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== | + | ==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. | ||
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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 22:07, 31 January 2020
DefineHow many integers are there such that ?
Solution 1
Notice that 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 for which Case 2: For there to be an odd number of negative factors, must be between an odd number squared and an even number squared. This means that there are total possible values of . Simplifying, there are possible numbers.
Summing, there are total possible values of .
Solution 2
Notice that is nonpositive when is between and , and , and (inclusive), which means that the amount of values equals .
This reduces to
~Zeric
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AMC 10 Problems and Solutions |
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