Determine the maximum value of a positive integer x that satisfies [x/2] + [x/3] + [x/11] + [x/13] < x, where [x] denotes the greatest integer ≤ x.
One possibility area of investigation are those numbers that leave a remainder of 1 when divided by 2, 2 when divided by 3, 10 when divided by 11, and 12 when divided by 13.
The smallest of these is 2*3*11*13 - 1 = 857
For x= 857, [x/2] + [x/3] + [x/11] + [x/13] - x = -2, so it works
The next largest in this family is reached by adding 858 (ie, 2*3*11*13).
For x = 1715, [x/2] + [x/3] + [x/11] + [x/13] - x = -1, so it works
x = 2573 works also, as [x/2] + [x/3] + [x/11] + [x/13] - x = 0
x = 3431 fails, as [x/2] + [x/3] + [x/11] + [x/13] - x = 1
So, 2573 is the best I have found.
I await a final answer from a computer program.
Possible answer (spoiler?)
