2008 iTest Problems/Problem 7

Revision as of 19:03, 25 February 2016 by Mewtwomew (talk | contribs) (Solution)

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