Question

97/19=a+1/b+1/c where a,b,c are positive integers find the a+b+c​

Answer 1

Answer:

Mean = 2

Proof:

Given a, b and c are positive integers and

1/(a+ (1/(b+ (1/ (c+1/2))))) = 16/23 ……………………………………………………………….(1)

Or, (a+ (1/(b+ (1/ (c+1/2))))) = 23/16 [Taking reciprocals on both sides]

Or, a+ (1/(b+ (1/ (c+1/2)))) = 1 + 7/16

Since a is a positive integer and the right-side consists of a positive integer and a fraction,

a = 1

∴ (1/(b+ (1/ (c+1/2)))) = 7/16

Again taking reciprocals on both sides,

(b+ (1/ (c+1/2))) = 16/7

Or, b+ (1/ (c+1/2)) = 2 + 2/7

By the same argument as before for a,

b = 2

∴ 1/ (c+1/2) = 2/7

Taking reciprocal for the last time,

c + 1/2 = 7/2

Or, c = 7/2 - 1/2 = (7–1)/2 = 6/2 = 3

Thus a = 1, b = 2, c = 3

and the mean of a, b and c

= (a+b+c)/3 = (1+2+3)/3

= 6/3

= 2