Question
Consider the set of all fractions of the form \frac{2^m}{3^n}, where m and n are integers such that 0 \le m \le n \le 3. For example, \frac{1}{9} and \frac{8}{27} would be included in this set, but not \frac{32}{81}. What is the sum of all the numbers in this set?
Answers
GPT-4o mini
We need to consider all fractions of the form \( \frac{2^m}{3^n} \), where \( m \) and \( n \) are integers satisfying \( 0 \leq m \leq n \leq 3 \).
Let's enumerate all the possible values of \( m \) and \( n \) under the constraints given.
1. **If \( n = 0 \)**:
- \( m = 0 \) (since \( 0 \leq m \leq 0 \)): \( \frac{2^0}{3^0} = 1 \)
2. **If \( n = 1 \)**:
- \( m = 0 \): \( \frac{2^0}{3^1} = \frac{1}{3} \)
- \( m = 1 \): \( \frac{2^1}{3^1} = \frac{2}{3} \)
3. **If \( n = 2 \)**:
- \( m = 0 \): \( \frac{2^0}{3^2} = \frac{1}{9} \)
- \( m = 1 \): \( \frac{2^1}{3^2} = \frac{2}{9} \)
- \( m = 2 \): \( \frac{2^2}{3^2} = \frac{4}{9} \)
4. **If \( n = 3 \)**:
- \( m = 0 \): \( \frac{2^0}{3^3} = \frac{1}{27} \)
- \( m = 1 \): \( \frac{2^1}{3^3} = \frac{2}{27} \)
- \( m = 2 \): \( \frac{2^2}{3^3} = \frac{4}{27} \)
- \( m = 3 \): \( \frac{2^3}{3^3} = \frac{8}{27} \)
Now, we list all the fractions we have computed:
- From \( n = 0 \): \( 1 \)
- From \( n = 1 \): \( \frac{1}{3}, \frac{2}{3} \)
- From \( n = 2 \): \( \frac{1}{9}, \frac{2}{9}, \frac{4}{9} \)
- From \( n = 3 \): \( \frac{1}{27}, \frac{2}{27}, \frac{4}{27}, \frac{8}{27} \)
Next, we add all the fractions together:
\[
1 + \frac{1}{3} + \frac{2}{3} + \frac{1}{9} + \frac{2}{9} + \frac{4}{9} + \frac{1}{27} + \frac{2}{27} + \frac{4}{27} + \frac{8}{27}
\]
**First, sum the fractions with common denominators.**
1. **For denominators \( 1 \), \( 3 \), \( 9 \), and \( 27 \)**, we can express all terms with a common denominator of \( 27 \):
- \( 1 = \frac{27}{27} \)
- \( \frac{1}{3} = \frac{9}{27} \)
- \( \frac{2}{3} = \frac{18}{27} \)
- \( \frac{1}{9} = \frac{3}{27} \)
- \( \frac{2}{9} = \frac{6}{27} \)
- \( \frac{4}{9} = \frac{12}{27} \)
- \( \frac{1}{27} = \frac{1}{27} \)
- \( \frac{2}{27} = \frac{2}{27} \)
- \( \frac{4}{27} = \frac{4}{27} \)
- \( \frac{8}{27} = \frac{8}{27} \)
Adding these all together gives:
\[
\frac{27}{27} + \frac{9}{27} + \frac{18}{27} + \frac{3}{27} + \frac{6}{27} + \frac{12}{27} + \frac{1}{27} + \frac{2}{27} + \frac{4}{27} + \frac{8}{27}
\]
Calculating the numerator:
\[
27 + 9 + 18 + 3 + 6 + 12 + 1 + 2 + 4 + 8 = 90
\]
Thus, we have:
\[
\frac{90}{27} = \frac{10}{3}
\]
Therefore, the sum of all the numbers in this set is:
\[
\boxed{\frac{10}{3}}
\]
Let's enumerate all the possible values of \( m \) and \( n \) under the constraints given.
1. **If \( n = 0 \)**:
- \( m = 0 \) (since \( 0 \leq m \leq 0 \)): \( \frac{2^0}{3^0} = 1 \)
2. **If \( n = 1 \)**:
- \( m = 0 \): \( \frac{2^0}{3^1} = \frac{1}{3} \)
- \( m = 1 \): \( \frac{2^1}{3^1} = \frac{2}{3} \)
3. **If \( n = 2 \)**:
- \( m = 0 \): \( \frac{2^0}{3^2} = \frac{1}{9} \)
- \( m = 1 \): \( \frac{2^1}{3^2} = \frac{2}{9} \)
- \( m = 2 \): \( \frac{2^2}{3^2} = \frac{4}{9} \)
4. **If \( n = 3 \)**:
- \( m = 0 \): \( \frac{2^0}{3^3} = \frac{1}{27} \)
- \( m = 1 \): \( \frac{2^1}{3^3} = \frac{2}{27} \)
- \( m = 2 \): \( \frac{2^2}{3^3} = \frac{4}{27} \)
- \( m = 3 \): \( \frac{2^3}{3^3} = \frac{8}{27} \)
Now, we list all the fractions we have computed:
- From \( n = 0 \): \( 1 \)
- From \( n = 1 \): \( \frac{1}{3}, \frac{2}{3} \)
- From \( n = 2 \): \( \frac{1}{9}, \frac{2}{9}, \frac{4}{9} \)
- From \( n = 3 \): \( \frac{1}{27}, \frac{2}{27}, \frac{4}{27}, \frac{8}{27} \)
Next, we add all the fractions together:
\[
1 + \frac{1}{3} + \frac{2}{3} + \frac{1}{9} + \frac{2}{9} + \frac{4}{9} + \frac{1}{27} + \frac{2}{27} + \frac{4}{27} + \frac{8}{27}
\]
**First, sum the fractions with common denominators.**
1. **For denominators \( 1 \), \( 3 \), \( 9 \), and \( 27 \)**, we can express all terms with a common denominator of \( 27 \):
- \( 1 = \frac{27}{27} \)
- \( \frac{1}{3} = \frac{9}{27} \)
- \( \frac{2}{3} = \frac{18}{27} \)
- \( \frac{1}{9} = \frac{3}{27} \)
- \( \frac{2}{9} = \frac{6}{27} \)
- \( \frac{4}{9} = \frac{12}{27} \)
- \( \frac{1}{27} = \frac{1}{27} \)
- \( \frac{2}{27} = \frac{2}{27} \)
- \( \frac{4}{27} = \frac{4}{27} \)
- \( \frac{8}{27} = \frac{8}{27} \)
Adding these all together gives:
\[
\frac{27}{27} + \frac{9}{27} + \frac{18}{27} + \frac{3}{27} + \frac{6}{27} + \frac{12}{27} + \frac{1}{27} + \frac{2}{27} + \frac{4}{27} + \frac{8}{27}
\]
Calculating the numerator:
\[
27 + 9 + 18 + 3 + 6 + 12 + 1 + 2 + 4 + 8 = 90
\]
Thus, we have:
\[
\frac{90}{27} = \frac{10}{3}
\]
Therefore, the sum of all the numbers in this set is:
\[
\boxed{\frac{10}{3}}
\]