Difference between revisions of "2003 AMC 10A Problems/Problem 20"
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The largest number in base-9 that has 3 digits in base-10 is <math>8\cdot9^2+8\cdot9^1+8\cdot9^0=888_{9}=728_{10}</math> | The largest number in base-9 that has 3 digits in base-10 is <math>8\cdot9^2+8\cdot9^1+8\cdot9^0=888_{9}=728_{10}</math> | ||
| − | Alternatively, you can do 9^3-1 | + | Alternatively, you can do <math>9^3-1</math> |
The smallest number in base-9 that has 3 digits in base-10 is <math>1\cdot9^2+2\cdot9^1+1\cdot9^0=121_{9}=100_{10}</math> | The smallest number in base-9 that has 3 digits in base-10 is <math>1\cdot9^2+2\cdot9^1+1\cdot9^0=121_{9}=100_{10}</math> | ||
Revision as of 17:35, 13 October 2019
Problem 20
A base-10 three digit number 
is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of are both three-digit numerals?
Solution
The smallest base-11 number that has 3 digits in base-10 is 
which is 
.
The largest number in base-9 that has 3 digits in base-10 is 
Alternatively, you can do 
The smallest number in base-9 that has 3 digits in base-10 is 
Hence, all numbers that will have 3 digits in base-9, 10, and 11 will be between 
and , thus the total amount of numbers that will have 3 digits in base-9, 10, and 11 is 
There are 900 possible 3 digit numbers in base 10.
Hence, the answer is 
See Also
| 2003 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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. 
