The sum of two numbers is 1215 and their HCF is 81. If the numbers lie between 500 and 700, then the sum of the reciprocals of the numbers is _____.

$\frac{5}{1512}$

Let the numbers be 81x and 81y, then

81x + 81y = 1215

x + y = 15

Possible pair of x + y are = (1, 14), (2, 13) and (4, 11) an (7, 8)

Number should be between 500 and 700, then

x = 7 and y = 8

First number = 81 × 7 = 567

Second number = 81 × 8 = 648

Sum of the reciprocals of the numbers is = \(\frac{1}{567}\) + \(\frac{1}{648}\) = \(\frac{1}{81}\)(\(\frac{1}{7}\) + \(\frac{1}{8}\)) = $\frac{5}{1512}$