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

Revision as of 20:59, 4 July 2024

wertesryrtutyrudtu

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 $98!+99!+100!$ ?

$\textbf{(A) }23 \qquad \textbf{(B) }24 \qquad \textbf{(C) }25 \qquad \textbf{(D) }26 \qquad \textbf{(E) }327$

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

Note: Can you say what formula this uses? most AMC 8 test takers won't know it. Also, can someone unvandalize this page?

Video Solution (Omega Learn)

https://www.youtube.com/watch?v=HISL2-N5NVg&t=817s

~ GeometryMystery

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 ~ GeometryMystery
+==See Also==
+{{AMC8 box|year=2017|num-b=18|num-a=20}}
+{{MAA Notice}}

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

Revision as of 20:59, 4 July 2024

Contents

Problem

Solution 1

Video Solution (Omega Learn)

See Also