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

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 $100_{11}$ which is $121_{10}$ . because 11^2

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 $9^3-1$

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, because it is 9 possibilities for the hundreds digit, 10 for the tens digit, and 10 fo the units digits, so its 9x10x10 which is $900$ .

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

~CharmaineMa07292010

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

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.

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