Difference between revisions of "1966 AHSME Problems/Problem 25"
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This means that for every single increment <math>n</math> goes up from <math>1</math>, <math>F(n)</math> will increase by <math>\frac{1}{2}.</math> Since <math>101</math> is <math>100</math> increments from <math>1</math>, <math>F(n)</math> will increase <math>\frac{1}{2}\times100=50.</math> | This means that for every single increment <math>n</math> goes up from <math>1</math>, <math>F(n)</math> will increase by <math>\frac{1}{2}.</math> Since <math>101</math> is <math>100</math> increments from <math>1</math>, <math>F(n)</math> will increase <math>\frac{1}{2}\times100=50.</math> | ||
| − | Since <math>F(1)=2,</math> <math>F(101)</math> will equal <math>2+50=\boxed{ | + | Since <math>F(1)=2,</math> <math>F(101)</math> will equal <math>2+50=\boxed{\text{(D)} \ 52}.</math> |
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| + | Solution by davidaops. | ||
== See also == | == See also == | ||
Revision as of 16:05, 5 December 2018
Problem
If 
for 
and 
, then 
equals:
Solution
Notice that 
This means that for every single increment 
goes up from 
, 
will increase by 
Since 
is 
increments from , will increase 
Since 
will equal 
Solution by davidaops.
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Problem 26 | |
| 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. 
