1989 AIME Problems/Problem 1: Difference between revisions
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== Solution == | == Solution == | ||
{ | Let's call our four consecutive integers <math>(n-1), n, (n+1), (n+2)</math>. Notice that <math>(n-1)(n)(n+1)(n+2)+1=(n^2+n)^2-2(n^2+n)+1 \Rightarrow (n^2+n-1)^2</math>. Thus, <math>\sqrt{(31)(30)(29)(28)+1} = (29^2+29-1) = 869</math> | ||
== See also == | == See also == | ||
* [[1989 AIME Problems/Problem 2|Next Problem]] | * [[1989 AIME Problems/Problem 2|Next Problem]] | ||
* [[1989 AIME Problems]] | * [[1989 AIME Problems]] | ||
Revision as of 21:32, 25 February 2007
Problem
Compute 
.
Solution
Let's call our four consecutive integers 
. Notice that 
. Thus, 
