Difference between revisions of "2020 AMC 10A Problems/Problem 17"
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<math>\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100</math> | <math>\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100</math> | ||
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| + | ==Solution== | ||
| + | |||
| + | Notice that <math>P(x)</math> 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 <math>n</math> for which <math>P(x)=0 | ||
| + | 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) }} 5100<math> total possible values of </math>n$. | ||
| + | |||
==See Also== | ==See Also== | ||
Revision as of 22:05, 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 
n
2+6+\dots+10
n
5000$possible numbers.
Summing, there are$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(E) }} 5100n$.
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 | ||
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