Question 407902
We want the sum of n consecutive two-digit numbers to be 600. Suppose that x, x+1, x+2, ..., x+n-1 are the two digit numbers, and


{{{sum(x+i, i=0, n-1) = 600}}}


This implies that {{{xn + sum(i, i=1, n-1) = 600}}} --> {{{xn + (n-1)(n)/2 = 600}}}, where x > 10 and x+n-1 < 100. 


Factoring our equation, this becomes {{{n(x + ((n-1)/2)) = 600}}}. Since {{{x + (n-1)/2 > 10}}}, it follows that {{{n < 60}}}. We can plug in odd numbers n (so that (n-1)/2 remains an integer) that are factors of 600 (note that 600 = 8*75, so we list all factors of 75):


n:------- x:
1 -------   600
3 -------   199
5 -------   118     
15-------   33
25-------   12
75-------   -29


The only cases that satisfy are {33, 34, 35, ..., 47} and {12, 14, 15, ..., 36}. It can be checked that their sums are 600.