2003 AMC 10A Problems/Problem 20

Problem 20

A base-10 three digit number $n$ is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of $n$ are both three-digit numerals?

$\mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7$

Solution

The smallest base-11 number that has 3 digits in base-10 is $100_{11}$ which is $121_{10}$.

The largest number in base-9 that has 3 digits in base-10 is $8\cdot9^2+8\cdot9^1+8\cdot9^0=888_{9}=728_{10}$ Alternatively, you can do $9^3-1$

The smallest number in base-9 that has 3 digits in base-10 is $1\cdot9^2+2\cdot9^1+1\cdot9^0=121_{9}=100_{10}$

Hence, all numbers that will have 3 digits in base-9, 10, and 11 will be between $728_{10}$ and $121_{10}$, thus the total amount of numbers that will have 3 digits in base-9, 10, and 11 is $728-121+1=608$

There are 900 possible 3 digit numbers in base 10.

Hence, the answer is $\frac{608}{900}\approx .675 \approx \boxed{0.7}$

Video Solution

https://youtu.be/YaV5oanhAlU

~IceMatrix

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 10 Problems and Solutions

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