Difference between revisions of "2002 AMC 10A Problems/Problem 22"
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Given <math>n^2</math> tiles, a step removes <math>n</math> tiles, leaving <math>n^2 - n</math> tiles behind. Now, <math>(n-1)^2 = n^2 - n + (1-n) < n^2 - n < n^2</math>, so in the next step <math>n-1</math> tiles are removed. This gives <math>(n^2 - n) - (n-1) = n^2 - 2n + 1 = (n-1)^2</math>, another perfect square. | Given <math>n^2</math> tiles, a step removes <math>n</math> tiles, leaving <math>n^2 - n</math> tiles behind. Now, <math>(n-1)^2 = n^2 - n + (1-n) < n^2 - n < n^2</math>, so in the next step <math>n-1</math> tiles are removed. This gives <math>(n^2 - n) - (n-1) = n^2 - 2n + 1 = (n-1)^2</math>, another perfect square. | ||
| − | Thus each two steps we cycle down a perfect square, and in <math>(10-1)\times 2 = 18</math> steps, we are left with <math>1</math> tile, hence our answer is <math>\boxed{\text{C } 18}</math>. | + | Thus each two steps we cycle down a perfect square, and in <math>(10-1)\times 2 = 18</math> steps, we are left with <math>1</math> tile, hence our answer is <math>\boxed{\text{(C) } 18}</math>. |
== See also == | == See also == | ||
Revision as of 17:19, 26 December 2008
Problem
A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
Solution
The pattern is quite simple to see after listing a couple of terms.
![\[\begin{tabular}{|r|r|r|} \hline \#&\text{Removed}&\text{Left}\\ \hline 1&10&90\\ 2&9&81\\ 3&9&72\\ 4&8&64\\ 5&8&56\\ 6&7&49\\ 7&7&42\\ 8&6&36\\ 9&6&30\\ 10&5&25\\ 11&5&20\\ 12&4&16\\ 13&4&12\\ 14&3&9\\ 15&3&6\\ 16&2&4\\ 17&2&2\\ \boxed{18}&1&1\\ \hline \end{tabular}\]](http://latex.artofproblemsolving.com/e/1/1/e11497d55171669f7da40e12ef943be39c3f038b.png)
Given 
tiles, a step removes 
tiles, leaving 
tiles behind. Now, 
, so in the next step 
tiles are removed. This gives 
, another perfect square.
Thus each two steps we cycle down a perfect square, and in 
steps, we are left with 
tile, hence our answer is 
.
See also
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
