Difference between revisions of "2020 AMC 10A Problems/Problem 17"
(→Solution 2) |
(→Solution 3 (end behavior)) |
||
Line 28: | Line 28: | ||
We know that <math>P(x)</math> is a <math>100</math>-degree function with a positive leading coefficient. That is, <math>P(x)=x^{100}+ax^{99}+bx^{98}+...+\text{(constant)}</math>. | We know that <math>P(x)</math> is a <math>100</math>-degree function with a positive leading coefficient. That is, <math>P(x)=x^{100}+ax^{99}+bx^{98}+...+\text{(constant)}</math>. | ||
+ | |||
+ | Since the degree of <math>P(x)</math> is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach <math>\infty</math> as <math>x</math> goes in either direction. | ||
+ | |||
+ | <cmath>\lim_{x\to-\infty} P(x)=\lim_{x\to\infty} P(x)=\infty</cmath> | ||
==Video Solution== | ==Video Solution== |
Revision as of 11:03, 1 February 2020
DefineHow many integers
are there such that
?
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. 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 amount 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.
Video Solution
~IceMatrix
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.