1989 AIME Problems/Problem 1

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 $868\cdot 870 + 1 = (869-1)(869+1) + 1 = 869^2 - 1^2 + 1 = 869^2$ .

1989 AIME (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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1989 AIME Problems/Problem 1

Contents

Problem

Solution

Solution 1

Solution 2

See also