A number is expressed as `2^mxx3^n` and the sum of all its factors is 124, find m and n.

Options

A. 2, 3
B. 3, 1
C. 4,1
D. 3, 2

Answer

To solve the problem, we need to find the values of \( m \) and \( n \) such that the number expressed as \( 2^m \times 3^n \) has a sum of all its factors equal to 124.

### Step-by-Step Solution:

1. **Understanding the Sum of Factors Formula**:
The sum of the factors of a number \( N = p_1^{k_1} \times p_2^{k_2} \) is given by the formula:
\[ \sigma(N) = (1 + p_1 + p_1^2 + \ldots + p_1^{k_1})(1 + p_2 + p_2^2 + \ldots + p_2^{k_2}) \]
For our case, where \( N = 2^m \times 3^n \), the sum of the factors can be expressed as:
\[ \sigma(N) = (1 + 2 + 2^2 + \ldots + 2^m)(1 + 3 + 3^2 + \ldots + 3^n) \]

2. **Calculating Each Part**:
The first part can be calculated using the formula for the sum of a geometric series:
\[ 1 + 2 + 2^2 + \ldots + 2^m = \frac{2^{m+1} - 1}{2 - 1} = 2^{m+1} - 1 \]
The second part is:
\[ 1 + 3 + 3^2 + \ldots + 3^n = \frac{3^{n+1} - 1}{3 - 1} = \frac{3^{n+1} - 1}{2} \]

3. **Setting Up the Equation**:
Now, we can set up the equation for the sum of the factors:
\[ \sigma(N) = (2^{m+1} - 1) \left( \frac{3^{n+1} - 1}{2} \right) = 124 \]
Multiplying both sides by 2 to eliminate the fraction:
\[ (2^{m+1} - 1)(3^{n+1} - 1) = 248 \]

4. **Finding Factor Pairs of 248**:
Next, we need to find pairs of factors of 248. The factor pairs of 248 are:
- (1, 248)
- (2, 124)
- (4, 62)
- (8, 31)
- (16, 15)

5. **Testing Each Factor Pair**:
We will test each factor pair to see if they can be expressed in the form \( (2^{m+1} - 1) \) and \( (3^{n+1} - 1) \).

- For the factor pair (8, 31):
\[ 2^{m+1} - 1 = 8 \implies 2^{m+1} = 9 \implies m+1 = 3 \implies m = 2 \]
\[ 3^{n+1} - 1 = 31 \implies 3^{n+1} = 32 \implies n+1 = 5 \implies n = 4 \]

- Check if \( m = 2 \) and \( n = 4 \) satisfy the original sum of factors:
\[ \sigma(N) = (2^{2+1} - 1)(3^{4+1} - 1) = (8 - 1)(243 - 1) = 7 \times 242 = 1694 \quad \text{(not valid)} \]

- For the factor pair (4, 62):
\[ 2^{m+1} - 1 = 4 \implies 2^{m+1} = 5 \implies m+1 = 3 \implies m = 2 \]
\[ 3^{n+1} - 1 = 62 \implies 3^{n+1} = 63 \implies n+1 = 3 \implies n = 2 \]

- Check if \( m = 2 \) and \( n = 2 \) satisfy the original sum of factors:
\[ \sigma(N) = (2^{2+1} - 1)(3^{2+1} - 1) = (8 - 1)(27 - 1) = 7 \times 26 = 182 \quad \text{(not valid)} \]

- For the factor pair (16, 15):
\[ 2^{m+1} - 1 = 16 \implies 2^{m+1} = 17 \implies m+1 = 5 \implies m = 4 \]
\[ 3^{n+1} - 1 = 15 \implies 3^{n+1} = 16 \implies n+1 = 4 \implies n = 3 \]

- Check if \( m = 4 \) and \( n = 3 \) satisfy the original sum of factors:
\[ \sigma(N) = (2^{4+1} - 1)(3^{3+1} - 1) = (32 - 1)(81 - 1) = 31 \times 80 = 248 \quad \text{(valid)} \]

6. **Conclusion**:
The values of \( m \) and \( n \) that satisfy the condition are:
\[ m = 4, n = 3 \]