2014 AMC 10B Problems/Problem 20
Contents
Problem
For how many integers is the number
negative?
Solution 1
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
integers
that satisfy this inequality,
.
Thus our answer is
Solution 2
Since the part of
has to be less than
(because we want
to be negative), we have the inequality
-->
.
has to be positive, so
is negative. Then we have
. We know that if we find a positive number that works, it's parallel negative will work. Therefore, we just have to find how many positive numbers work, then multiply that by
. If we try
, we get
, and
therefore doesn't work. Test two on your own, and then proceed. Since two works, all numbers above
that satisfy
work, that is the set {
}. That equates to
numbers. Since each numbers' negative counterparts work,
.
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 |
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