2020 AMC 10A Problems/Problem 17
Define How many integers are there such that ?
Solution 1 (Casework)
We perform casework on
In this case, there are such integers
There are factors in and we need an odd number of them to be negative. We construct the table below: Note that there are valid intervals of We count the integers in these intervals: In this case, there are such integers
Together, the answer is
Solution 2 (Casework)
Notice that is nonpositive when is between and and , and (inclusive), because there are an odd number of negatives, which means that the number of values equals This reduces to ~Zeric
~jesselan (Minor Edits)
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, from which 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 from Solution 2's result.
~Education, The Study of Everything
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