Difference between revisions of "1996 AIME Problems/Problem 9"
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A bored student walks down a hall that contains a row of closed lockers, numbered <math>1</math> to <math>1024</math>. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens? | A bored student walks down a hall that contains a row of closed lockers, numbered <math>1</math> to <math>1024</math>. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens? | ||
− | == | + | == Solutions == |
=== Solution 1 === | === Solution 1 === | ||
− | On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of <math>4</math>, leaving only lockers <math>2 \ | + | On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of <math>4</math>, leaving only lockers <math>2 \pmod{8}</math> and <math>6 \pmod{8}</math>. Then he goes ahead and opens all lockers <math>2 \pmod {8}</math>, leaving lockers either <math>6 \pmod {16}</math> or <math>14 \pmod {16}</math>. He then goes ahead and opens all lockers <math>14 \pmod {16}</math>, leaving the lockers either <math>6 \pmod {32}</math> or <math>22 \pmod {32}</math>. He then goes ahead and opens all lockers <math>6 \pmod {32}</math>, leaving <math>22 \pmod {64}</math> or <math>54 \pmod {64}</math>. He then opens <math>54 \pmod {64}</math>, leaving <math>22 \pmod {128}</math> or <math>86 \pmod {128}</math>. He then opens <math>22 \pmod {128}</math> and leaves <math>86 \pmod {256}</math> and <math>214 \pmod {256}</math>. He then opens all <math>214 \pmod {256}</math>, so we have <math>86 \pmod {512}</math> and <math>342 \pmod {512}</math>, leaving lockers <math>86, 342, 598</math>, and <math>854</math>, and he is at where he started again. He then opens <math>86</math> and <math>598</math>, and then goes back and opens locker number <math>854</math>, leaving locker number <math>\boxed{342}</math> untouched. He opens that locker. |
+ | |||
+ | -------------------------- | ||
+ | |||
+ | Lockers that are <math>2 \pmod{8}</math> and <math>6 \pmod{8}</math> are just lockers that are <math>2 \pmod{4}</math>. Edit by [[User: Yiyj1|Yiyj1]] | ||
=== Solution 2 === | === Solution 2 === | ||
We can also solve this with recursion. Let <math>L_n</math> be the last locker he opens given that he started with <math>2^n</math> lockers. Let there be <math>2^n</math> lockers. After he first reaches the end of the hallway, there are <math>2^{n-1}</math> lockers remaining. There is a correspondence between these unopened lockers and if he began with <math>2^{n-1}</math> lockers. The locker <math>y</math> (if he started with <math>2^{n-1}</math> lockers) corresponds to the locker <math>2^n+2-2y</math> (if he started with <math>2^n</math> lockers). It follows that <math>L_{n} = 2^{n} +2 -2L_{n-1}</math> as they are corresponding lockers. We can compute <math>L_1=2</math> and use the recursion to find <math>L_{10}=\boxed{342}</math> | We can also solve this with recursion. Let <math>L_n</math> be the last locker he opens given that he started with <math>2^n</math> lockers. Let there be <math>2^n</math> lockers. After he first reaches the end of the hallway, there are <math>2^{n-1}</math> lockers remaining. There is a correspondence between these unopened lockers and if he began with <math>2^{n-1}</math> lockers. The locker <math>y</math> (if he started with <math>2^{n-1}</math> lockers) corresponds to the locker <math>2^n+2-2y</math> (if he started with <math>2^n</math> lockers). It follows that <math>L_{n} = 2^{n} +2 -2L_{n-1}</math> as they are corresponding lockers. We can compute <math>L_1=2</math> and use the recursion to find <math>L_{10}=\boxed{342}</math> | ||
+ | |||
+ | === Solution 3 (bash)=== | ||
+ | List all the numbers from <math>1</math> through <math>1024</math>, then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get <math>\boxed{342}</math>. | ||
+ | |||
+ | (Note: If you try to do this, first look through all the problems! -Guy) | ||
+ | |||
+ | -------------------------- | ||
+ | Edit from EthanSpoon, Doing this with only even numbers will make it faster. | ||
+ | You'll see. | ||
+ | 02496 | ||
== See also == | == See also == | ||
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:51, 6 September 2023
Problem
A bored student walks down a hall that contains a row of closed lockers, numbered to
. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he encounters, and then alternates between skipping and opening each closed locker thereafter. The student continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?
Solutions
Solution 1
On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of , leaving only lockers
and
. Then he goes ahead and opens all lockers
, leaving lockers either
or
. He then goes ahead and opens all lockers
, leaving the lockers either
or
. He then goes ahead and opens all lockers
, leaving
or
. He then opens
, leaving
or
. He then opens
and leaves
and
. He then opens all
, so we have
and
, leaving lockers
, and
, and he is at where he started again. He then opens
and
, and then goes back and opens locker number
, leaving locker number
untouched. He opens that locker.
Lockers that are and
are just lockers that are
. Edit by Yiyj1
Solution 2
We can also solve this with recursion. Let be the last locker he opens given that he started with
lockers. Let there be
lockers. After he first reaches the end of the hallway, there are
lockers remaining. There is a correspondence between these unopened lockers and if he began with
lockers. The locker
(if he started with
lockers) corresponds to the locker
(if he started with
lockers). It follows that
as they are corresponding lockers. We can compute
and use the recursion to find
Solution 3 (bash)
List all the numbers from through
, then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get
.
(Note: If you try to do this, first look through all the problems! -Guy)
Edit from EthanSpoon, Doing this with only even numbers will make it faster. You'll see. 02496
See also
1996 AIME (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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