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HI Everyone...........
You have 12 coins one of them is defective by weight,all are identical...(not known heavier or lighter)....
Using weight balance 3 times only can you identify defective coin...........
IT is maths which has brought humanity from Ice Age to Present
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i'm not sure but i don't think you can do 12 coins in only 3 tries.
i'll have to think about how you could prove that though.
A logarithm is just a misspelled algorithm.
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There's a page on the site featuring that problem (with pool balls instead of coins, but exactly the same otherwise).
Here's one of the solutions on there:
This solution was provided by Charles Naumann. His method also solves it with just three weighings:
Label the balls 1-12
First Weighing:
Left: 1 2 3 4
Right: 5 6 7 8
Off: 9 10 11 12
Record the heavier side (L, R, or B)Second Weighing:
Left: 1 2 5 9
Right: 3 4 10 11
Off: 6 7 8 12
Record the heavier side (L, R or B)Third Weighing:
Left: 3 7 9 10
Right: 1 4 6 12
Off: 2 5 8 11
Record the heavier side (L, R, B)There are 27 (3^3) possible combination of scale readings. A complete sorted list of the scale reading appears below. Note that only 24 of the 27 readings should be possible given the original problem statement. The algorithm was designed so that if all three scale readings are the same, an error is flagged indicating that the scale is stuck.
BBB Error! There is not a single light or heavy ball (or scale is stuck).
BBL Ball #12 is light
BBR Ball #12 is heavy
BLB Ball #11 is light
BLL Ball #9 is heavy
BLR Ball #10 is light
BRB Ball #11 is heavy
BRL Ball #10 is heavy
BRR Ball #9 is light
LBB Ball #8 is light
LBL Ball #6 is light
LBR Ball #7 is light
LLL Error! Scale is stuck!
LLB Ball #2 is heavy
LLR Ball #1 is heavy
LRB Ball #5 is light
LRL Ball #3 is heavy
LRR Ball #4 is heavy
RBB Ball #8 is heavy
RBL Ball #7 is heavy
RBR Ball #6 is heavy
RLB Ball #5 is heavy
RLL Ball #4 is light
RLR Ball #3 is light
RRB Ball #2 is light
RRL Ball #1 is light
RRR Error! Scale is stuck!
Why did the vector cross the road?
It wanted to be normal.
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How about you put coins 1,2,3,4 on the left side 5,6,7,8 on the right. If the left is heavier, you know that the heavy coin is either 1,2,3 or 4. If it's the right then, 5,6,7 or 8. If neither then 9,10,11 and 12. No mater what, you end up with four options, a, b, c or d.
weigh ab against cd. if ab is heavier, weigh a against b. If cd is heavier, weigh c against d.
There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.
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That would work if we knew that the faulty coin was heavy, but it might be lighter than all the rest instead.
Why did the vector cross the road?
It wanted to be normal.
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