Difference between revisions of "2015 AIME I Problems/Problem 9"
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Therefore, if either <math>|y-x|</math> or <math>|z-y|</math> is less than or equal to 1, then that ordered triple meets the criteria. | Therefore, if either <math>|y-x|</math> or <math>|z-y|</math> 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. | Assume that to be the only way the criteria is met. | ||
− | To prove, let <math>|y-x|</math> | + | To prove, let <math>|y-x|>1</math>, and <math>|z-y|>1</math>. Then, <math>a_4 \ge 2z</math>, <math>a_5 \ge 4z</math>, and <math>a_6 \ge 4z</math>. |
However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be z=1, <math>|y-x|</math>=2. Again assume that any other scenario will not meet criteria. | However, since the minimum values of <math>a_5</math> and <math>a_6</math> are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be z=1, <math>|y-x|</math>=2. Again assume that any other scenario will not meet criteria. | ||
To prove, divide the other scenarios into two cases: z>1, <math>|y-x|</math>>1, and <math>|z-y|</math>>1; and z=1, <math>|y-x|</math>>2, and <math>|z-y|</math>>1. | To prove, divide the other scenarios into two cases: z>1, <math>|y-x|</math>>1, and <math>|z-y|</math>>1; and z=1, <math>|y-x|</math>>2, and <math>|z-y|</math>>1. |
Revision as of 17:22, 22 March 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 z=1,
=2. Again assume that any other scenario will not meet criteria.
To prove, divide the other scenarios into two cases: z>1,
>1, and
>1; and z=1,
>2, and
>1.
For the first one,
>=2z,
>=4z,
>=8z, and
>=16z, by which point we see that this function diverges.
For the second one,
>=3,
>=6,
>=18, and
>=54, 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. (14 options, none of which coincide with either option 1 or option 2)
Adding the total number of such ordered triples yields 280+280-80+14=494.
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.