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?
Suppose the answers to the two questions are "no" and "no".
Let the two numbers be abcd and efgh.
If d != 9 then we must have a = e, b = f, c = g, d + 1 = h, and then the sum of the digits of the second number equal 13.
Therefore, we know that d = 9. Suppose c != 9. Then it must be true that the (a + b + c + d) - (e + f + g + h) = 8, which implies that the sum of all eight digits must be even, which impossible. Therefore, if there exists a solution at all, we must have c = 9. However, we can't have b = 9, or else the sum is too great.
Therefore, we have: c = d = 9, g = h = 0, a = e and f = b + 1. Therefore, the sum of them all is 2a + 2b + 19. Therefore, we need a + b = 3.
So, the possibilities are: (i) a = 3, b = 0
(ii) a = 2, b = 1
(iii) a = 1, b = 2
(iv) a = 0, b = 3
We can rule out (iv) because then the numbers aren't four digit numbers at all. In case (i) the numbers are 3099 and 3100. Here, 0 appears three times, so we can rule it out.
In case (ii), we have 2199 and 2200, where 2 appears three times.
However, in case (iii) the numbers are 1299 and 1300, and no number appears more than twice. This is the unique solution.
Posted by Neil
on 2005-11-22 12:42:14