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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, no computer Comment 10 of 10 |
x + y = 62; where x and y, are sums of the 1st and 2nd set, respectively.

Because we know that the sum of the 1st set and 2nd set must equal 62, it is possible to discover a large set of numerical configurations which can be excluded.  With two sets of consecutive five digit numbers, there are 2 general patterns which are possible, with respect to their number structure.  1 of them cannot satisfy x + y = 62, and is immediately excluded.  And the other pattern, does satisfy x + y = 62, in certain instances.  Here is an evaluation of each pattern:

PATTERN CASE 1) The sum of the first 5, is 1 different, than the sum of the second five.  Example:  1st set: 12345  2nd set: 12346

In this case x and y, as defined above, are related such that:      y = (x-1), thus:  x + (x-1) = 62.  This simplifies to 2x=63, and further to, x=31.5.  At this point, it becomes apparent, that in this case, x and y require fractional sums to satisfy.  And this is impossible.  Thus, all spatial-numerical patterns defined by this case, can be excluded.

PATTERN CASE 2) In the alternative case, involving 2 sets of consecutive, 5 digit numbers, a 'turnover' is required.  By turnover, I mean that a 9 is part of one of the sets, which becomes a 0, in the other (consecutive) set.  Example: 1st set: 01119    2nd set:  01120.

In a large number of valid (with respect to x + y = 62) instances that I have come across, the sum of one set is 35, and the sum of the second set is 27 (the difference of the sums is 8).  Examples of this include 76769 and 76770, 98909 and 98910, and 99899 and 99900.  However, I did run across one instance, where the first set had a sum of 44, and the second set had a sum of 18. 

Because it was possible that your friend asked only one question, and could deduce both 5-digit numbers, and because this problem has one correct answer, I will assume these are the only two possible 'difference in sums', between two sets of consecutive 5 digit numbers, which can satisfy x + y = 62.  Therefore, my answer is:    98999 and 99000.

When your friend asked you if the sum of digits of either of the tickets equals your age, whether you said yes or no, your friend knew to either exclude the common set of initial possibilities, that include a sum of 35, for one set, and a sum of 27, for the other set.  Therefore your friend was left with the instance above, 98999 and 99000, with a sum of 44 and 18, respectively.  And whether you answered yes or no, your age at that time, could have only been either 35, 27, 44 or 18 years old.  Because 2 years have expired since, it means that today, your age could only be either 37,29,46, or 20 years old.  So you must be under 50 years old today.

  Posted by John zadeh on 2010-04-27 14:40:23
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