Difference between revisions of "2008 Grade 8 CEMC Gauss Problems/Problem 4"

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-==Problem==
+{{delete|moved}}
-The value of <math>(1 + 2)^2 - (1^2 + 2^2)</math> is
-<math> \text{ (A) }\ 14 \qquad\text{ (B) }\ 4 \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 1 </math>
-==Solution 1==
-Using order of operations, we get:
-<math>(1 + 2)^2 - (1^2 + 2^2) = (3)^2 - (1 + 4) = 9 - 5 = \boxed {\textbf {(B) } 4}</math>
-~anabel.disher
-==Solution 2==
-We can notice that <math>(x + y)^2 - (x^2 + y^2) = x^2 + 2xy + y^2 - x^2 - y^2 = 2xy</math>.
-We can now plug in <math>1</math> for <math>x</math> and <math>2</math> for <math>y</math> to get:
-<math>2xy = 2 \times 1 \times 2 = \boxed {\textbf {(B) } 4}</math>
-~anabel.disher
-==Solution 3 (answer choices)==
-Since <math>1^2 + 2^2 = 1 + 4 = 5</math> and all of the numbers involved in the equation are integers, we can notice that the number must be a perfect square after adding <math>5</math>.
-<math>14 + 5 = 19</math>, <math>4 + 5 = 9</math>, <math>2 + 5 = 7</math>, <math>12 + 5 = 17</math>, and <math>1 + 5 = 6</math>
-Of these numbers, the only number that was a perfect square after adding <math>5</math> was <math>\boxed {\textbf {(B) } 4}</math>
-~anabel.disher