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Difference between revisions of "2002 AMC 10A Problems/Problem 21"

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==Solution==
 
==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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Since the mean is 8, the sum of the eight numbers must be 64. Now we use the fact that the range is 8 to experiment. Letting the smallest number be 7, we find the largest is 15. Since the unique mode is 8, there must be atleast two eights. The sum of those four numbers is 38 so the other four numbers sum to 26. That means the average of the other four numbers is 6.5, which is below 7. Hence, atleast one of the numbers is below 7, which contradicts our assumption that the smallest is 7.  
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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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Now let's try with the smallest number equal to 6. The largest number is then 14. If we take two sixes and four eights, we get a sequence that fits all the requirements. Hence, our answer <math>\boxed{\text{(D)}\ 14 }</math>.
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2002|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2002|ab=A|num-b=20|num-a=22}}
  
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[[Category:Introductory Combinatorics Problems]]
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[[Category:Introductory Algebra Problems]]

Revision as of 10:25, 27 December 2008

Problem

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

$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution

Since the mean is 8, the sum of the eight numbers must be 64. Now we use the fact that the range is 8 to experiment. Letting the smallest number be 7, we find the largest is 15. Since the unique mode is 8, there must be atleast two eights. The sum of those four numbers is 38 so the other four numbers sum to 26. That means the average of the other four numbers is 6.5, which is below 7. Hence, atleast one of the numbers is below 7, which contradicts our assumption that the smallest is 7.

Now let's try with the smallest number equal to 6. The largest number is then 14. If we take two sixes and four eights, we get a sequence that fits all the requirements. Hence, our answer $\boxed{\text{(D)}\ 14 }$.

See Also

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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