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

Revision as of 15:37, 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

2017 AMC 8 (Problems • Answer Key • Resources)
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.

@@ Line 8: / Line 8: @@
 ==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.
+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>.
 == Video Solution ==

Page

Toolbox

Search