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

Revision as of 19:45, 20 April 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/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.

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 ==Solution 2==
-The number of <math>5</math>'s in the factorization of <math>98! + 99! + 100!</math> is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by <math>5</math>, until you can't divide by <math>5</math> anymore. Factorizing <math>98! + 99! + 100!</math>, you get <math>98!(1+99+9900)=98!(10000)</math>. To find the number of trailing zeroes in 98!, we do <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22</math>. Now since <math>10000</math> has 4 zeroes, we add <math>22 + 4</math>  to get <math>\boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
+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 ==

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