# Four digit Ticket Number

Question: I was sitting around with my friend Dave, Marty and James recently. I happened to have two tickets to a new movie in my pocket that I had just purchased. I mentioned that there were two four-digit numbers on the tickets and that the sum of all 8 digits was 25.

Dave asked if any digit appeared more out of the 8, which I answered. Then, Marty asked if the sum of the digits of either ticket was equal to 13, which I answered too. Much to my surprise James immediately told me what the two numbers were!
What were they?

The first thing James realized was that the tickets were consecutively numbered. There are 4 possible cases:

1. ABCD and ABC(D+1)
2. ABC9 and AB(C+1)0
3. AB99 and A(B+1)00
4. A999 and (A+1)000

If the tickets were numbered ABCD and ABC(D+1) and my answer to Marty's question had been "YES", the only conclusion James could have reached would have been that A+B+C+D=12, and regardless of my answer to Dave's question there would not have been a unique solution. So
my answer to Marty's question must have been "NO". It follows that the tickets could not have been numbered in this manner.

If the numbers were ABC9 and AB(C+1)0, we would have 2A+2B+2C+10=25, and so 2(A+B+C)=15, which is impossible.

If the numbers were AB99 and A(B+1)00, we would have 2A+2B+19=25 or A+B=3, leading to the four possibilities (0399 and 0400), (1299 and 1300), (2199 and 2200), (3099 and 3100). Of these, three of them would have had me answer "YES" to Dave's question (if any digit
appeared more?), and only the pair (1299 and 1300) would have had me answer "NO". It follows that these were my ticket numbers i.e. (1299 and 1300)

Similarly, if the numbers were A999 and (A+1)000, we would have 2A + 28 = 25, which is again impossible.

Thus, my ticket numbers were 1299 and 1300.