Difference between revisions of "1975 AHSME Problems/Problem 1"
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Calculating, we find that <math>\frac {1}{2 - \frac {1}{2 - \frac {1}{2 - \frac12}}} = \frac {1}{2 - \frac {1}{2 - \frac {2}{3}}} = \frac {1}{2 - \frac {3}{4}} = \frac {1}{\frac {5}{4}} = \boxed{\textbf{(B) } \dfrac{4}{5}}</math>. | Calculating, we find that <math>\frac {1}{2 - \frac {1}{2 - \frac {1}{2 - \frac12}}} = \frac {1}{2 - \frac {1}{2 - \frac {2}{3}}} = \frac {1}{2 - \frac {3}{4}} = \frac {1}{\frac {5}{4}} = \boxed{\textbf{(B) } \dfrac{4}{5}}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1975|before=First Problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 15:46, 19 January 2021
Problem
The value of 
is
Solution
Solution by e_power_pi_times_i
Calculating, we find that 
.
See Also
| 1975 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
