Difference between revisions of "2020 AMC 10A Problems/Problem 17"
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− | Define<cmath>P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).</cmath>How many integers <math>n</math> are there such that <math>P(n)\leq 0</math>? | + | Define <cmath>P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).</cmath> How many integers <math>n</math> are there such that <math>P(n)\leq 0</math>? |
<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> | ||
− | ==Solution 1== | + | == Solutions == |
+ | === Solution 1 === | ||
+ | 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</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+\dots+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 | |
− | Notice that <math>P(x)</math> is nonpositive when <math>x</math> is between <math>100^2</math> and <math>99^2</math>, <math>98^2</math> and <math>97^2 \ldots</math> , <math>2^2</math> and <math>1^2</math> (inclusive), which means that the | + | === Solution 2 === |
+ | Notice that <math>P(x)</math> is nonpositive when <math>x</math> is between <math>100^2</math> and <math>99^2</math>, <math>98^2</math> and <math>97^2 \ldots</math> , <math>2^2</math> and <math>1^2</math> (inclusive), which means that the number of values equals <math>((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \ldots + ((2+1)(2-1)+1)</math>. | ||
− | This reduces to <math>200 + 196 + 192 + \ldots + 4 = 4(1+2+\ldots + 50) = 4 \frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}</math> | + | This reduces to <math>200 + 196 + 192 + \ldots + 4 = 4(1+2+\ldots + 50) = 4 \cdot\frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}</math> |
~Zeric | ~Zeric | ||
− | |||
+ | === Solution 3 (end behavior) === | ||
+ | |||
+ | 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> | ||
+ | |||
+ | So the first time <math>P(x)</math> is going to be negative is when it intersects the <math>x</math>-axis at an <math>x</math>-intercept and it's going to dip below. This happens at <math>1^2</math>, 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 <math>2^2</math>. And when it hits <math>3^2</math>, it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until <math>100^2</math>. | ||
+ | |||
+ | To get the amount of integers below and/or on the <math>x</math>-axis, we simply need to count the integers. For example, the amount of integers in between the <math>[1^2,2^2]</math> interval we got earlier, we subtract and add one. <math>(2^2-1^2+1)=4</math> integers, so there are four integers in this interval that produce a negative result. | ||
+ | |||
+ | Doing this with all of the other intervals, we have | ||
+ | |||
+ | <math>(2^2-1^2+1)+(4^2-3^2+1)+...+(100^2-99^2+1)</math>. Proceed with Solution 2. ~quacker88 | ||
+ | |||
+ | === Video Solution === | ||
+ | |||
+ | Education, The Study of Everything | ||
+ | |||
+ | https://youtu.be/zl5rtHnk0rY | ||
+ | |||
+ | |||
+ | https://youtu.be/RKlG6oZq9so | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | == See Also == | ||
{{AMC10 box|year=2020|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2020|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:53, 7 November 2020
Define How many integers are there such that ?
Contents
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. 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
Education, The Study of Everything
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
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Followed by Problem 18 | |
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