2021 GMC 12B Problems/Problem 3

Problem 3

The expression $\frac{100!+99!}{99!+98!}$ can be written as $\frac{a}{b}$ which $a$ and $b$ are natural numbers and they are relatively prime. Find $a+b$ .

$\textbf{(A)} ~9999 \qquad\textbf{(B)} ~10009 \qquad\textbf{(C)} ~10099 \qquad\textbf{(D)} ~10999 \qquad\textbf{(E)} ~11009$

Solution

Simplifying, $\begin{align*} \dfrac{100! + 99!}{99! + 98!} &= \dfrac{99!(1 + 100)}{98!(1+99)} \\ &= \dfrac{99(101)}{100} \\ &= \dfrac{9999}{100}. \end{align*}$

Thus, since $a = 9999$ and $b = 100$ , $a + b = \boxed{\textbf{(C)}~10099}$ .

~pineconee

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2021 GMC 12B Problems/Problem 3

Problem 3

Solution