2005 CEMC Pascal Problems/Problem 2

Problem

The expression $6a - 5a + 4a - 3a + 2a - a$ is equal to

$\text{ (A) }\ 3a \qquad\text{ (B) }\ 3a^{6} \qquad\text{ (C) }\ 3 \qquad\text{ (D) }\ -21a \qquad\text{ (E) }\ -21a^{6}$

$6a - 5a + 4a - 3a + 2a - a = a + 4a - 3a + 2a - a$

$=5a - 3a + 2a - a = 2a + 2a - a = 4a - a = \boxed {\textbf {(A) } 3a}$

~anabel.disher

We can notice that we can pair $6a - 5a$ , $4a - 3a$ , and $2a - a$ to get $a$ for each. This means that we have:

$a + a + a = \boxed {\textbf {(A) } 3a}$

~anabel.disher