2017 AMC 8 Problems/Problem 19

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

Solution 3

We can first factor a $98!$ out of the $98! + 99! + 100!$ to get $98! ( 1 + 99 + 100*99 ),$ Simplify to get $98! (10,000)$ .

Let's first find how many factors of $5 10,000$ has. $10,000$ is $(2*5)^4$ because $10,000$ is $(10)^4$ . After we remove the brackets, we get $2^4$ , and $5^4$ . We only care about the latter (second one), because the problem only ask's for the power of $5$ . We get $4$

Next, we can look at the multiples of 5 in $98!$ . $98/5 = 19$ so there is 19 multiples of 5. We get $19$

But we cannot forget the multiples of $5$ with $2$ fives in it. Multiples of $25$ . How many multiples of $25$ are between $1$ and $98$ ? $3$ . $25,50,75,$ and that's it. We get $3$

Finally, we add all of the numbers (powers of $5$ ) up. That is $4 + 19 + 3$ , which is just $26$

So the answer is $26$ . Which is answer choice D $\boxed{\textbf{(D)}\ 26}$ .

~CHECKMATE2021

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/WKux87BEO1U

~Education, the Study of Everything

Video Solution by OmegaLearn

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

~ pi_is_3.14

Video Solution

https://youtu.be/alj9Y8jGNz8

https://youtu.be/meEuDzrM5Ac

~savannahsolver

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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