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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>
  
−
== Solution 1 ==
+
== Solution 1 (Casework) ==
−
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 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
 
−
 
 
−
== 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), because there are an odd number of negatives, 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 \cdot\frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}</math>
 
−
 
 
−
~Zeric
 
−
-jesselan made minor edits
 
−
 
 
−
== 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
 
−
 
 
−
== Solution 4 (Similar to Solution 1: Casework) ==
 
 
We perform casework on <math>P(n)\leq0:</math>
 
We perform casework on <math>P(n)\leq0:</math>
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 
   <li><math>P(n)=0</math></li><p>
 
   <li><math>P(n)=0</math></li><p>
−
In this case, there are <math>100</math> such integers <math>n:</math> <cmath>1^2,2^2,3^2,\cdots,100^2.</cmath>
+
In this case, there are <math>100</math> such integers <math>n:</math> <cmath>1^2,2^2,3^2,\ldots,100^2.</cmath>
 
   <li><math>P(n)<0</math></li><p>
 
   <li><math>P(n)<0</math></li><p>
 
There are <math>100</math> factors in <math>P(x),</math> and we need an odd number of them to be negative. We construct the table below:
 
There are <math>100</math> factors in <math>P(x),</math> and we need an odd number of them to be negative. We construct the table below:
Line 57: Line 23:
 
\left(4^2,5^2\right) & 96 & \\ [0.5ex]
 
\left(4^2,5^2\right) & 96 & \\ [0.5ex]
 
\left(5^2,6^2\right) & 95 & \checkmark \\ [0.5ex]
 
\left(5^2,6^2\right) & 95 & \checkmark \\ [0.5ex]
−
\left(6^2,7^2\right) & 94 & \\ [0.5ex]
+
\left(6^2,7^2\right) & 94 & \\
−
\cdots & \cdots & \cdots \\ [0.5ex]
+
\vdots & \vdots & \vdots \\ [0.75ex]
 
\left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex]
 
\left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex]
 
\left(100^2,\infty\right) & 0 &  \\ [0.5ex]
 
\left(100^2,\infty\right) & 0 &  \\ [0.5ex]
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Note that there are <math>50</math> valid intervals of <math>x.</math> We count the integers in these intervals:  
 
Note that there are <math>50</math> valid intervals of <math>x.</math> We count the integers in these intervals:  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\left(2^2-1^2-1\right)+\left(4^2-3^2-1\right)+\left(6^2-5^2-1\right)+\cdots+\left(100^2-99^2-1\right)&=\underbrace{\bigl(2^2-1^2\bigr)}_{(2+1)(2-1)}+\underbrace{\bigl(4^2-3^2\bigr)}_{(4+3)(4-3)}+\underbrace{\bigl(6^2-5^2\bigr)}_{(6+5)(6-5)}+\cdots+\underbrace{\left(100^2-99^2\right)}_{(100+99)(100-99)}-50 \\
+
\left(2^2-1^2-1\right)+\left(4^2-3^2-1\right)+\left(6^2-5^2-1\right)+\cdots+\left(100^2-99^2-1\right)&=\underbrace{\left(2^2-1^2\right)}_{(2+1)(2-1)}+\underbrace{\left(4^2-3^2\right)}_{(4+3)(4-3)}+\underbrace{\left(6^2-5^2\right)}_{(6+5)(6-5)}+\cdots+\underbrace{\left(100^2-99^2\right)}_{(100+99)(100-99)}-50 \\
 
&=\underbrace{(2+1)+(4+3)+(6+5)+\cdots+(100+99)}_{1+2+3+4+5+6+\cdots+99+100}-50 \\
 
&=\underbrace{(2+1)+(4+3)+(6+5)+\cdots+(100+99)}_{1+2+3+4+5+6+\cdots+99+100}-50 \\
 
&=\frac{101(100)}{2}-50 \\
 
&=\frac{101(100)}{2}-50 \\
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Together, the answer is <math>100+5000=\boxed{\textbf{(E) } 5100}.</math>
 
Together, the answer is <math>100+5000=\boxed{\textbf{(E) } 5100}.</math>
  
−
~MRENTHUSIASM
+
~PCChess (Solution)
 +
 
 +
~MRENTHUSIASM (Reformatting)
 +
 
 +
== Solution 2 (Casework) ==
 +
Notice that <math>P(x)</math> is nonpositive when <math>x</math> is between <math>100^2</math> and <math>99^2, 98^2</math> and <math>97^2, \cdots</math> , <math>2^2</math> and <math>1^2</math> (inclusive), because there are an odd number of negatives, which means that the number of values equals <cmath>((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \cdots + ((2+1)(2-1)+1).</cmath>
 +
This reduces to <cmath>200 + 196 + 192 + \cdots + 4 = 4(1+2+\cdots + 50) = 4 \cdot\frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}.</cmath>
 +
~Zeric
 +
 
 +
~jesselan (Minor Edits)
 +
 
 +
== 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, from which <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 <cmath>(2^2-1^2+1)+(4^2-3^2+1)+\cdots+(100^2-99^2+1)=\boxed{\textbf{(E) } 5100}</cmath> from Solution 2's result.
 +
 
