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 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 number of <math>5</math>’s in <math>98!(10,000)</math> is the same as the number of trailing zeros. Number of zeros is <math>98!</math> 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 number of <math>5</math>’s in <math>98!</math> which 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 = 26</math>.
  
==Solution 3==
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== Video Solution ==
We can rewrite the expression as <math>98!+99!+100!=98!(1+99+99\cdot100)=98!(100+99\cdot100)=98!\cdot10,000</math>.
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https://youtu.be/alj9Y8jGNz8
The exponent of <math>5</math> in <math>10,000</math> is <math>4</math>. Onto the <math>98!</math> part.
 
  
Remember that <math>98!</math> is the product of the integers from 1 to 98. Among these, there are multiples of <math>5</math> and multiples of <math>25</math>.
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https://youtu.be/HISL2-N5NVg?t=817
The number of multiples of <math>5</math> below or equal to <math>98</math> is <math>\left\lfloor\frac{98}{5}\right\rfloor</math> (try to see why), which is <math>19</math>. Every such number has a factor of <math>5</math>, so they contribute <math>1</math> to the total each. So these numbers contribute <math>19</math> to the exponent of <math>5</math> in <math>98!</math>.
 
  
However, we forgot about multiples of <math>25</math>. Using similar logic, we have <math>\left\lfloor\frac{98}{25}\right\rfloor=3</math>, so there are 3 multiples of 25 in this range. Multiples of 25 contribute 2 to the total each. We already counted 1 of 2 contributions while counting multiples of 5. So we need to add another 1 to the exponent of 5 for every multiple of 25. So these numbers contribute 3 to the exponent of 5 in <math>98!</math>.
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~ pi_is_3.14
  
 
We thus have that the exponent of <math>5</math> in <math>98!</math> is <math>19 + 3 = 22</math>, and so our answer is <math>22 + 4 = 26 \longrightarrow \boxed{(D)26}</math>.
 
 
~Math4Life2020
 
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2017|num-b=18|num-a=20}}
 
{{AMC8 box|year=2017|num-b=18|num-a=20}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:37, 28 June 2021

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 $5$. Now $10,000$ has $4$ factors of $5$, so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$.

Solution 2

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 $22 + 4 = 26$.

Video Solution

https://youtu.be/alj9Y8jGNz8

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

~ pi_is_3.14

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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