Difference between revisions of "2017 AMC 8 Problems/Problem 19"
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==Solution 3== | ==Solution 3== | ||
| − | We can first factor a <math>98!</math> out of the | + | We can first factor a <math>98!</math> out of the <math>98! + 99! + 100!</math> to get <math>98! ( 1 + 99 + 100*99 ),</math> Simplify to get <math>98! (10,000)</math>. |
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ||
Revision as of 17:05, 21 January 2024
Contents
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 .The 
is because of all the multiples of 
. Now, 
has 
factors of , so there are a total of 
factors of .
~CHECKMATE2021
Solution 2
Also, keep in mind that the number of ’s in 
is the same as the number of trailing zeros. The number of zeros is , which means we need pairs of ’s and 
’s; we know there will be many more ’s, so we seek to find the number of ’s in , which the solution tells us. And, that is 
factors of . has trailing zeros, so it has factors of and 
.
~CHECKMATE2021
Solution 3
We can first factor a out of the 
to get 
Simplify to get .
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/HISL2-N5NVg?t=817
~ pi_is_3.14
Video Solution
~savannahsolver
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. 
