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

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

Revision as of 18:03, 25 February 2016

Problem

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

Solution

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

$(n+5)^2 < 1983$

The integers from $-49$ to $39$ satisfy this equation, so the answer is $39- (-49)+1 = 89$

See also