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

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==Problem 19==
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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!(10,000)</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>. 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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The number of <math>5</math>'s in the factorization of <math>98! + 99! + 100!</math> is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by <math>5</math>, until you can't divide by <math>5</math> anymore. Factorizing <math>98! + 99! + 100!</math>, you get <math>98!(1+99+9900)=98!(10000)</math>. To find the number of trailing zeroes in 98!, we do <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22</math>. Now since <math>10000</math> has 4 zeroes, we add <math>22 + 4</math>  to get <math>\boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
 
  
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==See Also==
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~CHECKMATE2021
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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 20:38, 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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