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Difference between revisions of "2002 AMC 10A Problems/Problem 22"

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<math>\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20</math>
 
<math>\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20</math>
  
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== Solution ==
+
==Solution 1==
−
===Solution 1===
 
 
The pattern is quite simple to see after listing a couple of terms.
 
The pattern is quite simple to see after listing a couple of terms.
  
Line 32: Line 31:
 
\hline
 
\hline
 
\end{tabular} </cmath>
 
\end{tabular} </cmath>
 +
 
Thus, the answer is <math>\boxed{\text{(C) } 18}</math>.
 
Thus, the answer is <math>\boxed{\text{(C) } 18}</math>.
  
−
===Solution 2===
+
==Solution 2==
 
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.
  

Revision as of 14:52, 30 April 2021

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?

$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$

Solution 1

The pattern is quite simple to see after listing a couple of terms.

\[\begin{tabular}{|r|r|r|} \hline \#&\text{Removed}&\text{Left}\\ \hline 1&10&90\\ 2&9&81\\ 3&9&72\\ 4&8&64\\ 5&8&56\\ 6&7&49\\ 7&7&42\\ 8&6&36\\ 9&6&30\\ 10&5&25\\ 11&5&20\\ 12&4&16\\ 13&4&12\\ 14&3&9\\ 15&3&6\\ 16&2&4\\ 17&2&2\\ 18&1&1\\ \hline \end{tabular}\]

Thus, the answer is $\boxed{\text{(C) } 18}$.

Solution 2

Given $n^2$ tiles, a step removes $n$ tiles, leaving $n^2 - n$ tiles behind. Now, $(n-1)^2 = n^2 - n + (1-n) < n^2 - n < n^2$, so in the next step $n-1$ tiles are removed. This gives $(n^2 - n) - (n-1) = n^2 - 2n + 1 = (n-1)^2$, another perfect square.

Thus each two steps we cycle down a perfect square, and in $(10-1)\times 2 = 18$ steps, we are left with $1$ tile, hence our answer is $\boxed{\text{(C) } 18}$.

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

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