swap 3 balls from the4 with three ok balls,,

and switch the fourth ball with a ball from the high pan,,

the 3 ay deductive logic explains itself,gareth stephen colton

]]>60 = 10 + 50 = burning both ends: 5 + 25

After 5 minutes, you cut the remaining rope in two: (25 – 5) * 2 = 40 minutes of single-burn-time rope

cut that in twain: 40 = 12 + 28 = burn both ends: 6 + 14

after 6 minutes of burning 14-6 = 8

cut the 8 x 2 = 16

16 = 8 + 8 = burn both ends: 4 + 4

done burning

Elapsed time 5+6+4 = 15 ]]>

• Cut the 60 minute rope into 2 pieces: unknown to us they turned out to burn up in 10 minutes and 50 minutes

• Burn both of those at both ends: 10/2 and 50/2 = 5 minutes and 25 minutes

• At the end of 5 minutes, the first rope is burned up, and there is 20 minutes left of the 25 min rope (Elaped 5 minutes)

• cut the remaining 20 minute rope in 2: unknown to us, these 2 new ropes would would burn up in 8 minutes and 12 minutes

• burn both ends of both ropes: (8/2) 4 minutes and (12/2) 6 minutes

• at end of 4 minutes the first rope is burned up and there are 2 minutes left of the 6 minute rope (Elapsed 5+4=9 minutes)

• cut that in 2 min rope in half: 1 min and 1 min

• burn both ends of both ropes: .5 minutes each

• both are completly burned at the end of .5 minutes so no more iterations are possible/needed

• Elapsed time 9.5 minutes!

If anyone can point out where I messed up, let me know. I’ve done a few different sets of burn-lengths and I get a different answer every time. ]]>

So your solution does not work.

]]>Divide the balls in 3 sets of 3 balls, say A, B & C. And select a weight machine randomly, say M having two plates P1 and P2.

Step 1: Weigh the sets A & B in M. (A in P1 and B in P2)

Step 2: Weigh the sets A & B in same M, but exchanging the plates.( A in P2 and B in P1).

case 1: If the M is not defective and A=B. Plates are balanced both the times. Hence C has the lighter ball and can be found in next 2 steps.

case 2: If the M is not defective and A has lighter ball. Balance favors A both the times and Lighter ball can be identified in next one step.

Case 3: If the M is defective and A = B. Plates tilt towards same side both the times. hence C has the lighter ball and can be found in next 2 steps.

Case 4: If M is defective and A has lighter ball. Balance tilts toward A( or equally balanced) and second time it doesn’t. Using the other balance lighter ball can be identified in next 2 steps.

Let me know what you think.

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