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

(→‎Solution 1 (Casework))
(→‎Solution 3 (End Behavior))
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~quacker88
 
~quacker88
 +
 +
== Solution 4 (Fast) ==
 +
We know <math>P(x) \leq 0</math> when an odd number of its factors are negative. The only way to do this is to make an odd number of positive factors. For example, to make the first factor positive, <math>x \in [1^2, 2^3]</math>. The reason we stop at <math>2^2</math> is if it is anything higher, 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 &= \\
 +
x^2 - (x^2 - 2x + 1) &= \\
 +
x^2 - x^2 + 2x - 1 + 1 &= 2x \\
 +
\end{align*}</cmath>
 +
 +
But remember this is 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 just have to solve:
 +
<cmath>\begin{align*}
 +
\sum_{x=2}^{x = 100}2x &=
 +
2\sum_{x=2}^{x = 100}x &=
 +
2 \cdot \dfrac{(50)(102)}{}
 +
\end{align*}</cmath>
  
 
== Video Solutions ==
 
== Video Solutions ==

Revision as of 15:18, 1 October 2023

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 negative. The only way to do this is to make an odd number of positive factors. For example, to make the first factor positive, $x \in [1^2, 2^3]$. The reason we stop at $2^2$ is if it is anything higher, 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 &= \\ x^2 - (x^2 - 2x + 1) &= \\ x^2 - x^2 + 2x - 1 + 1 &= 2x \\ \end{align*}

But remember this is 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 just have to solve: \begin{align*} \sum_{x=2}^{x = 100}2x &= 2\sum_{x=2}^{x = 100}x &= 2 \cdot \dfrac{(50)(102)}{} \end{align*}

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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