Difference between revisions of "2020 AMC 10A Problems/Problem 17"

m (Solution 4 (Casework))
m (Use \ldots instead of \cdots for lists.)
(10 intermediate revisions by 2 users not shown)
Line 5: Line 5:
 
<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.
+
We perform casework on <math>P(n)\leq0:</math>
 +
<ol style="margin-left: 1.5em;">
 +
  <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,\ldots,100^2.</cmath>
 +
  <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:
 +
<cmath>\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 & \\ [0.5ex]
 +
\cdots & \cdots & \cdots \\ [0.5ex]
 +
\left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex]
 +
\left(100^2,\infty\right) & 0 &  \\ [0.5ex]
 +
\end{array}</cmath>
 +
Note that there are <math>50</math> valid intervals of <math>x.</math> We count the integers in these intervals:
 +
<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{\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*}</cmath>
 +
In this case, there are <math>5000</math> such integers <math>n.</math>
 +
</ol>
 +
Together, the answer is <math>100+5000=\boxed{\textbf{(E) } 5100}.</math>
  
Case 1: There are 100 integers <math>n</math> for which <math>P(x)=0</math>
+
~PCChess (Solution)
  
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.
+
~MRENTHUSIASM (Reformatting)
 
 
Summing, there are <math>\boxed{\textbf{(E) } 5100}</math> total possible values of <math>n</math>. ~PCChess
 
  
 
== Solution 2 ==
 
== 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), 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>
 
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
 +
-jesselan made minor edits
  
 
== Solution 3 (End Behavior) ==
 
== Solution 3 (End Behavior) ==
Line 37: Line 67:
  
 
<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>
 
There are <math>100</math> such integers <math>n:</math> <cmath>1^2,2^2,3^2,\cdots,100^2.</cmath>
 
  <li><math>P(n)<0</math></li><p>
 
Note that <math>P(x)</math> has <math>100</math> factors, and we need an odd number of negative factors. We construct the table below:
 
<cmath>\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 &  \\ [0.5ex]
 
\left(2^2,3^2\right) & 98 & \checkmark \\ [0.5ex]
 
\left(3^2,4^2\right) & 97 &  \\ [0.5ex]
 
\left(4^2,5^2\right) & 96 & \checkmark \\ [0.5ex]
 
\left(5^2,6^2\right) & 95 &  \\ [0.5ex]
 
\left(6^2,7^2\right) & 94 & \checkmark \\ [0.5ex]
 
\cdots & \cdots &  \\ [0.5ex]
 
\left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex]
 
\left(100^2,\infty\right) & 0 &  \\ [0.5ex]
 
\end{array}</cmath>
 
</ol>
 
 
~MRENTHUSIASM
 
  
 
== Video Solutions ==
 
== Video Solutions ==

Revision as of 16:26, 8 September 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 (Casework)

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

  1. $P(n)=0$
  2. In this case, there are $100$ such integers $n:$ \[1^2,2^2,3^2,\ldots,100^2.\]

  3. $P(n)<0$
  4. There are $100$ factors in $P(x),$ and we need an odd number of them to be negative. We construct the table below: \[\begin{array}{c|c|c} & & \\ [-2.5ex] \textbf{Interval of }\boldsymbol{x} & \boldsymbol{\#}\textbf{ of Negative Factors} & \textbf{Valid?} \\ [0.5ex] \hline & & \\ [-2ex] \left(-\infty,1^2\right) & 100 & \\ [0.5ex] \left(1^2,2^2\right) & 99 & \checkmark \\ [0.5ex] \left(2^2,3^2\right) & 98 & \\ [0.5ex] \left(3^2,4^2\right) & 97 & \checkmark \\ [0.5ex] \left(4^2,5^2\right) & 96 & \\ [0.5ex] \left(5^2,6^2\right) & 95 & \checkmark \\ [0.5ex] \left(6^2,7^2\right) & 94 & \\ [0.5ex] \cdots & \cdots & \cdots \\ [0.5ex] \left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex] \left(100^2,\infty\right) & 0 &  \\ [0.5ex] \end{array}\] 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

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), because there are an odd number of negatives, which means that the number of values equals $((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \ldots + ((2+1)(2-1)+1)$.

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 -jesselan made 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.

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

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

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 (ProblemsAnswer KeyResources)
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. AMC logo.png