Difference between revisions of "1989 AIME Problems/Problem 1"

Revision as of 14:22, 27 December 2008

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

@@ Line 7: / Line 7: @@
 === Solution 2===
-Note that the four numbers to multiply are symmetric with the center at <math>29.5</math>. Multiply the symmetric pairs to get <math>31\cdot 28=868</math> and <math>30\cdot 29=870</math>. Now clearly <math>868\cdot 870 + 1 = (869-1)(869+1) + 1 = 869^2 - 1 + 1 = 869^2</math>.
+Note that the four numbers to multiply are symmetric with the center at <math>29.5</math>. Multiply the symmetric pairs to get <math>31\cdot 28=868</math> and <math>30\cdot 29=870</math>. Now clearly <math>868\cdot 870 + 1 = (869-1)(869+1) + 1 = 869^2 - 1^2 + 1 = 869^2</math>.
 == See also ==

1989 AIME (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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