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

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==Problem==
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wertesryrtutyrudtu
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For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through <math>M</math>. What is the largest integer <math>n</math> for which <math>5^n</math> is a factor of the sum <math>98!+99!+100!</math> ?
 
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<math>\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math>
 
  
 
==Solution 1==
 
==Solution 1==
 
Factoring out <math>98!+99!+100!</math>, we have <math>98! (1+99+99*100)</math>, which is <math>98! (10000)</math>. Next, <math>98!</math> has <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22</math> factors of <math>5</math>. The <math>19</math> is because of all the multiples of <math>5</math>.The <math>3</math> is because of all the multiples of <math>25</math>. Now, <math>10,000</math> has <math>4</math> factors of <math>5</math>, so there are a total of <math>22 + 4 = \boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
 
Factoring out <math>98!+99!+100!</math>, we have <math>98! (1+99+99*100)</math>, which is <math>98! (10000)</math>. Next, <math>98!</math> has <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22</math> factors of <math>5</math>. The <math>19</math> is because of all the multiples of <math>5</math>.The <math>3</math> is because of all the multiples of <math>25</math>. Now, <math>10,000</math> has <math>4</math> factors of <math>5</math>, so there are a total of <math>22 + 4 = \boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
  
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==Solution 2==
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~CHECKMATE2021
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Also, keep in mind that the number of <math>5</math>’s in <math>98! (10,000)</math> is the same as the number of trailing zeros. The number of zeros is <math>98!</math>, which means we need pairs of <math>5</math>’s and <math>2</math>’s; we know there will be many more <math>2</math>’s, so we seek to find the number of <math>5</math>’s in <math>98!</math>, which the solution tells us. And, that is <math>22</math> factors of <math>5</math>. <math>10,000</math> has <math>4</math> trailing zeros, so it has <math>4</math> factors of <math>5</math> and <math>22 + 4 = \boxed{26}</math>.
 
  
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
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https://youtu.be/WKux87BEO1U
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https:/90ijn bidxrfgv
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~Education, the Study of Everything
 
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== Video Solution by OmegaLearn==
 
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https://youtu.be/HISL2-N5NVg?t=817
 
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~ pi_is_3.14
 
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== Video Solution ==
 
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https://youtu.be/alj9Y8jGNz8
 
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https://youtu.be/meEuDzrM5Ac
 
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~savannahsolver
 
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==See Also==
 
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{{AMC8 box|year=2017|num-b=18|num-a=20}}
 
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{{MAA Notice}}
 

Latest revision as of 20:38, 10 June 2024

wertesryrtutyrudtu

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

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https:/90ijn bidxrfgv

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