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> | ||
− | == Solution == | + | ==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. | ||
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16&2&4\\ | 16&2&4\\ | ||
17&2&2\\ | 17&2&2\\ | ||
− | + | 18&1&1\\ | |
+ | |||
+ | |||
+ | |||
+ | |||
\hline | \hline | ||
\end{tabular} </cmath> | \end{tabular} </cmath> | ||
+ | Thus, the answer is <math>\boxed{\text{(C) } 18}</math>. | ||
+ | |||
+ | ==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. | ||
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>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | We start of with <math>100 = 10 \cdot 10</math> numbers. When we use the certain operation, call if <math>P(x)</math>, have <math>100 - 10 = 90 = 10 \cdot 9</math>. | ||
+ | Then we do <math>P(x)</math> again, to subtract <math>9</math> numbers and get <math>9 \cdot 9</math>. In the end, we will want <math>1 = 1 \cdot 1</math>. We can say we have to use <math>P(x)</math> once to make <math>n \cdot n</math> into <math>n \cdot (n-1)</math>. Thus we must use it twice to get from <math>n \cdot n</math> to <math>(n-1)(n-1)</math>. For example, it takes us <math>2</math> of <math>P(x)</math> to get from <math>10 \cdot 10</math> to <math>9 \cdot 9</math>. Then <math>2</math> of <math>P(x)</math> to get from <math>9 \cdot 9</math> to <math>8 \cdot 8</math>. You should try this with <math>7</math> and <math>6</math>, and see it works. This means we can have <math>n</math> be the number we start with, and <math>1</math> be the number we want. Then we would use <math>P(x)</math>, <math>2(n - 1)</math> times to get <math>1 \cdot 1</math>. Substituting <math>n</math> for <math>10</math> we get <math>2(10-1) = 2 \cdot 9 = \boxed{18}</math>. | ||
+ | - Wiselion | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://www.youtube.com/watch?v=CuKko0JpIdQ ~David | ||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 23:13, 13 September 2023
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 1
The pattern is quite simple to see after listing a couple of terms.
Thus, the answer is .
Solution 2
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
.
Solution 3
We start of with numbers. When we use the certain operation, call if
, have
.
Then we do
again, to subtract
numbers and get
. In the end, we will want
. We can say we have to use
once to make
into
. Thus we must use it twice to get from
to
. For example, it takes us
of
to get from
to
. Then
of
to get from
to
. You should try this with
and
, and see it works. This means we can have
be the number we start with, and
be the number we want. Then we would use
,
times to get
. Substituting
for
we get
.
- Wiselion
Video Solution
https://www.youtube.com/watch?v=CuKko0JpIdQ ~David
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.