Turnip for the Books

‘It’s been a good year for turnips,’ farmer Hogswill remarked to his neighbour, Farmer Suticle.
‘Yup, that it has,’ the other replied. ‘How many did you grow?’
‘Well . . . I don’t exactly recall, but I do remember that when I took the turnips to market, I sold six-sevenths of them, plus one-seventh of a turnip, in the first hour.’
‘Must’ve been tricky cuttin’ ’em up.’
‘No, it was a whole number that I sold. I never cuts ’em.’
‘If’n you say so, Hogswill. Then what?’
‘I sold six-sevenths of what was left, plus one-seventh of a turnip, in the second hour. Then I sold six-sevenths of what was left, plus one-seventh of a turnip, in the third hour. And finally I sold six-sevenths of what was left, plus one-seventh of a turnip, in the fourth hour. Then I went home.’
‘Why?’
‘’Cos I’d sold the lot.’

How many turnips did Hogswill take to market?

(Source: Professor Stewart's Cabinet of Mathematical Curiosities, Ian Stewart)

Initially I was going to attempt to do this with a really horrible-looking single equation but there'd be no way I have enough of an attention span to hold it all together and solve it without errors. The solution I eventually realized was the right way to go, and the one the answers section suggests, is doing things recurrently, starting from the last, fourth hour. Let's state what Farmer Hogswill did in that hour algebraically:

\[\frac{6}{7}x_4 + \frac{1}{7} = x_4 \]

This states that he sold all of what remained, $x_4$, the subscript standing for the fourth hour, and that the sale was partitioned into six-sevenths of this unknown quantity and one-seventh of a turnip. The answer is one. The trick then is to take things back to the third hour. That is done accordingly:

\[ \frac{6}{7}x_3 + \frac{1}{7} + x_4 = x_3 \]

This states that the quantity of turnips Farmer Hogswill had at the beginning of the third hour was equal to six-sevenths of that quantity, plus one-seventh of a turnip as before, in addition to whatever is left behind for the next and last hour. Solving this equation, $x_3$ is equal to eight. Solving the remaining two equations:

\[ \frac{6}{7}x_2 + \frac{1}{7} + x_3 = x_2 \]

and:

\[ \frac{6}{7}x_1 + \frac{1}{7} + x_2 = x_1 \]

...it can be known that the final answer is 400.