Difference between revisions of "2017 AMC 8 Problems/Problem 19"
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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>. | ||
==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. | ||
==See Also== | ==See Also== | ||
Revision as of 14:33, 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.
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. 
