Difference between revisions of "2003 AMC 10A Problems/Problem 20"
(→Problem 20) |
(→Problem 20) |
||
Line 3: | Line 3: | ||
<math> \mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7 </math> | <math> \mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7 </math> | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | The smallest base-11 number that has 3 digits in base-10 is <math>100_{11}</math> which is <math>121_{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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Hence, all numbers that will have 3 digits in base-9, 10, and 11 will be between <math>728_{10} and </math>121_{10}, thus the total amount of numbers that will have 3 digits in base-9, 10, and 11 is <math>728-121+1=608</math> | ||
+ | |||
+ | There are 900 possible 3 digit numbers in base 10. | ||
+ | |||
+ | Hence, the answer is <math>\frac{608}{900}\approx .675 \approx \boxed{0.7}</math> | ||
== See Also == | == See Also == |
Revision as of 15:31, 14 June 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
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 121_{10}, 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.