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1989 AIME Problems/Problem 1

From AoPSWiki

Problem

Compute .

Solution

Let's call our four consecutive integers . Notice that (n-1)(n)(n+1)(n+2)+1=(n^2+n)^2-2(n^2+n)+1 \Rightarrow (n^2+n-1)^2. Thus, \sqrt{(31)(30)(29)(28)+1} = (29^2+29-1) = \boxed{869}.

See also

1989 AIME (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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