Difference between revisions of "2003 AMC 10A Problems/Problem 20"
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== Solution == | == Solution == | ||
+ | To be a three digit number in base-10: | ||
− | + | <math>10^{2} \leq n \leq 10^{3}-1</math> | |
− | + | <math>100 \leq n \leq 999</math> | |
− | |||
− | + | Thus there are <math>900</math> three-digit numbers in base-10 | |
− | + | To be a three-digit number in base-9: | |
− | + | <math>9^{2} \leq n \leq 9^{3}-1</math> | |
− | + | <math>81 \leq n \leq 728</math> | |
+ | |||
+ | To be a three-digit number in base-11: | ||
+ | |||
+ | <math>11^{2} \leq n \leq 11^{3}-1</math> | ||
+ | |||
+ | <math>121 \leq n \leq 1330</math> | ||
+ | |||
+ | So, <math>121 \leq n \leq 728</math> | ||
+ | |||
+ | Thus, there are <math>608</math> base-10 three-digit numbers that are three digit numbers in base-9 and base-11. | ||
+ | |||
+ | Therefore the desired probability is <math>\frac{608}{900}\approx 0.7 \Rightarrow\boxed{\mathrm{(E)}\ 0.7}</math>. | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/SCGzEOOICr4?t=596 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/YaV5oanhAlU | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/-ei6Ni-jnlc | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Latest revision as of 23:04, 31 July 2023
Contents
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
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 .
Video Solution by OmegaLearn
https://youtu.be/SCGzEOOICr4?t=596
~ pi_is_3.14
Video Solution
~IceMatrix
Video Solution by WhyMath
~savannahsolver
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.