Difference between revisions of "1966 AHSME Problems/Problem 25"
(→Solution) |
m (→Solution) |
||
(2 intermediate revisions by the same user not shown) | |||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | <math>\ | + | Notice that <math>\frac{2F(n)+1}{2}=F(n)+\frac{1}{2}.</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{\text{(D)} \ 52}.</math> | ||
+ | |||
+ | |||
+ | Solution by davidaops. | ||
== See also == | == See also == |
Latest revision as of 17:06, 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.