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

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==Problem 19==
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==Problem==
 
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> ?
 
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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==Solution 1==
 
==Solution 1==
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{(E)}\ 27}</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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~CHECKMATE2021
  
 
==Solution 2==
 
==Solution 2==
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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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>.
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~CHECKMATE2021
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==Solution 3==
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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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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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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 18:18, 21 January 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

Solution 2

Also, keep in mind that the number of $5$’s in $98! (10,000)$ is the same as the number of trailing zeros. The number of zeros is $98!$, which means we need pairs of $5$’s and $2$’s; we know there will be many more $2$’s, so we seek to find the number of $5$’s in $98!$, which the solution tells us. And, that is $22$ factors of $5$. $10,000$ has $4$ trailing zeros, so it has $4$ factors of $5$ and $22 + 4 = \boxed{26}$.

~CHECKMATE2021


Solution 3

We can first factor a $98!$ out of the $98! + 99! + 100!$ to get $98! ( 1 + 99 + 100*99 ),$ Simplify to get $98! (10,000)$.

Let's first find how many factors of $5 10,000$ has. $10,000$ is $(2*5)^4$ because $10,000$ is $(10)^4$. After we remove the brackets, we get $2^4$, and $5^4$. We only care about the latter (second one), because the problem only ask's for the power of $5$. We get $4$

Next, we can look at the multiples of 5 in $98!$. $98/5 = 19$ so there is 19 multiples of 5. We get $19$

But we cannot forget the multiples of $5$ with $2$ fives in it. Multiples of $25$. How many multiples of $25$ are between $1$ and $98$? $3$. $25,50,75,$ and that's it. We get $3$

Finally, we add all of the numbers (powers of $5$) up. That is $4 + 19 + 3$, which is just $26$

So the answer is $26$. Which is answer choice D $\boxed{\textbf{(D)}\ 26}$.

~CHECKMATE2021

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/WKux87BEO1U

~Education, the Study of Everything

Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=817

~ pi_is_3.14

Video Solution

https://youtu.be/alj9Y8jGNz8

https://youtu.be/meEuDzrM5Ac

~savannahsolver

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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