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Difference between revisions of "2021 GMC 12B Problems/Problem 3"

(Created page with "==Problem 3== The expression <math>\frac{100!+99!}{99!+98!}</math> can be written as <math>\frac{a}{b}</math> which <math>a</math> and <math>b</math> are natural numbers and t...")
 
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Latest revision as of 18:44, 25 April 2022

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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