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

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 $P(n)\leq0:$

$P(n)=0$

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

$P(n)<0$

There are $100$ factors in $P(x),$ and we need an odd number of them to be negative. We construct the table below: $\begin{array}{c|c|c} & & \\ [-2.5ex] \textbf{Interval of }\boldsymbol{x} & \boldsymbol{\#}\textbf{ of Negative Factors} & \textbf{Valid?} \\ [0.5ex] \hline & & \\ [-2ex] \left(-\infty,1^2\right) & 100 & \\ [0.5ex] \left(1^2,2^2\right) & 99 & \checkmark \\ [0.5ex] \left(2^2,3^2\right) & 98 & \\ [0.5ex] \left(3^2,4^2\right) & 97 & \checkmark \\ [0.5ex] \left(4^2,5^2\right) & 96 & \\ [0.5ex] \left(5^2,6^2\right) & 95 & \checkmark \\ [0.5ex] \left(6^2,7^2\right) & 94 & \\ \vdots & \vdots & \vdots \\ [0.75ex] \left(99^2,100^2\right) & 1 & \checkmark \\ [0.5ex] \left(100^2,\infty\right) & 0 & \\ [0.5ex] \end{array}$ 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

@@ Line 23: / Line 23: @@
 \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]
+\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(100^2,\infty\right) & 0 &  \\ [0.5ex]
@@ Line 63: / Line 63: @@
 ~quacker88
-==Solution 4==
+== 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:
-It is clear to see that if an odd number of factors of this are negative or <math>0</math>, then <math>P(x)</math> is also negative or <math>0</math>. Using a parity argument, we see that if an odd amount is POSITIVE, it also satisfies, which lets us do casework:
+<cmath>\begin{align*}
+x^2 - (x-1)^2 + 1 &= 2x \\
-From the interval <math>{1,2,…4}</math> we see that the first term is negative or 0 and the rest is positive.
+x^2 - (x^2 - 2x + 1) + 1 &= 2x \\
+x^2 - x^2 + 2x - 1 + 1 &= 2x. \\
-From the interval <math>{9,10,…16}</math> we see that the first 3 terms are negative or 0.
+\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:
-As a general rule, the satisfying condition is <math>{1,2…4}, {9,10,…16}, {25,26,…36},…{n^2,(n+1)^2}, {(n+3)^2,(n+4)^2}</math>.
+<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
-Using difference of squares, we get that it is the sum of all integers from 1 to 100, making it <math>5050</math>. Adding the other <math>50</math> <math>0</math> cases gets us to <math>5050+50=5100= {\boxed{\textbf{(E)} 5100</math>
--dragoon
 == Video Solutions ==

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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