Difference between revisions of "2003 AMC 10A Problems/Problem 20"
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== Problem20 == | == Problem20 == | ||
− | A base-10 three digit number <math>n</math> is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of <math>n</math> are both | + | A base-10 three digit number <math>n</math> is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of <math>n</math> are both three-digit numerals? |
<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> |
Revision as of 23:07, 7 February 2010
Problem20
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
To be a three digit number in base-10:
Thus there are three-digit numbers in base-10
To be a three-digit number in base-9:
To be a three-digit number in base-11:
So,
Thus, there are base-10 three-digit numbers that are three digit numbers in base-9 and base-11.
Therefore the desired probability 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 |