Find the last three digits of 79999
Please find the solution,
r + (1/r) = 2007 + (1/2007) and
s + (1/s) = 2006 + (1/2006)
=> r = 2007 or (1/2007) and
s = 2006 or (1/2006)
We can verify these results by solving the two quadratic equations also
2007r2 - (20072 + 1)r + 2007 = 0 and
2006s2 - (20062 + 1)s + 2006 = 0
Now (r - s) is maximum when r is maximum and s is smallest.
So (r - s)maximum = rmax - smin = 2007 - (1/2006) = (4026041/2006) = 2006.9999
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