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Difference between revisions of "2017 AMC 8 Problems/Problem 19"

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~CHECKMATE2021
 
~CHECKMATE2021
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Note: Can you say what formula this uses? most AMC 8 test takers won't know it. Also, can someone unvandalize this page?
  
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 
https:/90ijn bidxrfgv
 
https:/90ijn bidxrfgv

Latest revision as of 20:23, 13 June 2024

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Solution 1

Factoring out $98!+99!+100!$, we have $98! (1+99+99*100)$, which is $98! (10000)$. Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$. The $19$ is because of all the multiples of $5$.The $3$ is because of all the multiples of $25$. Now, $10,000$ has $4$ factors of $5$, so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$.

~CHECKMATE2021

Note: Can you say what formula this uses? most AMC 8 test takers won't know it. Also, can someone unvandalize this page?

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https:/90ijn bidxrfgv

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