A Pre-RMO question! -19

From the conditions given: $f(1) = 1, f(2) = 2 \ldots f(99) = 99$ , we can generate a polynomial from the given conditions whose degree is 99.

$f(x) = a(x-1)(x-2)(x-3)\cdots(x-99) + x$

With the extra condition given that $f(0) = 1$ , we can find the value of $a$

$\begin{aligned} f(0) &= a(0-1)(0-2)(0-3)\cdots(0-99) + 0 \\ 1 &= a \times -99! \\ a &=- \dfrac{1}{99!} \end{aligned}$

From here, we get the exact polynomial, and we can substitue $-1 = x$ to find the value of $f(-1)$

$\begin{aligned} f(x) &= -\dfrac{1}{99!}(x-1)(x-2)(x-3)\cdots(x-99) + x \\ &= -\dfrac{1}{99!}(-1-1)(-1-2)(-1-3)\cdots(-1-99) -1 \\ &= -\dfrac{1}{99!}(-2)(-3)(-4)\cdots(-100) -1 \\ &= -\dfrac{1}{99!} \cdot (-100!) -1 \\ &= 100 -1 \\ f(-1) &= \boxed{99} \end{aligned}$