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

m (Video Solution)
(Video Solution)
 
(18 intermediate revisions by 10 users not shown)
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
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>.
+
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>.
  
==Solution 2==
+
~CHECKMATE2021
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 ==
+
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
https://youtu.be/alj9Y8jGNz8
+
https://youtu.be/WKux87BEO1U
  
https://youtu.be/HISL2-N5NVg?t=817
+
~Education, the Study of Everything
 
 
~ pi_is_3.14
 
  
 
==See Also==
 
==See Also==

Latest revision as of 22:34, 24 May 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

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/WKux87BEO1U

~Education, the Study of Everything

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png