 +
~quacker88
 +
 
 +
== Solution 4 (Fast) ==
 +
We know <math>P(x) \leq 0</math> when an odd number of its factors are positive and negative. For example, to make the first factor positive, <math>x \in [1^2, 2^2]</math>. then there will be a even number of positive factors. We would do <math>2^2 - 1^2 + 1 (\text{inclusive})</math> to find all integers that work. In short we can generalize too:
 +
<cmath>\begin{align*}
 +
x^2 - (x-1)^2 + 1 &= 2x \\
 +
x^2 - (x^2 - 2x + 1) + 1 &= 2x \\
 +
x^2 - x^2 + 2x - 1 + 1 &= 2x. \\
 +
\end{align*}</cmath>
 +
But remember this only works when <math>x \in \{2, 4, 6, 8 \cdots 98, 100\}</math> because only then will there be a odd amount of positive and negative factors. So we can set <math>x = 2k</math>, for <math>k \in \{1, 2, 3, 4, \cdots 49, 50\}</math> Now we only have to solve:
 +
<cmath>\sum_{k=1}^{k=50}2(2k) = 2\sum_{k = 1}^{k = 50}2k = 4\sum_{k = 1}^{k = 50}k = 4 \cdot \dfrac{(50)(51)}{2} = 2 \cdot (50)(51) = \boxed{\textbf{(E) } 5100}.</cmath>
 +
~Wiselion
  
 
== Video Solutions ==
 
== Video Solutions ==

Revision as of 14:27, 4 October 2024

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 (Casework)

We perform casework on $P(n)\leq0:$

  1. $P(n)=0$

    2. In this case, there are $100$ such integers $n:$ \[1^2,2^2,3^2,\ldots,100^2.\]

  2. $P(n)<0$

    3. There are $100$ factors in $P(x),$ and we need an odd number of them to be negative. We construct the table below: \[\begin{array}{c|c|c} & & \\ [-2.5ex] \textbf{Interval of }\boldsymbol{x} & \boldsymbol{\#}\textbf{ of Negative Factors} & \textbf{Valid?} \\ [0.5ex] \hline & & \\ [-2ex] \left(-\infty,1^2\right) & 100 & \\ [0.5ex] \left(1^2,2^2\right) & 99 & \checkmark \\ [0.5ex] \left(2^2,3^2\right) & 98 & \\ [0.5ex] \left(3^2,4^2\right) & 97 & \checkmark \\ [0.5ex] \left(4^2,5^2\right) & 96 & \\ [0.5ex] \left(5^2,6^2\right) & 95 & \checkmark \\ [0.5ex] \left(6^2,7^2\right) & 94 & \\ \vdots & \vdots & \vdots \\ [0.75ex] \left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex] \left(100^2,\infty\right) & 0 & \\ [0.5ex] \end{array}\] Note that there are $50$ valid intervals of $x.$ We count the integers in these intervals: \begin{align*} \left(2^2-1^2-1\right)+\left(4^2-3^2-1\right)+\left(6^2-5^2-1\right)+\cdots+\left(100^2-99^2-1\right)&=\underbrace{\left(2^2-1^2\right)}_{(2+1)(2-1)}+\underbrace{\left(4^2-3^2\right)}_{(4+3)(4-3)}+\underbrace{\left(6^2-5^2\right)}_{(6+5)(6-5)}+\cdots+\underbrace{\left(100^2-99^2\right)}_{(100+99)(100-99)}-50 \\ &=\underbrace{(2+1)+(4+3)+(6+5)+\cdots+(100+99)}_{1+2+3+4+5+6+\cdots+99+100}-50 \\ &=\frac{101(100)}{2}-50 \\ &=5000. \end{align*} In this case, there are $5000$ such integers $n.$

Together, the answer is $100+5000=\boxed{\textbf{(E) } 5100}.$

~PCChess (Solution)

~MRENTHUSIASM (Reformatting)

Solution 2 (Casework)

Notice that $P(x)$ is nonpositive when $x$ is between $100^2$ and $99^2, 98^2$ and $97^2, \cdots$ , $2^2$ and $1^2$ (inclusive), because there are an odd number of negatives, which means that the number of values equals \[((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \cdots + ((2+1)(2-1)+1).\] This reduces to \[200 + 196 + 192 + \cdots + 4 = 4(1+2+\cdots + 50) = 4 \cdot\frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}.\] ~Zeric

~jesselan (Minor Edits)

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, from which \[\lim_{x\to-\infty} P(x)=\lim_{x\to\infty} P(x)=\infty.\] 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 \[(2^2-1^2+1)+(4^2-3^2+1)+\cdots+(100^2-99^2+1)=\boxed{\textbf{(E) } 5100}\] from Solution 2's result.

~quacker88

Solution 4 (Fast)

We know $P(x) \leq 0$ when an odd number of its factors are positive and negative. For example, to make the first factor positive, $x \in [1^2, 2^2]$. then there will be a even number of positive factors. We would do $2^2 - 1^2 + 1 (\text{inclusive})$ to find all integers that work. In short we can generalize too: \begin{align*} x^2 - (x-1)^2 + 1 &= 2x \\ x^2 - (x^2 - 2x + 1) + 1 &= 2x \\ x^2 - x^2 + 2x - 1 + 1 &= 2x. \\ \end{align*} But remember this only works when $x \in \{2, 4, 6, 8 \cdots 98, 100\}$ because only then will there be a odd amount of positive and negative factors. So we can set $x = 2k$, for $k \in \{1, 2, 3, 4, \cdots 49, 50\}$ Now we only have to solve: \[\sum_{k=1}^{k=50}2(2k) = 2\sum_{k = 1}^{k = 50}2k = 4\sum_{k = 1}^{k = 50}k = 4 \cdot \dfrac{(50)(51)}{2} = 2 \cdot (50)(51) = \boxed{\textbf{(E) } 5100}.\] ~Wiselion

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)
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Problem 16 		Followed by
Problem 18
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All AMC 10 Problems and Solutions

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