Difference between revisions of "2003 AMC 10A Problems/Problem 20"
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== Solution == | == Solution == | ||
− | The smallest base-11 number that has 3 digits in base-10 is <math>100_{11}</math> which is <math>121_{10}</math>. | + | If we explore a similar problem: |
+ | Which positive intergers have 3 digits in base 10? | ||
+ | The smallest one ranges from 100-999, or 10^2 --> 10^3-1 | ||
+ | Therefore, | ||
+ | The smallest base-11 number that has 3 digits in base-10 is <math>100_{11}</math> which is <math>121_{10}</math>. because 11^2 | ||
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 <math>9^3-1 | + | Alternatively, you can do <math>9^3-1</math> |
− | |||
− | |||
Hence, all numbers that will have 3 digits in base-9, 10, and 11 will be between <math>728_{10}</math> and <math>121_{10}</math>, thus the total amount of numbers that will have 3 digits in base-9, 10, and 11 is <math>728-121+1=608</math> | Hence, all numbers that will have 3 digits in base-9, 10, and 11 will be between <math>728_{10}</math> and <math>121_{10}</math>, thus the total amount of numbers that will have 3 digits in base-9, 10, and 11 is <math>728-121+1=608</math> | ||
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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> | ||
+ | |||
+ | ~CharmaineMa07292010 | ||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Revision as of 07:33, 27 February 2023
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
If we explore a similar problem: Which positive intergers have 3 digits in base 10? The smallest one ranges from 100-999, or 10^2 --> 10^3-1 Therefore, The smallest base-11 number that has 3 digits in base-10 is which is . because 11^2
The largest number in base-9 that has 3 digits in base-10 is Alternatively, you can do
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
~CharmaineMa07292010
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.