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

Revision as of 10:18, 15 January 2008

Problem20

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 thee-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 $\frac{608}{900}\approx 0.7 \Rightarrow E$ .

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

@@ Line 32: / Line 32: @@
 == See Also ==
-*[[2003 AMC 10A Problems]]
+{{AMC10 box|year=2003|ab=A|num-b=19|num-a=21}}
-*[[2003 AMC 10A Problems/Problem 19|Previous Problem]]
-*[[2003 AMC 10A Problems/Problem 21|Next Problem]]
 [[Category:Introductory Number Theory Problems]]

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

Revision as of 10:18, 15 January 2008

Problem20

Solution

See Also