Cannot figure this out

[gamaz321]

I have no idea how to get started with the below problem. Any help or advice is appreciated.
Thank you.

If 139/22 = a + 1/(b + 1/c) where a, b and c are positive intergers.
Find the value of (a + b + c)

 

[Dr.Peterson]

I have no idea how to get started with the below problem. Any help or advice is appreciated.
Thank you.

If 139/22 = a + 1/(b + 1/c) where a, b and c are positive integers.
Find the value of (a + b + c)

Try long division. What mixed number do you get when you divide 139 by 22?

Then look at the fraction part. What do you have to do to make it look like 1.(b + 1/c)? (Hint: think about reciprocals.)

This is related to the subject of continued fractions, but you don't need to know anything about them to solve it.

 

[Steven G]

Is it really 139/22=a+ 1/(b + 1/c)?
Or is it 139/22=a+ 1/b + 1/c?

 

[Steven G]

I too thought of long division. It gives you a immediately

 

[gamaz321]

Try long division. What mixed number do you get when you divide 139 by 22?

Then look at the fraction part. What do you have to do to make it look like 1.(b + 1/c)? (Hint: think about reciprocals.)

This is related to the subject of continued fractions, but you don't need to know anything about them to solve it.

Thanks for your help. Yes, by following your advice I could solve the problem.