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A streetcar named Enigma (Posted on 2010-02-10) Difficulty: 4 of 5
Two years ago, traveling with my friend (a logic wiz) I have purchased two streetcar tickets, consecutively numbered. I told my friend that the sum of all ten digits equals 62, which was his age. He than asked me whether the sum of digits (s.o.d) of either of the tickets equals my age (which he knew) and upon getting my answer quoted exactly both 5-digit numbers.

What were the numbers?
Am I over 50 today?

See The Solution Submitted by Ady TZIDON    
Rating: 4.6667 (3 votes)

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Solution re: almost empiric solution | Comment 6 of 10 |
(In reply to almost empiric solution by Elisabeth)

Let a, b, c, d and e be the digits of the first number. So, these are the steps to my solution:
1) e < 9 and the digits of the second number are a, b, c, d and e+1.
So, we get 2(a+b+c+d+e) + 1 = 62, which is impossible. We get e = 9.
2) e = 9, d < 9 and the digits of the second number are a, b, c, d+1 and 0 (zero).
So, we get 2(a+b+c+d) + 1 = 62 - 9, then a+b+c+d=26. We would have a lot of combinations yielding this. The sums of the digits would  be, respectively, 35 and 27.
3) e = 9, d = 9, c < 9 and the digits of the second number are a, b, c+1, 0 (zero) and 0 (zero).
So, we get 2(a+b+c) + 1 = 62 - 18, which is impossible. If e = 9 and d = 9, we must have c = 9.
4) e = 9, d = 9, c = 9, b < 9; and the digits of the second number are a, b+1, 0 (zero), 0 (zero) and 0 (zero).
So, we get 2(a+b) + 1 = 62 - 27, then a+b=17. Since we put b < 9, the numbers would be 98999 and 99000. The sum of digits would be, respectively, 44 and 18.
5) e = 9, d = 9, c = 9, b = 9, a < 9; and the digits of the second number are a+1, 0 (zero), 0 (zero), 0 (zero) and 0 (zero).
So, we get 2a+1 = 62 - 36, then 2a = 25, which is impossible.

If the buyer answered "Yes", in the second case (his age equals 27 or 35), his friend would not be able to get the numbers without ambiguity. So, if he answered "Yes" and his friend had an immediate solution, we must assume that the buyer was 44 or 18 years old at that time.
If the buyer was not 18, 27, 35 or 42 (his answer would be "no"), his friend would be at a total loss in finding the numbers.

Conclusion: the numbers are 98999 and 99000, and the buyer's is NOT over 50 today (he is now 20 or 46 years old).

 


  Posted by Elisabeth on 2010-02-12 14:11:25
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