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Pick two numbers! (Posted on 2005-11-22) Difficulty: 4 of 5
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?

See The Solution Submitted by e.g.    
Rating: 3.8889 (9 votes)

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Solution Solution | Comment 4 of 15 |

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
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