2008 iTest Problems/Problem 7

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:

$(n+5)^2 < 2033$

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

See Also

2008 iTest (Problems)
Preceded by:
Problem 6
Followed by:
Problem 8
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