# Sub-13 table-tennis there is a different than the other 12 (or light or heavy) with Ping said that s

2010-02-12 02:01:09 +0000 #1
To be a detailed explanation of the process!
2010-02-12 02:03:05 +0000 #2
Method:

1. The ball numbered 1,2,3,4,5,6,7,8,9,10,11,12,13; to 1,2,3,4, on the left side; 5,6,7,8 placed on the right weighing; if there is no pros and cons, defects in 9,10,11,12,13 (this leave you to continue the discussion) If there are pros and cons, defects in the balance on the eight ball;

2. The 1,2,5,6, on the left; 3,7,9,10, weighing on the right side;

2-1 if there is no light and heavy, defective in 4,8, middle; 3. To 4, on the left; 5, weighing on the right side; if not the severity of defects is 8, if there are pros and cons, then the defect is 4,

2-2 if there are pros and cons, (Note: This is the key) To the tendency to look at balance ,2-2-1, if with the first one. Times the same defects in the 1,2,7, China; 3. To 1, on the left; 2, weighing in on the right; if there is no light and heavy, defective is 7, if there are pros and cons, look at the tendency of scale; constant, defective is 1, otherwise it is 2,

2-2 -- 2 If with the first one. Times reverse, defective in the 3,5,6, China, and the same token, can be called 5,6, the method to identify defects. This is not to solve it? This method can find the defective ball 13
2010-02-12 02:35:18 +0000 #3
The first six on each side of the same so if the weight of the rest of a failure

second time if the first time in more light Well, from the lighter side of the years out of six balls on each side of 2

4 Third, if both sides are equal then the second time the remaining two and then weigh the lighter failed

If the second does not equal then the light out of that two re - said about the failure of light
2010-02-12 02:51:24 +0000 #4
The first four on each side of the same so if the weight of the remaining five on each side if the two are equal then the rest of the weight of a failed

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