TURKZEKA 2010 – STRUGGLE WITH NUMBERS (By Erol Kasapoglu)
Solution 1: Scales (8 Points)
3 of the balls are a grams, 3 are b grams, 3 are c grams and 1 of the balls is d grams.Because of we don’t know that which weight is only one, let’s find the possible double balances, by assuming that each weight are 3 times.
28+29=28+29 28+30=29+29 28+30=28+30 28+33=28+33 29+30=29+30 29+33=29+33 30+33=30+33
As seen, there are 7 possibilities.
Now, if we find out and analyze the triple balance possibilities;
28+28+30 = 28+29+29
Possible balls that remains 29 30 30 33 or 29 33 33 33 double balance is not possible
28+28+33 = 29+30+30
Possible double balance with remaining balls: 28+30 = 29+29 33 remains alone.
28+29+30 = 28+29+30
Possible balls that remains 28 29 30 33 double balance is not possible
28+29+33 = 28+29+33
Possible balls that remains 28 29 30 33 double balance is not possible
28+29+33 = 30+30+30
Possible double balance with remaining balls: 28+29 = 28+29 33 remains alone.
28+33 = 28+33 29 remains alone.
29+33 = 29+33 28 remains alone.
28+30+30 = 29+29+30
Possible balls that remains 28 28 29 33 or 29 33 33 33 double balance is not possible
28+30+33 = 28+30+33
Possible balls that remains 28 29 30 33 double balance is not possible
28+30+33 = 29+29+33
Possible balls that remains 29 30 30 33 or 28 28 29 33 double balance is not possible
29+30+33 = 29+30+33
Possible balls that remains 28 29 30 33 double balance is not possible
In case of that each of 2.8, 2.9 and 3.3 gr. balls are unique, we can see that both of the balances is performed.
In case of that 3 gr. ball is unique, we can see that both of the ba ances can not be performed.
So, the answer is “3”.
Solution 2: Harvesting the Colours (10 Points)
Solution 3: KakKuro (12 Points)
Solution 4: Colour Operation (14 Points)
Solution 5: Sudoku (16 Points)
Solution 6: Black Box (20 Points)
Solution 7: Sliding Puzzle Optimization
The highest score belongs to Meng-Hsuan Wu from Taiwan.
Solution 8: KakKuro Optimization
The highest score belongs to Rara Hartono from Indonesia.
