Difference between revisions of "2017 AMC 8 Problems/Problem 19"
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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
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 
. Next, has 
factors of 
. Now 
has 
factors of , so there are a total of 
factors of .
Solution 2
The number of 5's in the factorization of 
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 , you get 
. To find the number of trailing seroes in 98!, we do
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
