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

Revision as of 19:59, 5 June 2019

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?

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

it's A lmaooooo xd lel

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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@@ Line 4: / Line 4: @@
 <math> \mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7 </math>
-== Solution ==
+it's A lmaooooo xd lel
-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>.
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Difference between revisions of "2003 AMC 10A Problems/Problem 20"

Revision as of 19:59, 5 June 2019

Problem 20

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