Question 361865
Let x = amount of provision for each man for each day.
Then {{{5700x + 5680x + 5660x}}}+... +{{{(5700 + (n-1)*(-20))x = 66*5700x}}}.
Dividing both sides by x, we get
{{{5700 + 5680 + 5660}}}+... +{{{(5700 + (n-1)*(-20)) = 66*5700}}},

{{{5700 + 5680 + 5660}}}+... +{{{(5720-20n) = 66*5700}}}.
The left side is the sum of an arithmetic sequence, whose sum is 
{{{S = (n/2)*(5700+5720 - 20n) = n(5710-10n)}}}.  Thus 
{{{n(5710-10n) = 66*5700}}}, or
{{{n(571 - n) = 66*570}}}, or
{{{n^2 - 571n+37620 = 0}}}, or
{{{(n-495)(n-76) = 0}}}.
Thus n = 495, or n = 76.  The first value won't satisfy the 1st or the 2nd equation at the start of this solution, so n = 76.  Therefore the garrison can hold out for 76 days.