Difference between revisions of "2007 Alabama ARML TST Problems/Problem 1"

Latest revision as of 08:26, 18 June 2008

Compute: $\sqrt{2000\cdot 2007\cdot 2008\cdot 2015+784}.$

$\sqrt{2000(2000+7)(2000+8)(2000+15)+784}$

$=\sqrt{(2000^2+15\cdot 2000+56)(2000^2+15\cdot 2000)+28^2}$

$=\sqrt{(2000^2+15\cdot 2000+28)(2000^2+15\cdot 2000+28)-28^2+28^2}$

$=2000^2+15\cdot 2000+28=\boxed{4030028}$

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 ==Solution==
-<math>2000(2000+7)(2000+8)(2000+15)+784=(2000^2+15\cdot 2000+56)(2000^2+15\cdot 2000)+28^2=(2000^2+15\cdot 2000+28)(2000^2+15\cdot 2000+28)-28^2+28^2</math>
+<center><math>\sqrt{2000(2000+7)(2000+8)(2000+15)+784}</math>
-Thus <math>\sqrt{2000\cdot 2007\cdot 2008\cdot 2015+784}=\boxed{4030028}</math>
+<math>=\sqrt{(2000^2+15\cdot 2000+56)(2000^2+15\cdot 2000)+28^2}</math>
+<math>=\sqrt{(2000^2+15\cdot 2000+28)(2000^2+15\cdot 2000+28)-28^2+28^2}</math>
+<math>=2000^2+15\cdot 2000+28=\boxed{4030028}</math></center>
 ==See also==
+{{ARML box|year=2007|state=Alabama|before=First Question|num-a=2}}