1951 AHSME Problems/Problem 24

Revision as of 11:24, 5 July 2013 by Nathan wailes (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

$\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}$ when simplified is:

$\textbf{(A)}\ 2^{n+1}-\frac{1}{8}\qquad\textbf{(B)}\ -2^{n+1}\qquad\textbf{(C)}\ 1-2^{n}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{7}{4}$

Solution

We have $2(2^n)=2^{n+1}$, and $2(2^{n+3})=2^{n+4}$. Thus, $\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}=\dfrac{2^{n+4}-2^{n+1}}{2^{n+4}}$. Factoring out a $2^{n+1}$ in the numerator, we get $\dfrac{2^{n+1}(2^3-1)}{2^{n+4}}=\dfrac{8-1}{2^3}=\boxed{\textbf{(D)}\ \frac{7}{8}}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png