Difference between revisions of "2014 AMC 10B Problems/Problem 20"
Claudeaops (talk | contribs) (Simplifying the solution) |
Claudeaops (talk | contribs) m (→Solution) |
||
Line 7: | Line 7: | ||
First, note that <math>50+1=51</math>, which motivates us to factor the polynomial as <math>(x^2-50)(x^2-1)</math>. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so <math>x^2-50<0<x^2-1</math>. Solving this inequality, we find <math>1<x^2<50</math>. There are exactly 12 integers <math>x</math> that satisfy this inequality, <math>\pm 2,3,4,5,6,7</math>. | First, note that <math>50+1=51</math>, which motivates us to factor the polynomial as <math>(x^2-50)(x^2-1)</math>. Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so <math>x^2-50<0<x^2-1</math>. Solving this inequality, we find <math>1<x^2<50</math>. There are exactly 12 integers <math>x</math> that satisfy this inequality, <math>\pm 2,3,4,5,6,7</math>. | ||
− | Thus our answer is <math> | + | Thus our answer is <math>\boxed{\textbf {(C) } 12}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2014|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:31, 21 February 2014
Problem
For how many integers is the number negative?
Solution
First, note that , which motivates us to factor the polynomial as . Since this expression is negative, one term must be negative and the other positive. Also, the first term is obviously smaller than the second, so . Solving this inequality, we find . There are exactly 12 integers that satisfy this inequality, .
Thus our answer is
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.