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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

593.0. "Ticket numbers" by MODEL::YARBROUGH () Mon Oct 06 1986 17:15

This one's from "The Sunday Times Book of Brain Teasers" by Bryant and 
Postill:

On a recent bus journey I purchased the tickets for my wife and myself. On 
each was a four-figure [serial] number, and the sum of all eight digits was 
25. I remarked upon this to my wife, who thereupon asked if any digit 
appeared more than twice [among the eight] and whether the sum of the 
digits on either ticket was 13. I answered both questions, and she was able 
to deduce the two numbers on the tickets. What were the two numbers?
T.RTitleUserPersonal
Name		DateLines
593.1Partial solution?CHOVAX::YOUNGI think we're all bozos on this BUS.Tue Oct 07 1986 00:2071
    If all 8 digits were different then the lowest total would be:
    
    		0 + 1 + 2 + 3 + 4 + 5 + 6 + 7   =    28
    
    	since this is still to high ( > 25 ) we may then conclude that
    there must be some duplicate digits, and the first answer was "yes".
    
    Unstated is the implication that the tickets must have sequential
    serial numbers.  There are 2 possible kinds of pairs of numbers:
    
    	1.	abc0  to  abc8		and	2.	abc9
    		abc1  to  abc9                          def0
    
    Type (1) pairs must have sums of 12, and 13 seperatly.  Type (2)
    pairs must not have these sums seperatly.  Type (1) pairs are so
    numerous that we could not deduce them from the available information.
    Since his wife WAS able to deduce the numbers from the available
    information, we can then conclude that we are dealing with a type
    (2) pair, and that the answer to the second question was "no".
    
    Now a pair of the form:
    
    		abc9
    		def0
    
    			accounts for 9 of the total of 25, leaving a
    sum of 16 still unaccounted for.  Once again we find that there
    are 2 kinds of pairs that we may be dealing with:
    
    	2a.	ab09  to  ab89		2b.	ab99
    		ab10  to  ab90                  cd00
                              
    	If we are dealing with type (2a) then the sum of the digits
    is equal to 2*(a+b+x)+1 + 9 where 'x' is the lessor tens digit.  Notice
    however that all possible values of a,b, and x result in an even
    sum, whereas 25 is odd.  We may therefore conclude that we are dealing
    with a type (2b) pair.
    
    Now with
    		ab99
    		cd00
    			we have accounted for 18 out of 25 leaving 7
    more.
    
    Now a pair like
    			a999
    			b000
    			    	has a total of at least 27 which is
    greater than 25.  We may therefore conclude that we are dealing with a 
    pair of the form:
    
    		a099  to  a899
    		a100  to  a900
    				call the lessor hundreds digit 'h'.  Now we
    know that the total must be:
    
    		2*(a+h)+1 + 18  =  25
    Reducing:
    		a + h  =  3
    
    So either   a = 0, h = 3
    		a = 1, h = 2
    		a = 2, h = 1
    		a = 3, h = 0
    
    
    HMMM...
    		it seems that I am stuck here however.  Any help?
    
    
    --  Barry
593.2CLT::GILBERTeager like a childTue Oct 07 1986 01:555
    "...who thereupon asked if any digit appeared more than twice..."

    If no digit appeared more than twice, the lowest total would be:
    
    		0 + 0 + 1 + 1 + 2 + 2 + 3 + 3   =    12
593.3How's this?SMURF::DIKETue Oct 07 1986 10:2541
    As already noted, the two numbers must have the form 
    	    1)	abcd and abc(d+1) or
    	    2)	ab99 and a(b+1)00
    
    There are 4 possibilities for case #2:
    		a) 0399 0400
    		b) 1299 1300
    		c) 2199 2200
    		d) 3099 3100
    
    There are numerous possibilities for case #1.
    However, we know that given the information about either ticket
    totalling 13 and one digit appearing more than twice, his wife was
    able to deduce the numbers.  In case #1, it is guaranteed that one
    of the tickets totalled 13.  If in case #1, there exist two sets
    of numbers, one of which has a digit appearing more than twice and
    the other not having any digit appear more than twice, then she
    wouldn't be able to uniquely identify the numbers, and case #2 must
    contain the numbers.
    
    So, case #1 contains the following sets of possible numbers:
    		a) 0093 0094 and 0048 0049
    		b) 0246 0247 and 0345 0346
    Now, both members of (a) contain the digit '0' 4 times, while neither
    member of (b) contain any digit more than twice.  So, if the actual numbers
    were in case #1, then the information would be insufficient to
    determine the numbers.  Hence the actual numbers must be in case
    #2.
    
    Now the possibilities are:
    		0399 0400
    		1299 1300
    		2199 2200
    		3099 3100
    
    All of these except the second contain a digit more than twice.
    Since his wife was able to solve the problem, the numbers must be
    1299 and 1300.
    
    				Jeff