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2020 AMC 10A Problems/Problem 17: Difference between revisions

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Case 2: For there to be an odd number of negative factors, <math>n</math> must be between an odd number squared and an even number squared. This means that there are <math>2+6+\dots+10</math> total possible values of <math>n</math>. Simplifying, there are <math>5000</math> possible numbers.
Case 2: For there to be an odd number of negative factors, <math>n</math> must be between an odd number squared and an even number squared. This means that there are <math>2+6+\dots+10</math> total possible values of <math>n</math>. Simplifying, there are <math>5000</math> possible numbers.


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>.




Revision as of 21:07, 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

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 $n$.


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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