Difference between revisions of "2017 AMC 8 Problems/Problem 19"
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==Solution 1== | ==Solution 1== | ||
| − | Factoring out <math>98!+99!+100!</math>, we have <math>98!( | + | Factoring out <math>98!+99!+100!</math>, we have <math>98!(1+99+99*100)</math> which is <math>98!(1000)</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>. The <math>19</math> is because of all the multiples 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>. |
==Solution 2== | ==Solution 2== | ||
Revision as of 19:30, 29 October 2020
Contents
Problem 19
For any positive integer 
, the notation 
denotes the product of the integers 
through . What is the largest integer 
for which 
is a factor of the sum 
?
Solution 1
Factoring out , we have 
which is 
Next, 
has 
factors of 
. The 
is because of all the multiples of . Now 
has 
factors of , so there are a total of 
factors of .
Solution 2
The number of 's in the factorization of 
is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by , until you can't divide by anymore. Factorizing , you get 
. To find the number of trailing zeroes in 98!, we do 
. Now since 
has 4 zeroes, we add 
to get 
factors of .
See Also
| 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 | ||
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