Difference between revisions of "1998 AHSME Problems/Problem 5"

 
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== Problem 5 ==
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== Problem==
 
If <math>2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},</math> what is the value of <math>k</math>?
 
If <math>2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},</math> what is the value of <math>k</math>?
  
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==Solution 2==
 
==Solution 2==
Divide both sides of the original equation by <math>2^1995</math>, giving:
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Divide both sides of the original equation by <math>2^{1995}</math>, giving:
  
 
<math>2^3 - 2^2 - 2^1 + 1 = k</math>
 
<math>2^3 - 2^2 - 2^1 + 1 = k</math>
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== See also ==
 
== See also ==
 
{{AHSME box|year=1998|num-b=4|num-a=6}}
 
{{AHSME box|year=1998|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 13:28, 5 July 2013

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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