Difference between revisions of "2017 AMC 8 Problems/Problem 19"
m (→Solution 3) |
m (→Solution 2) |
||
Line 8: | Line 8: | ||
==Solution 2== | ==Solution 2== | ||
− | + | Also keep in mind that number of 5’s in 98!(10,000) is the same as the number of trailing zeros. Number of zeros is 98! means we need pairs of 5’s and 2’s; we know there will be many more 2’s, so we seek to find number of 5’s in 98! which solution tells us and that is 22 factors of 5. 10,000 has 4 trailing zeros, so it has 4 factors of 5 and 22 + 4 = 26. | |
== Video Solution == | == Video Solution == |
Revision as of 19:45, 20 April 2021
Problem
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
Also keep in mind that number of 5’s in 98!(10,000) is the same as the number of trailing zeros. Number of zeros is 98! means we need pairs of 5’s and 2’s; we know there will be many more 2’s, so we seek to find number of 5’s in 98! which solution tells us and that is 22 factors of 5. 10,000 has 4 trailing zeros, so it has 4 factors of 5 and 22 + 4 = 26.
Video Solution
https://youtu.be/HISL2-N5NVg?t=817
~ pi_is_3.14
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.