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| − | ==Problem==
| + | #redirect [[2002 AMC 12A Problems/Problem 15]] |
| − | The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
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| − | <math>\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15</math>
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| − | ==Solution==
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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, and the process repeats.
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| − | 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.
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=A|num-b=20|num-a=22}}
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| − | [[Category:Introductory Combinatorics Problems]] | |
