Difference between revisions of "2008 iTest Problems/Problem 7"

Revision as of 00:21, 8 March 2018

Find the number of integers $n$ for which $n^2 + 10n < 2008$ .

First, we complete the square of the left side of the equation, giving us:

$(n+5)^2 < 2033$

The integers from $-50$ to $40$ satisfy this equation, so the answer is $40- (-50)+1 = 91$

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 First, we complete the square of the left side of the equation, giving us:
-<math>(n+5)^2 < 1983</math>
+<math>(n+5)^2 < 2033</math>
-The integers from <math>-49</math> to <math>39</math> satisfy this equation, so the answer is <math> 39- (-49)+1 = 89</math>
+The integers from <math>-50</math> to <math>40</math> satisfy this equation, so the answer is <math> 40- (-50)+1 = 91</math>
 ==See also==