Difference between revisions of "2020 AMC 10A Problems/Problem 17"
| Line 9: | Line 9: | ||
Case 2: For there to be an odd number of negative factors, <math>n</math> must be between an odd number squared and an even number squared. This means that there are <math>2+6+\dots+10</math> total possible values of <math>n</math>. Simplifying, there are <math>5000</math> possible numbers. | Case 2: For there to be an odd number of negative factors, <math>n</math> must be between an odd number squared and an even number squared. This means that there are <math>2+6+\dots+10</math> total possible values of <math>n</math>. Simplifying, there are <math>5000</math> possible numbers. | ||
| − | Summing, there are <math>\boxed{\textbf{(E) }} | + | Summing, there are <math>\boxed{\textbf{(E) } 5100}</math> total possible values of <math>n</math>. |
Revision as of 21:07, 31 January 2020
Define![\[P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).\]](http://latex.artofproblemsolving.com/4/4/6/4468a133d4086845edcd539fce61e77b481d271b.png)
How many integers 
are there such that 
?
Solution
Notice that 
is a product of many integers. We either need one factor to be 0 or an odd number of negative factors.
Case 1: There are 100 integers for which 
Case 2: For there to be an odd number of negative factors, must be between an odd number squared and an even number squared. This means that there are 
total possible values of . Simplifying, there are 
possible numbers.
Summing, there are 
total possible values of .
See Also
| 2020 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
