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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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== Solutions ==
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== Solution 1 ==
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=== 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.
 
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.
  
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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
  
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=== Solution 2 ===
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== 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>.
 
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>.
  
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~Zeric
 
~Zeric
  
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=== Solution 3 (end behavior) ===
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== Solution 3 (End Behavior) ==
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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>.
  
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<math>(2^2-1^2+1)+(4^2-3^2+1)+...+(100^2-99^2+1)</math>. Proceed with Solution 2. ~quacker88
 
<math>(2^2-1^2+1)+(4^2-3^2+1)+...+(100^2-99^2+1)</math>. Proceed with Solution 2. ~quacker88
 +
 +
== Solution 4 (Casework) ==
 +
We apply casework to <math>P(n)\leq0:</math>
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>P(n)=0</math></li><p>
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There are <math>100</math> such integers <math>n:</math> <cmath>1^2,2^2,3^2,\cdots,100^2.</cmath>
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  <li><math>P(n)<0</math></li><p>
 +
</ol>
 +
 +
~MRENTHUSIASM
  
 
== Video Solutions ==
 
== Video Solutions ==

Revision as of 15:04, 24 April 2021

Problem

Define \[P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).\] How many integers $n$ are there such that $P(n)\leq 0$?

$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

Solution 1

Notice that $P(x)$ 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 $n$ for which $P(x)=0$

Case 2: For there to be an odd number of negative factors, $n$ 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 $2+6+10+\dots+194+198$ total possible values of $n$. Simplifying, there are $5000$ possible numbers.

Summing, there are $\boxed{\textbf{(E) } 5100}$ total possible values of $n$. ~PCChess

Solution 2

Notice that $P(x)$ is nonpositive when $x$ is between $100^2$ and $99^2$, $98^2$ and $97^2, \ldots$ , $2^2$ and $1^2$ (inclusive), which means that the number of values equals $((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \ldots + ((2+1)(2-1)+1)$.

This reduces to $200 + 196 + 192 + \ldots + 4 = 4(1+2+\ldots + 50) = 4 \cdot\frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}$

~Zeric

Solution 3 (End Behavior)

We know that $P(x)$ is a $100$-degree function with a positive leading coefficient. That is, $P(x)=x^{100}+ax^{99}+bx^{98}+...+\text{(constant)}$.

Since the degree of $P(x)$ is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach $\infty$ as $x$ goes in either direction.

\[\lim_{x\to-\infty} P(x)=\lim_{x\to\infty} P(x)=\infty\]

So the first time $P(x)$ is going to be negative is when it intersects the $x$-axis at an $x$-intercept and it's going to dip below. This happens at $1^2$, 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 $2^2$. And when it hits $3^2$, it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until $100^2$.

To get the amount of integers below and/or on the $x$-axis, we simply need to count the integers. For example, the amount of integers in between the $[1^2,2^2]$ interval we got earlier, we subtract and add one. $(2^2-1^2+1)=4$ integers, so there are four integers in this interval that produce a negative result.

Doing this with all of the other intervals, we have

$(2^2-1^2+1)+(4^2-3^2+1)+...+(100^2-99^2+1)$. Proceed with Solution 2. ~quacker88

Solution 4 (Casework)

We apply casework to $P(n)\leq0:$

  1. $P(n)=0$

    2. There are $100$ such integers $n:$ \[1^2,2^2,3^2,\cdots,100^2.\]

  2. $P(n)<0$

~MRENTHUSIASM

Video Solutions

https://youtu.be/3dfbWzOfJAI?t=4026

~ pi_is_3.14

https://youtu.be/zl5rtHnk0rY

~Education, The Study of Everything

https://youtu.be/RKlG6oZq9so

~IceMatrix

https://www.youtube.com/watch?v=YDMMhSguq0w&list=PLeFyQ1uCoINM4D5Lgi5Y3KkfvQuYuIbj

-Walt S.

https://youtu.be/chDmeTQBxq8

~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

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