Difference between revisions of "2015 AIME I Problems/Problem 9"
(→Solution) |
(→Solution) |
||
Line 16: | Line 16: | ||
<math>|y-x|</math><2 (280 options) | <math>|y-x|</math><2 (280 options) | ||
<math>|z-y|</math><2 (280 options, 80 of which coincide with option 1) | <math>|z-y|</math><2 (280 options, 80 of which coincide with option 1) | ||
− | z=1, <math>|y-x|</math>=2. ( | + | z=1, <math>|y-x|</math>=2. (16 options, 2 of which coincide with either option 1 or option 2) |
− | Adding the total number of such ordered triples yields <math>280+280-80+ | + | Adding the total number of such ordered triples yields <math>280+280-80+16-2=494</math>. |
==See Also== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=8|num-a=10}} | {{AIME box|year=2015|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:58, 5 April 2015
Problem
Let be the set of all ordered triple of integers
with
. Each ordered triple in
generates a sequence according to the rule
for all
. Find the number of such sequences for which
for some
.
Solution
Let . First note that if any absolute value equals 0, then
=0.
Also note that if at any position,
, then
.
Then, if any absolute value equals 1, then
=0.
Therefore, if either
or
is less than or equal to 1, then that ordered triple meets the criteria.
Assume that to be the only way the criteria is met.
To prove, let
, and
. Then,
,
, and
.
However, since the minimum values of
and
are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be
,
. Again assume that any other scenario will not meet criteria.
To prove, divide the other scenarios into two cases:
,
, and
; and
,
, and
.
For the first one,
>=2z,
>=4z,
>=8z, and
>=16z, by which point we see that this function diverges.
For the second one,
,
,
, and
, by which point we see that this function diverges.
Therefore, the only scenarios where
=0 is when any of the following are met:
<2 (280 options)
<2 (280 options, 80 of which coincide with option 1)
z=1,
=2. (16 options, 2 of which coincide with either option 1 or option 2)
Adding the total number of such ordered triples yields
.
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.