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2008 iTest Problems/Problem 7

Revision as of 00:21, 8 March 2018 by Bawbdmo (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:

$(n+5)^2 < 2033$

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

See also

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