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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==
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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>. 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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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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Also keep in mind that number of 5’s in 98!(10,000) is the same as the number of trailing zeros. Number of zeros is 98! means we need pairs of 5’s and 2’s; we know there will be many more 2’s, so we seek to find number of 5’s in 98! which solution tells us and that is 22 factors of 5. 10,000 has 4 trailing zeros, so it has 4 factors of 5 and 22 + 4 = 26.
 
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== Video Solution ==
 
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https://youtu.be/alj9Y8jGNz8
 
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https://youtu.be/HISL2-N5NVg?t=817
 
  
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~ pi_is_3.14
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~CHECKMATE2021
  
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==See Also==
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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?
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{{AMC8 box|year=2017|num-b=18|num-a=20}}
 
  
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{{MAA Notice}}
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==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
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https:/90ijn bidxrfgv

Latest revision as of 20:23, 13 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

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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