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

Revision as of 12:54, 14 July 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), 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

Solution 4 (Similar to Solution 1: Casework)

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

$P(n)=0$

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

$P(n)<0$

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 $100+5000=\boxed{\textbf{(E) } 5100}.$

~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

@@ Line 64: / Line 64: @@
 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{\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 \\
 &=\frac{101(100)}{2}-50 \\

2020 AMC 10A (Problems • Answer Key • Resources)
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