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

Revision as of 14:45, 6 June 2011

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

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

$2^{1998} - 2^{1997} - 2^{1996} = 2^{1996}$ . $2^{1996} + 2^{1995} = 2^{1995}(2 + 1) = 3 \cdot 2^{1995}$ . So, the answer is $\text{(C)}.$

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

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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>
+== Problem 5 ==
+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>
+== Solution ==
+<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>
+== See also ==
+{{AHSME box|year=1998|num-b=4|num-a=6}}