Difference between revisions of "2003 AMC 10A Problems/Problem 20"
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Hence, the answer is <math>\frac{608}{900}\approx .675 \approx \boxed{0.7}</math> | Hence, the answer is <math>\frac{608}{900}\approx .675 \approx \boxed{0.7}</math> | ||
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| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/SCGzEOOICr4?t=596 | ||
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| + | ~ pi_is_3.14 | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 10:08, 4 November 2022
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 
Video Solution by OmegaLearn
https://youtu.be/SCGzEOOICr4?t=596
~ pi_is_3.14
Video Solution
~IceMatrix
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. 
