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

Latest revision as of 23:04, 31 July 2023

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 $\frac{608}{900}\approx 0.7 \Rightarrow\boxed{\mathrm{(E)}\ 0.7}$ .

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
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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 5: / Line 5: @@
 == Solution ==
+To be a three digit number in base-10:
-If we explore a similar problem:
+<math>10^{2} \leq n \leq 10^{3}-1</math>
-Which positive integers 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 <math>100_{11}</math> which is <math>121_{10}</math>. because 11^2
-The largest number in base-9 that has 3 digits in base-10 is <math>8\cdot9^2+8\cdot9^1+8\cdot9^0=888_{9}=728_{10}</math>
+<math>100 \leq n \leq 999</math>
-Alternatively, you can do <math>9^3-1</math>
-Hence, all numbers that will have 3 digits in base-9, 10, and 11 will be between <math>728_{10}</math> and <math>121_{10}</math>, thus the total amount of numbers that will have 3 digits in base-9, 10, and 11 is <math>728-121+1=608</math>
+Thus there are <math>900</math> three-digit numbers in base-10
-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 <math>900</math>.
+To be a three-digit number in base-9:
-Hence, the answer is <math>\frac{608}{900}\approx .675 \approx \boxed{(E)\;0.7}</math>
+<math>9^{2} \leq n \leq 9^{3}-1</math>
-~CharmaineMa07292010
+<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>.
 == Video Solution by OmegaLearn ==

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