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==
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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.
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Case 1: There are 100 integers <math>n</math> for which <math>P(x)=0
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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.
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Summing, there are </math>\boxed{\textbf{(E) }} 5100<math> total possible values of </math>n$.
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==See Also==
 
==See Also==

Revision as of 21:05, 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$ (Error compiling LaTeX. ! Missing $ inserted.)\boxed{\textbf{(E) }} 5100$total possible values of$n$.


See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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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