Difference between revisions of "2003 AMC 10A Problems/Problem 20"

Revision as of 22:50, 16 July 2014

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

To be a three digit number in base-10:

$10^{2} \leq n \leq 10^{3}-1$

$100 \leq n \leq 999$

Thus there are $900$ three-digit numbers in base-10

To be a three-digit number in base-9:

$9^{2} \leq n \leq 9^{3}-1$

$81 \leq n \leq 728$

To be a three-digit number in base-11:

$11^{2} \leq n \leq 11^{3}-1$

$121 \leq n \leq 1330$

So, $121 \leq n \leq 728$

Thus, there are $608$ base-10 three-digit numbers that are three digit numbers in base-9 and base-11.

Therefore the desired probability is $\frac{608}{900}\approx 0.7 \Rightarrow\boxed{\mathrm{(E)}\ 0.7}$ .

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 29: / Line 29: @@
 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 E</math>.
+Therefore the desired probability is <math>\frac{608}{900}\approx 0.7 \Rightarrow\boxed{\mathrm{(E)}\ 0.7}</math>.
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Difference between revisions of "2003 AMC 10A Problems/Problem 20"

Revision as of 22:50, 16 July 2014

Problem 20

Solution

See Also