Difference between revisions of "1951 AHSME Problems/Problem 24"
Yankeesfan (talk | contribs) (Created page with "== Problem == <math> \frac{2^{n+4}-2(2^{n})}{2(2^{n+3})} </math> when simplified is: <math> \textbf{(A)}\ 2^{n+1}-\frac{1}{8}\qquad\textbf{(B)}\ -2^{n+1}\qquad\textbf{(C)}\ 1-2^...") |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 7: | Line 7: | ||
We have <math>2(2^n)=2^{n+1}</math>, and <math>2(2^{n+3})=2^{n+4}</math>. Thus, <math> \frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}=\dfrac{2^{n+4}-2^{n+1}}{2^{n+4}}</math>. Factoring out a <math>2^{n+1}</math> in the numerator, we get <math>\dfrac{2^{n+1}(2^3-1)}{2^{n+4}}=\dfrac{8-1}{2^3}=\boxed{\textbf{(D)}\ \frac{7}{8}}</math>. | We have <math>2(2^n)=2^{n+1}</math>, and <math>2(2^{n+3})=2^{n+4}</math>. Thus, <math> \frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}=\dfrac{2^{n+4}-2^{n+1}}{2^{n+4}}</math>. Factoring out a <math>2^{n+1}</math> in the numerator, we get <math>\dfrac{2^{n+1}(2^3-1)}{2^{n+4}}=\dfrac{8-1}{2^3}=\boxed{\textbf{(D)}\ \frac{7}{8}}</math>. | ||
| − | == See | + | == See Also == |
| − | {{AHSME box|year=1951|num-b=23|num-a=25}} | + | {{AHSME 50p box|year=1951|num-b=23|num-a=25}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:24, 5 July 2013
Problem
when simplified is:
Solution
We have 
, and 
. Thus, 
. Factoring out a 
in the numerator, we get 
.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| 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. 
