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

Revision as of 19:53, 9 March 2022

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

https://youtu.be/meEuDzrM5Ac

~savannahsolver

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.

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