Difference between revisions of "1989 AIME Problems/Problem 1"
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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> | + | Let's call our four [[consecutive]] integers <math>(n-1), n, (n+1), (n+2)</math>. Notice that <math>\displaystyle (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 == | ||
| − | + | {{AIME box|year=1989|before=First Question|num-a=2}} | |
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. Notice that 
. Thus, 
