Yesterday I bought two tickets to a movie, which logically had consecutive numbers. I examined the numbers, which had four digits, and mentioned that the sum of the eight digits was 25.

A friend asked if any digit appeared more than twice out of the eight, and I answered him.

Other friend asked if the sum of the digits of either ticket was equal to 13, which I also answered.

And then, to my surprise, my daughter told me what the two numbers were!

What were they?

We have two possibilities:

a) The girl that was selling the tickets is e.g.īs daughter.

b) She is not.

Forgeting the first one (e.g. wonīt do this to us), letīs analyze the second possibility:

It is said that the numbers were consecutively numbered, so they could only be:

I) abcd and abc(d+1)
II) abc9 and ab(c+1)0
III) ab99 and a(b+1)00.
We canīt have a999 and (a+1)000 because the sum of digits wouldnīt be 25.

Analyzing (I) - abcd and abc(d+1): if e.g.īs answer to his second friend had been YES ("YES, the sum of digits of either ticket is 13"), the only conclusion that e.g.īs daughter could have reached would have been that a+b+c+d = 12, and no matter what was his answer to his first friend, there would not have been a unique solution. So, e.g.īs answer to his second friend must have been NO, and it follows that the tickets could not have been numbered in this way.

Analyzing II - abc9 and ab(c+1)0: we have 2a + 2b + 2c + 10 = 25, and so 2(a+b+c) = 15, wich is impossible. Discharged.

So, the hypothesis in III is the true one, the tickets were numbered ab99 and a(b+1)00, and we have:

2a + 2b + 19 = 25, and a + b = 3, leading to 4 possibilities: (0399, 0400), (1299, 1300), (2199, 2200), and (3099, 3100).

Of these, three of them would have had e.g. answer "YES" to his first friend ("YES, a digit appears more than twice") - (0399, 0400), (2199, 2200), and (3099, 3100) - and only the pair (1299, 1300) would have had e.g. answer "NO".

So the tickets numbers are 1299 and 1300.

Posted by pcbouhid on 2005-11-22 11:48:19