Question

If 37/13 = 2+1/(x+1/(y+1/z)),where x,y,z are natural number..what are x,y,z=?

Answer 1

In such sums where a lot of variables are represented in a form of  '1 by' or '1 upon' it is important to remove this form. This can be only done using reciprocals on both the sides. Thus, always bring the left hand side ( L.H.S ) and ( R.H.S ) in a form that allows you to reciprocate. Initially this '2' on the R.H.S is stopping us from reciprocating thus our initial task is to try and remove this 2.

Thus simply taking the 2 on LHS and taking LCM we get,

11/13 = 1/(x+1/(y+1/z))

Now reciprocating on both sides,

13/11 = x+1/(y+1/z)

Now we must find the value of x to reciprocate again, this can be done by bringing 13/11 to a suitable form.
13/11 can be written as 1+ 2/11,

1 + 2/11 = x + 1/(y+1/z) 

Thus x = 1

Now,

2/11 = 1/(y+1/z)

Reciprocating again we get,

11/2 = y+1/z

11/2 can be written as 5 + 1/2 

5+1/2 = y+1/z

Thus y = 5

Now 1/2 = 1/z

Reciprocating again,
 
Z = 2

Answer 2

Answer:

Step-by-step explanation:

Given that-- 37/13=2+11/13=2+1/13/11

13/11=1+2/11

2+1/1+1/11/2

11/2=5+1/2

2+1/1+1/5+1/2

Therefore,a=2,b=1,c=5,d=2

So,a+b+c+d=?

Let us think answer be X,,

a+b+c+d=X

2+1+5+2=X

>>X=10<<