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

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==Problem==
 
==Problem==
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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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For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through
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<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>  
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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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~CHECKMATE2021
 
~CHECKMATE2021
  
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==Solution 3==
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Note: Can you say what formula this uses? most AMC 8 test takers won't know it.  
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We can first factor a <math>98!</math> out of the <math>98! + 99! + 100!</math> to get <math>98! ( 1 + 99 + 100*99 ),</math> Simplify to get <math>98! (10,000)</math>.
 
  
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Let's first find how many factors of <math>5 10,000</math> has. <math>10,000</math> is <math>(2*5)^4</math> because <math>10,000</math> is <math>(10)^4</math>. After we remove the brackets, we get <math>2^4</math>, and <math>5^4</math>. We only care about the latter (second one), because the problem only ask's for the power of <math>5</math>. We get <math>4</math>
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==Video Solution (Omega Learn)==
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https://www.youtube.com/watch?v=HISL2-N5NVg&t=817s
  
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Next, we can look at the multiples of 5 in <math>98!</math>. <math>98/5 = 19</math> so there is 19 multiples of 5. We get <math>19</math>
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~ GeometryMystery
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But we cannot forget the multiples of <math>5</math> with <math>2</math> fives in it. Multiples of <math>25</math>. How many multiples of <math>25</math> are between <math>1</math> and <math>98</math>? <math>3</math>. <math>25,50,75,</math> and that's it. We get <math>3</math>
 
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Finally, we add all of the numbers (powers of <math>5</math>) up. That is <math>4 + 19 + 3</math>, which is just <math>26</math>
 
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So the answer is <math>26</math>. Which is answer choice D <math>\boxed{\textbf{(D)}\ 26}</math>.
 
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~CHECKMATE2021
 
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==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 
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https://youtu.be/WKux87BEO1U
 
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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
 
  
 
==See Also==
 
==See Also==

Latest revision as of 03:20, 13 October 2024

Problem

For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?

$\textbf{(A) }23 \qquad \textbf{(B) }24 \qquad \textbf{(C) }25 \qquad \textbf{(D) }26 \qquad \textbf{(E) }27$

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.

Video Solution (Omega Learn)

https://www.youtube.com/watch?v=HISL2-N5NVg&t=817s

~ GeometryMystery

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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