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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>?
  
 
<math> \mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5 </math>
 
<math> \mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5 </math>
 
   
 
   
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== Solution ==
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== Solution 1==
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<math>2^{1998} - 2^{1997} - 2^{1996} = 2^{1996}</math>. <math>2^{1996} + 2^{1995} = 2^{1995}(2 + 1) = 3 \cdot 2^{1995}</math>. So, the answer is <math>\text{(C)}.</math>
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<math>2^{1998} - 2^{1997} - 2^{1996} + 2^{1995}</math>
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Factor out <math>2^{1995}</math>:
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<math>2^{1995}(2^{3} - 2^{2} - 2^{1} + 1)</math>
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Simplify:
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<math>2^{1995}\cdot (8 - 4 - 2 + 1)</math>
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<math>2^{1995}\cdot (3)</math>
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By comparing the answer with the original equation, <math>k=3</math>, and the answer is <math>\text{(C)}.</math>
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==Solution 2==
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Divide both sides of the original equation by <math>2^{1995}</math>, giving:
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<math>2^3 - 2^2 - 2^1 + 1 = k</math>
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<math>k = 3</math>, and the answer is <math>\boxed{C}</math>
  
 
== 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 (Problems • Answer Key • Resources)
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
All AHSME Problems and Solutions

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