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Difference between revisions of "2020 AMC 10A Problems/Problem 17"
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==Video Solution==
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https://youtu.be/RKlG6oZq9so
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~IceMatrix
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==See Also==
 
==See Also==
  
 
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Revision as of 22:20, 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

Video Solution

https://youtu.be/RKlG6oZq9so

~IceMatrix

See Also

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

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