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

(Solution 2)
(Solution 1)
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==Solution 1==
 
==Solution 1==
 
Factoring out <math>98!</math>, we have <math>98!(10,000)</math>. Next, <math>98!</math> has <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22</math> factors of <math>5</math>. Now <math>10,000</math> has <math>4</math> factors of <math>5</math>, so there are a total of <math>22 + 4 = \boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
 
Factoring out <math>98!</math>, we have <math>98!(10,000)</math>. Next, <math>98!</math> has <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22</math> factors of <math>5</math>. Now <math>10,000</math> has <math>4</math> factors of <math>5</math>, so there are a total of <math>22 + 4 = \boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
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==Solution 2==
 
==Solution 2==
 
The number of 5's in the factorization of <math>98! + 99! + 100!</math> is the same as the number fo trailing zeroes. The number of zeroes is taken by the floor value of each number divided by 5, until you can't divide by 5 anymore. Factorizing <math>98! + 99! + 100!</math>, you get <math>98!(1+99+9900)=98!(1000)</math>. To find the number of trailing seroes in 98!, we do
 
The number of 5's in the factorization of <math>98! + 99! + 100!</math> is the same as the number fo trailing zeroes. The number of zeroes is taken by the floor value of each number divided by 5, until you can't divide by 5 anymore. Factorizing <math>98! + 99! + 100!</math>, you get <math>98!(1+99+9900)=98!(1000)</math>. To find the number of trailing seroes in 98!, we do

Revision as of 14:36, 23 November 2017

Problem 19

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 $98!$, we have $98!(10,000)$. Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors 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

The number of 5's in the factorization of $98! + 99! + 100!$ is the same as the number fo trailing zeroes. The number of zeroes is taken by the floor value of each number divided by 5, until you can't divide by 5 anymore. Factorizing $98! + 99! + 100!$, you get $98!(1+99+9900)=98!(1000)$. To find the number of trailing seroes in 98!, we do

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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

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