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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>
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== Solution Visual Aid ==
 +
Visualization that makes it easier to see solutions:
 +
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<asy>
 +
size(600);
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 +
// Draw the main number line (x-axis)
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draw((-5, 0) -- (5, 0), Arrow);  // Line from -10 to 10 with an arrow at the end
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draw((5, 0) -- (-5, 0), Arrow);  // Line from -10 to 10 with an arrow at the end
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// Add ticks and labels at each integer point from -10 to 10
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for (int i = -4; i <= 4; ++i) {
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    if (i != 0)
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    draw((i, -0.1) -- (i, 0.1));  // Tick mark
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    // label(string(i), (i, -0.5), S);  // Label each tick below the line
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}
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label("$1^2$", (-4, -0.5), S);  // Label each tick below the line
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label("$2^2$", (-3, -0.5), S);  // Label each tick below the line
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label("$3^2$", (-2, -0.5), S);  // Label each tick below the line
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label("$4^2$", (-1, -0.5), S);  // Label each tick below the line
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label("\dots", (0, -0.5), S);  // Label each tick below the line
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label("$97^2$", (1, -0.5), S);  // Label each tick below the line
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label("$98^2$", (2, -0.5), S);  // Label each tick below the line
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label("$99^2$", (3, -0.5), S);  // Label each tick below the line
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label("$100^2$", (4, -0.5), S);  // Label each tick below the line
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label("$0$", (-4, 0.1), N);  // Label each tick below the line
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label("$0$", (-3, 0.1), N);  // Label each tick below the line
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label("$0$", (-2, 0.1), N);  // Label each tick below the line
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label("$0$", (-1, 0.1), N);  // Label each tick below the line
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label("$0$", (1, 0.1), N);  // Label each tick below the line
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label("$0$", (2, 0.1), N);  // Label each tick below the line
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label("$0$", (3, 0.1), N);  // Label each tick below the line
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label("$0$", (4, 0.1), N);  // Label each tick below the line
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label("$+$", (-4.5, 0), N);  // Label each tick below the line
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label("$-$", (-3.5, 0), N);  // Label each tick below the line
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label("$+$", (-2.5, 0), N);  // Label each tick below the line
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label("$-$", (-1.5, 0), N);  // Label each tick below the line
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label("$+$", (4.5, 0), N);  // Label each tick below the line
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label("$-$", (3.5, 0), N);  // Label each tick below the line
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label("$+$", (2.5, 0), N);  // Label each tick below the line
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label("$-$", (1.5, 0), N);  // Label each tick below the line
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</asy>
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=== Note ===
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Clearly, there are a finite amount of integers that result in <math>P(x)</math> being nonpositive. This is because there are an even number of factors of <math>P(x).</math> From here, you can read other solutions that count the amount of negative numbers of <math>P(x).</math> ~jasmineLOVER7254
  
 
== Solution 1 (Casework) ==
 
== Solution 1 (Casework) ==
Line 73: Line 123:
 
<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>
 
<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
 
~Wiselion
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== Video Solution by Pi Academy ==
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https://youtu.be/hqdnNqds2mw?si=dHhmbLrh3pWWIG9T
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 +
~ Pi Academy
  
 
== Video Solutions ==
 
== Video Solutions ==

Latest revision as of 23:47, 3 November 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 Visual Aid

Visualization that makes it easier to see solutions:

[asy] size(600); // Draw the main number line (x-axis) draw((-5, 0) -- (5, 0), Arrow); // Line from -10 to 10 with an arrow at the end draw((5, 0) -- (-5, 0), Arrow); // Line from -10 to 10 with an arrow at the end // Add ticks and labels at each integer point from -10 to 10 for (int i = -4; i <= 4; ++i) { if (i != 0) draw((i, -0.1) -- (i, 0.1)); // Tick mark // label(string(i), (i, -0.5), S); // Label each tick below the line } label("$1^2$", (-4, -0.5), S); // Label each tick below the line label("$2^2$", (-3, -0.5), S); // Label each tick below the line label("$3^2$", (-2, -0.5), S); // Label each tick below the line label("$4^2$", (-1, -0.5), S); // Label each tick below the line label("\dots", (0, -0.5), S); // Label each tick below the line label("$97^2$", (1, -0.5), S); // Label each tick below the line label("$98^2$", (2, -0.5), S); // Label each tick below the line label("$99^2$", (3, -0.5), S); // Label each tick below the line label("$100^2$", (4, -0.5), S); // Label each tick below the line label("$0$", (-4, 0.1), N); // Label each tick below the line label("$0$", (-3, 0.1), N); // Label each tick below the line label("$0$", (-2, 0.1), N); // Label each tick below the line label("$0$", (-1, 0.1), N); // Label each tick below the line label("$0$", (1, 0.1), N); // Label each tick below the line label("$0$", (2, 0.1), N); // Label each tick below the line label("$0$", (3, 0.1), N); // Label each tick below the line label("$0$", (4, 0.1), N); // Label each tick below the line label("$+$", (-4.5, 0), N); // Label each tick below the line label("$-$", (-3.5, 0), N); // Label each tick below the line label("$+$", (-2.5, 0), N); // Label each tick below the line label("$-$", (-1.5, 0), N); // Label each tick below the line label("$+$", (4.5, 0), N); // Label each tick below the line label("$-$", (3.5, 0), N); // Label each tick below the line label("$+$", (2.5, 0), N); // Label each tick below the line label("$-$", (1.5, 0), N); // Label each tick below the line [/asy]

Note

Clearly, there are a finite amount of integers that result in $P(x)$ being nonpositive. This is because there are an even number of factors of $P(x).$ From here, you can read other solutions that count the amount of negative numbers of $P(x).$ ~jasmineLOVER7254

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 Solution by Pi Academy

https://youtu.be/hqdnNqds2mw?si=dHhmbLrh3pWWIG9T

~ Pi Academy

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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