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1997 CEMC Pascal Problems/Problem 4

Problem

$(1)^10 + (-1)^8 + (-1)^7 + (1)^5$ equals

$\text{ (A) }\ 0 \qquad\text{ (B) }\ 1 \qquad\text{ (C) }\ 2 \qquad\text{ (D) }\ 16 \qquad\text{ (E) }\ 4$

Solution

We can use the fact that $(1)^n = 1$ for any value of $n$, and $(-1)^n = 1$ when $n$ is even, but $(-1)^n = -1$ when $n$ is odd:

$(1)^10 + (-1)^8 + (-1)^7 + (1)^5 = 1 + 1 - 1 + 1 = 2 - 1 + 1 = 1 + 1 = \boxed {\textbf {(C) } 2}$

~anabel.disher

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