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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>
  
−
==Solution==
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==Solution 1==
  
 
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.
 
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.
Line 11: Line 11:
 
Summing, there are <math>\boxed{\textbf{(E) } 5100}</math> total possible values of <math>n</math>.
 
Summing, there are <math>\boxed{\textbf{(E) } 5100}</math> total possible values of <math>n</math>.
  
 +
==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 amount 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 \frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}</math>
 +
 +
~Zeric
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2020|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2020|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:07, 31 January 2020

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. This means that there are $2+6+\dots+10$ total possible values of $n$. Simplifying, there are $5000$ possible numbers.

Summing, there are $\boxed{\textbf{(E) } 5100}$ total possible values of $n$.

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), which means that the amount of values equals $((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \ldots + ((2+1)(2-1)+1)$.

This reduces to $200 + 196 + 192 + \ldots + 4 = 4(1+2+\ldots + 50) = 4 \frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}$

~Zeric

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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