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!(1+99+99*100)</math> which is <math>98!( | + | Factoring out <math>98!+99!+100!</math>, we have <math>98!(1+99+99*100)</math> which is <math>98!(10000)</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 20:32, 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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