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#1




Puzzle 1: Same sum
Fill in the digits from 1 to 9 into the circles, so that the sum of the numbers in each straight line is the same.
This puzzle appeared in yesterday's "Appointments" section in Times of India as a recruitment ad of Geometric Software. People applying to the positions needed to solve this puzzle and send the solution along with their CVs! Enjoy! 
#2




Re: Puzzle 1: Same sum
What about this ?
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#3




Re: Puzzle 1: Same sum
Quote:
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#4




Re: Puzzle 1: Same sum
Nice try, but you used 8 twice and skipped 4.
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#5




Re: Puzzle 1: Same sum
Quote:
in particular, one can fill in the numbers via magic square (3 x 3) to get sum 15 in diagonals/edges/ midpoint connexions as follows 2 9 4 7 5 3 6 1 8 I believe (have yet to prove) that the figure provided in the problem statement can be collapsed into a square with 9 circles at each corner/midpoint/diagonal at its intersection. As usual, I am making leap of faith and hope not to fall flat on my face But the line segment connected by the two circles stump me What say guys?
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#6




Re: Puzzle 1: Same sum
the line with 2 circles is the one which was causing all the problems.......I too tried for a sum of 15....the problem is stated weakly...not clear if one can use a number again.....

#7




Re: Puzzle 1: Same sum
Quote:
I think the problem is sufficiently clearly stated, unless one is bent upon confusing himself 
#8




Re: Puzzle 1: Same sum
It is quite easy.. I found the solution for this problem in 15 minutes...
It needs a little algebra and some trial and error... Hint 1: try to get a equation involving only those values on a single line. Play with some values on this line and propagate the values to other lines satisfying the same sum rule. Hint 2: This problem has several solutions involving a duplicated value, but has just one solution with all values from 1 to 9. 
#9




Re: Puzzle 1: Same sum
Quote:
The best way to clarify is that you tell us exactly how many lines are there in the figure that you uploaded at the first post. Then the problem becomes worth attempting...else its a waste of time
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#10




Re: Puzzle 1: Same sum
For your convenience:

#11




Re: Puzzle 1: Same sum
Quote:
There are 7 lines in all namely ABC AE BFH DEF DGH AGI IH Now the sum of values in circles on a line need to be same, i.e., A + B + C = X A + E = X B + F + H = X D + E + F = X D + G + H = X A + G + I = X I + H = X Now don't tell me that it is still unclear! 
#12




Re: Puzzle 1: Same sum
Spoiler warning  if you still wish to solve it yourself don't peep down! The puzzle is solved below!
1. A + B + C = X 2. A + E = X 3. B + F + H = X 4. D + E + F = X 5. D + G + H = X 6. A + G + I = X 7. I + H = X Combing 2 and 4 8. A  D  F = 0 Combining 6 and 7 9. A + G  H = 0 Combining 3 and 5 10. B + F  D  G = 0 Combining 8 and 9 11. G  H + D + F = 0 Combining 10 and 11 12. H  B = 2F Now we have a equation (not containing X) but having values all on a single line HFB There are a limited number of triplet integers satisfying this HFB: 941 933 925 917 832 824 816 731 723 715 622 614 521 513 412 311 (Here I actually used trial and error to solve the problem, but I give a more exact procedure to solve this below) Also the sum of the triplets should at least be greater than 12 (Why? because to apply same sum rule on the whole figure, some triplet involving 9 puts a restriction on the sum that the other two numbers should be 1 and 2 only, i.e., the sum should at least be 12 coz  9+1+2 = 12 and any lower sum is thus impossible). Also the sum of the triplets should at most be 14, because sums above 14 could not be achieved using 9 integers in the given line set up. Also there should not be duplicates in any triplet. Also, If the sum of numbers is 12, there will definitely be a triplet 912, so 9, 1, 2 cannot be present in any other triplet who some is 12. Also if the sum of numbers is 13, there will definitely be a triplet 913, so 9, 1, 3 cannot be present in any other triplet who some is 13. Also if the sum of numbers is 14, there will definitely be any of the two triplets 913 or 932, 3 cannot be present in any other triplet who some is 14. So the prospective triplets for HFB line set now reduces to just: 941 So H = 9, F = 4, B = 1 Solving for other variables we have the unique solution A = 6, B = 1, C = 7, D = 2, E = 8, F = 4, G = 3, H = 9, I = 5. Last edited by 2gud; November 22nd, 2005 at 03:54 AM. 
#13




Re: Puzzle 1: Same sum
Quote:
I am finally relieved that I could solve this one. keep them coming
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