1998 AHSME Problems/Problem 5

Problem

If $2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},$ what is the value of $k$?

$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5$

Solution 1

$2^{1998} - 2^{1997} - 2^{1996} + 2^{1995}$

Factor out $2^{1995}$:

$2^{1995}(2^{3} - 2^{2} - 2^{1} + 1)$

Simplify:

$2^{1995}\cdot (8 - 4 - 2 + 1)$

$2^{1995}\cdot (3)$

By comparing the answer with the original equation, $k=3$, and the answer is $\text{(C)}.$

Solution 2

Divide both sides of the original equation by $2^{1995}$, giving:

$2^3 - 2^2 - 2^1 + 1 = k$

$k = 3$, and the answer is $\boxed{C}$

See also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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