To solve the problem step-by-step, we will follow these steps:
### Step 1: Define the numbers using HCF
Let the two numbers be \( a \) and \( b \). Given that their HCF is 81, we can express the numbers as:
\[
a = 81x \quad \text{and} \quad b = 81y
\]
where \( x \) and \( y \) are co-prime integers.
### Step 2: Set up the equation based on the sum of the numbers
We know that the sum of the two numbers is 1215:
\[
a + b = 1215
\]
Substituting the expressions for \( a \) and \( b \):
\[
81x + 81y = 1215
\]
Factoring out 81 gives:
\[
81(x + y) = 1215
\]
### Step 3: Solve for \( x + y \)
Dividing both sides by 81:
\[
x + y = \frac{1215}{81}
\]
Calculating the right side:
\[
x + y = 15
\]
### Step 4: Find pairs of \( (x, y) \)
Now we need to find pairs of integers \( (x, y) \) such that:
\[
x + y = 15
\]
The possible pairs are:
- \( (1, 14) \)
- \( (2, 13) \)
- \( (3, 12) \)
- \( (4, 11) \)
- \( (5, 10) \)
- \( (6, 9) \)
- \( (7, 8) \)
### Step 5: Calculate the corresponding numbers and check the range
We need to check which pairs give numbers between 500 and 700:
- For \( (1, 14) \): \( 81 \times 1 = 81 \) and \( 81 \times 14 = 1134 \) (not valid)
- For \( (2, 13) \): \( 81 \times 2 = 162 \) and \( 81 \times 13 = 1053 \) (not valid)
- For \( (3, 12) \): \( 81 \times 3 = 243 \) and \( 81 \times 12 = 972 \) (not valid)
- For \( (4, 11) \): \( 81 \times 4 = 324 \) and \( 81 \times 11 = 891 \) (not valid)
- For \( (5, 10) \): \( 81 \times 5 = 405 \) and \( 81 \times 10 = 810 \) (not valid)
- For \( (6, 9) \): \( 81 \times 6 = 486 \) and \( 81 \times 9 = 729 \) (not valid)
- For \( (7, 8) \): \( 81 \times 7 = 567 \) and \( 81 \times 8 = 648 \) (valid)
### Step 6: Calculate the sum of the reciprocals
Now we have the valid numbers \( 567 \) and \( 648 \). We need to find the sum of their reciprocals:
\[
\frac{1}{567} + \frac{1}{648}
\]
Finding a common denominator:
\[
\text{LCM of } 567 \text{ and } 648 = 3672
\]
Calculating the individual fractions:
\[
\frac{1}{567} = \frac{648}{3672} \quad \text{and} \quad \frac{1}{648} = \frac{567}{3672}
\]
Adding these:
\[
\frac{648 + 567}{3672} = \frac{1215}{3672}
\]
### Step 7: Simplify the fraction
Now we simplify \( \frac{1215}{3672} \):
\[
\frac{1215 \div 81}{3672 \div 81} = \frac{15}{45.5} = \frac{5}{1224}
\]
### Final Answer
Thus, the sum of the reciprocals of the two numbers is:
\[
\frac{5}{1512}
\]
