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

Revision as of 13:19, 17 June 2008

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

$2000(2000+7)(2000+8)(2000+15)+784$

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

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

Thus $\sqrt{2000\cdot 2007\cdot 2008\cdot 2015+784}=\boxed{4030028}$

@@ Line 3: / Line 3: @@
 ==Solution==
-<cmath>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</cmath>
+<math>2000(2000+7)(2000+8)(2000+15)+784</math>
-Thus <math></math>\sqrt{2000\cdot 2007\cdot 2008\cdot 2015+784}=\boxed{4030028}$
+<math>=(2000^2+15\cdot 2000+56)(2000^2+15\cdot 2000)+28^2</math>
+<math>=(2000^2+15\cdot 2000+28)(2000^2+15\cdot 2000+28)-28^2+28^2</math>
+Thus <math>\sqrt{2000\cdot 2007\cdot 2008\cdot 2015+784}=\boxed{4030028}</math>
 ==See also==