Difference between revisions of "2018 AMC 10B Problems/Problem 20"
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<cmath>f(n)=f(n-1)-f(n-2)+n.</cmath> | <cmath>f(n)=f(n-1)-f(n-2)+n.</cmath> | ||
<cmath>f(n-1)=f(n-2)-f(n-3)+n-1.</cmath> | <cmath>f(n-1)=f(n-2)-f(n-3)+n-1.</cmath> |
Revision as of 22:23, 18 July 2019
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 +3, +2, +0, -1, +0.
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 1 (mod 6) and less
is
, so we have
Solution 1 (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.