Difference between revisions of "2018 AMC 10B Problems/Problem 20"
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<math>\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}</math> | <math>\textbf{(A)} \text{ 2016} \qquad \textbf{(B)} \text{ 2017} \qquad \textbf{(C)} \text{ 2018} \qquad \textbf{(D)} \text{ 2019} \qquad \textbf{(E)} \text{ 2020}</math> | ||
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==Solution 2 (A Bit Bashy)== | ==Solution 2 (A Bit Bashy)== |
Revision as of 13:14, 24 November 2020
Contents
Problem
A function is defined recursively by
and
for all integers
. What is
?
Solution 2 (A Bit Bashy)
Start out by listing some terms of the sequence.
Notice that
whenever
is an odd multiple of
, and the pattern of numbers that follow will always be
,
,
,
,
.
The largest odd multiple of
smaller than
is
, so we have
Solution 3 (Bashy Pattern Finding)
Writing out the first few values, we get:
. Examining, we see that every number
where
has
,
, and
. The greatest number that's
and less
is
, so we have
Solution 4 (Algebra)
Adding the two equations, we have that
Hence,
.
After plugging in
to the equation above and doing some algebra, we have that
.
Consequently,
Adding these
equations up, we have that
and
.
~AopsUser101
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 | ||
All AMC 10 Problems and Solutions |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.