(I) Each of x and y is a positive integer with x < y such that, reading from left to right, the first two digits in the base ten expansions of 1978^{x} and 1978^{y} are congruent.

Determine the minimum value of x+y.

(II) What is the minimum value of x+y - if, keeping all the other conditions in (i) unaltered, the first three digits in the base ten expansions of 1978^{x} and 1978^{y} are congruent?

__Note__: None of the expansions of 1978^{x} and 1978^{y} can contain any leading zero.

For 1978^x and 1978^y to start with the same digits then 1978^(y-x) will start with 99... or 100...

1978^27 starts with 995... Then 1978^(x+27) will have similar first digits to 1978^x.

x=1 makes 1978^1 equals 1978 and 1978^28 begin with 1969.... These match for the first two digits, so x+y=29 for part 1.

x=17 makes 1978^17 begin with 1086... and 1978^44 begin with 1081... These match for the first three digits, so x+y=61 for part 2.