Difference between revisions of "2015 AIME I Problems/Problem 9"
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<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 |
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