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1989 AIME Problems/Problem 1

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Contents

Problem

Compute \sqrt{(31)(30)(29)(28)+1}.

Solution

Solution 1

Let's call our four consecutive integers (n-1), n, (n+1), (n+2). 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}.

Solution 2

Note that the four numbers to multiply are symmetric with the center at 29.5. Multiply the symmetric pairs to get 31\cdot 28=868 and 30\cdot 29=870. Now clearly 868\cdot 870 + 1 = (869-1)(869+1) + 1 = 869^2 - 1^2 + 1 = 869^2.

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
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