed{203}.$