Difference between revisions of "2020 AMC 10A Problems/Problem 17"
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Case 1: There are 100 integers <math>n</math> for which <math>P(x)=0</math> | Case 1: There are 100 integers <math>n</math> for which <math>P(x)=0</math> | ||
− | 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+10+\dots+194+198</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 (the odd number squared is smaller than the even number squared). This means that there are <math>2+6+10+\dots+194+198</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</math>. ~PCChess | Summing, there are <math>\boxed{\textbf{(E) } 5100}</math> total possible values of <math>n</math>. ~PCChess |
Revision as of 00:07, 1 February 2021
Contents
Problem
Define How many integers are there such that ?
Solutions
Solution 1
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 (the odd number squared is smaller than the even number squared). This means that there are total possible values of . Simplifying, there are possible numbers.
Summing, there are total possible values of . ~PCChess
Solution 2
Notice that is nonpositive when is between and , and , and (inclusive), which means that the number of values equals .
This reduces to
~Zeric
Solution 3 (end behavior)
We know that is a -degree function with a positive leading coefficient. That is, .
Since the degree of is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach as goes in either direction.
So the first time is going to be negative is when it intersects the -axis at an -intercept and it's going to dip below. This happens at , which is the smallest intercept.
However, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at . And when it hits , it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until .
To get the amount of integers below and/or on the -axis, we simply need to count the integers. For example, the amount of integers in between the interval we got earlier, we subtract and add one. integers, so there are four integers in this interval that produce a negative result.
Doing this with all of the other intervals, we have
. Proceed with Solution 2. ~quacker88
Video Solution
https://youtu.be/3dfbWzOfJAI?t=4026
~ pi_is_3.14
~Education, The Study of Everything
~IceMatrix
https://www.youtube.com/watch?v=YDMMhSguq0w&list=PLeFyQ1uCoINM4D5Lgi5Y3KkfvQuYuIbj
-Walt S.
~savannahsolver
https://youtu.be/R220vbM_my8?t=463
~ amritvignesh0719062.0
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.