Difference between revisions of "2002 AMC 12B Problems/Problem 1"
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− | == Problem == | + | {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #1]] and [[2002 AMC 10B Problems|2002 AMC 10B #3]]}}== Problem == |
The [[arithmetic mean]] of the nine numbers in the set <math>\{9, 99, 999, 9999, \ldots, 999999999\}</math> is a <math>9</math>-digit number <math>M</math>, all of whose digits are distinct. The number <math>M</math> does not contain the digit | The [[arithmetic mean]] of the nine numbers in the set <math>\{9, 99, 999, 9999, \ldots, 999999999\}</math> is a <math>9</math>-digit number <math>M</math>, all of whose digits are distinct. The number <math>M</math> does not contain the digit | ||
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\qquad\mathrm{(E)}\ 8</math> | \qquad\mathrm{(E)}\ 8</math> | ||
== Solution == | == Solution == | ||
− | + | We wish to find <math>\frac{9+99+\cdots +999999999}{9}</math>, or <math>\frac{9(1+11+111+\cdots +111111111)}{9}=123456789</math>. This does not have the digit 0, so the answer is <math>\boxed{\mathrm{(A)}\ 0}</math> | |
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== See also == | == See also == | ||
+ | {{AMC10 box|year=2002|ab=B|num-b=2|num-a=4}} | ||
{{AMC12 box|year=2002|ab=B|before=First question|num-a=2}} | {{AMC12 box|year=2002|ab=B|before=First question|num-a=2}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 16:19, 28 July 2011
- The following problem is from both the 2002 AMC 12B #1 and 2002 AMC 10B #3, so both problems redirect to this page.== Problem ==
The arithmetic mean of the nine numbers in the set is a -digit number , all of whose digits are distinct. The number does not contain the digit
Solution
We wish to find , or . This does not have the digit 0, so the answer is
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 |
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by First question |
Followed by Problem 2 |
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 12 Problems and Solutions |