Mathematics in the Story of Farmer's 3 Sons

What was the story?

Let X be the number of chapatis that farmer's wife made.

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First son gave 1 chapati to dog. Chapatis left = (X - 1)

Made 3 equal parts of remaining. Chapatis in each part = (X - 1)/3

He Ate one part of it. Chapatis left = 2(X - 1)/3.

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Second son gave 1 chapati to dog. Chapatis left = 2(X - 1)/3 - 1

Made 3 equal parts of remaining. Chapatis in each part = [2(X - 1)/3 - 1]/3

He Ate one part of it. Chapatis left = 2[2(X - 1)/3 - 1]/3 

                                                  = [4(X-1) – 6]/9

                                                  = (4X - 10)/9

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Third son gave 1 chapati to dog. Chapatis left = (4x - 10)/9 - 1

Made 3 equal parts of remaining. Chapatis in each part = [(4X - 10)/9 - 1]/3

He Ate one part of it. Chapatis left = 2[(4X - 10)/9 - 1]/3

                                                  = (8X – 20 – 18)/27 
       
                                                  = (8X – 38)/27 

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At that night, one of them gave 1 chapati to dog.

Chapatis left = (8X – 38)/27 - 1

After that they made 3 equal parts of remaining parts and each son ate one part.

If we assume that each son ate Y chapatis then,

Y = [(8X – 38)/27 - 1]/3

Y = (8X - 38 – 27)/81

Y =  (8X – 65)/81

81Y= 8X - 65

X = (81Y + 65)/8 

X = 81Y/8 + 65/8

X = 10Y + Y/8 + 8 + 1/8

X = 10y + 8 + (y+1)/8

For X to be integer, Y has to be 7,15,23.....

So possible values of X are - 79, 160, 241.

Since, question asks minimum number, the farmer's wife must have made 79 chapatis in total.