2008 CEMC Gauss (Grade 8) Problems/Problem 4

Problem

The value of $(1 + 2)^2 - (1^2 + 2^2)$ is

$\text{ (A) }\ 14 \qquad\text{ (B) }\ 4 \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 1$

Solution 1

Using order of operations, we get:

$(1 + 2)^2 - (1^2 + 2^2) = (3)^2 - (1 + 4) = 9 - 5 = \boxed {\textbf {(B) } 4}$

~anabel.disher

Solution 2

We can notice that $(x + y)^2 - (x^2 + y^2) = x^2 + 2xy + y^2 - x^2 - y^2 = 2xy$ .

We can now plug in $1$ for $x$ and $2$ for $y$ to get:

$2xy = 2 \times 1 \times 2 = \boxed {\textbf {(B) } 4}$

~anabel.disher

Solution 3 (answer choices)

Since $1^2 + 2^2 = 1 + 4 = 5$ and all of the numbers involved in the equation are integers, we can notice that the number must be a perfect square after adding $5$ .

$14 + 5 = 19$ , $4 + 5 = 9$ , $2 + 5 = 7$ , $12 + 5 = 17$ , and $1 + 5 = 6$

Of these numbers, the only number that was a perfect square after adding $5$ was $\boxed {\textbf {(B) } 4}$

~anabel.disher

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2008 CEMC Gauss (Grade 8) Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

Solution 3 (answer choices)