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

Revision as of 16:54, 8 August 2011

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

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

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

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 $\text{(C)}.$

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 $\boxed{C}$

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

@@ Line 4: / Line 4: @@
 <math> \mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5 </math>
-== Solution ==
+== Solution 1==
-<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>
+<math>2^{1998} - 2^{1997} - 2^{1996} + 2^{1995}</math>
+Factor out <math>2^{1995}</math>:
+<math>2^{1995}(2^{3} - 2^{2} - 2^{1} + 1)</math>
+Simplify:
+<math>2^{1995}\cdot (8 - 4 - 2 + 1)</math>
+<math>2^{1995}\cdot (3)</math>
+By comparing the answer with the original equation, <math>k=3</math>, and the answer is <math>\text{(C)}.</math>
+==Solution 2==
+Divide both sides of the original equation by <math>2^1995</math>, giving:
+<math>2^3 - 2^2 - 2^1 + 1 = k</math>
+<math>k = 3</math>, and the answer is <math>\boxed{C}</math>
 == See also ==
 {{AHSME box|year=1998|num-b=4|num-a=6}}