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Posted by: ★Vinåyåkåm★ chitt♥♥r At: 9, Nov 2006 8:34:36 PM IST
My solution may look a little complicated, may be there's an easy way too! Weighing 1 Weigh RB----BG, keep the remaining 'GR' as the 3rd pair. If they balance, then swap the B's. Then one side should become heavy, so you have 2 different color heavy balls, pick the remaining color from the third pair of 'GR'. If they dont balance, then it has to be one the 3 possibilities....(assuming left side shows heavier) RB----BG a. HH---LL b. HH---LH c. LH---LL Weighing 2 Swap the B from the left side with the G from the other side. e.g. If you are weighing RB----BG and if left side shows heavy, then swap the left B with right G. Once you do this, one of 3 things will happen a. HL---LH - it balances b. HH---LH - no change, status quo c. LL---LH - right side shows heavy From the above a- means left side RB are heavier, pick the remaining color from the 3rd group b- means left side RB and the right side G are all heavier, bingo! c- means the left B is heavy and the rest are all light, bingo!

Posted by: చంటోడు   At: 9, Nov 2006 7:29:50 AM IST
Well, but you wouldnt know if its heaviest(all 3 heavy) or heavier(2 heavy 1 light) no? :P

Posted by: చంటోడు   At: 9, Nov 2006 3:35:36 AM IST
From step 1, the heavier set could also have all 3 heavy no? Then in step 2, it would be difficult to sort no?

Posted by: చంటోడు   At: 9, Nov 2006 3:24:38 AM IST
hmmmm.... akkaDa kooDu ettukeLLipOyaaru.. ika nEnu Emi tinaali? elaa nidrapOvaali?? :(

Posted by: Du At: 9, Nov 2006 0:15:44 AM IST
ammo, aite cheppanu! :P em ready ga levu mari :(

Posted by: చంటోడు   At: 9, Nov 2006 0:12:15 AM IST
oohhh..aitE cheppEsi next question aDigEyamDi ...

Posted by: Du At: 9, Nov 2006 0:08:51 AM IST
Think I got it! :-)

Posted by: చంటోడు   At: 8, Nov 2006 11:55:05 PM IST
nope.. u can weigh as many as u can...

Posted by: Du At: 8, Nov 2006 11:32:13 PM IST
Any restrictions in the number of balls to weigh?

Posted by: మోహన రాగం At: 8, Nov 2006 11:28:59 PM IST
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