VICTORY!!!!!!!!!!!!!!!!!!!!!!!!
Sweet victory... months of agony and mental torture finally comes to an end... if you people are still wondering well Im just talkin about the puzzle involving 12 coins having 1 defective coin ( for details luk at d q... its below this one only)... well ive cracked it... finally!!!!!
alrite here goes the soln
let the 12 coins be divided into 3 groups of 4 coins each such that
group A - A1, A2, A3, A4 COINS
group B - B1, B2, B3, B4 COINS AND SO ON FOR GROUP C
WEIGHING (1)
MEASURE A AGAINST B
1) A = B
WELL THIS ONE IS SIMPLE...
THE WRONG ONE LIES IN GROUP C...
TAKE C1 , C2..
WEIGH AGAINST ANY 2 COINS OF GROUP A AND B ( SINCE THEY ARE ALL NORMAL) { WEIGHING 2}
3 RD WEIGHIN :
IF UNEQUAL THAT MEANS EITHER C1 OR C2 IS DEFECTIVE..... AGAIN WEIGH OF THEM AGAINST 1 NORMAL COIN AND YOULL GET THE DEFECTIVE COIN.....
IF EQUAL THEN DEFCTIVE COIN LIES IN OTHER 2 C COINS IE C3 OR C4... FOLLOW THE SAME PROCEDURE ... WIEGHING AGANIST 1 RIGHT COIN ETC ETC
2) A > B ...( CAN BE THE OTHER WAY ALSO... BUT JUST FOR INSTANCE)....
WELL THIS MEANS C1- C4 ARE ALL NORMAL COINS
WEIGHING (2)
PAN 1 CONSISTS OF - A1 , C1 , C2 , C3
PAN 2 CONSISTS OF - B1 , B2 , A2 , A3
SO WE 'RE LEFT WITH ( THE OTHER POSSIBLE DEFCTIVE COINS) - A4 , B3 , B4
AFTER THIS THERE ARE 3 CASES
A) PAN 1 = PAN 2
THAT MEANS DEFCTIVE IS AMONG A4 , B3 , B4 ... OUT OF WHICH EITHER A4 IS HEAVY OR ONE OF B3 AND B4 IS LIGHT SINCE WE ALREADY KNOW ( BY 1 WEIGHING ) THAT A > B
SO 3RD WEIGHING WILL BE AS FOLLOWS...
PAN 1 - A4 , B3
PAN 2 - C1, C2 ( IE ANY OF THE NORMAL ONES)
AGAIN 3 POSSIBLITIES,,,,
1) PAN 1 = PAN 2..
SO A4 AND B3 ARE NORMAL... IE B4 IS DEFCTIVE
2) PAN 1 > PAN 2
SINCE WE KNOW THAT OUT OF A4 AND B3 ... A4 CAN POSSIBLY BE THE HEAVY ONE ( AS BELONGS TO A... WHICH WAS HEAVIER)... HENCE A4 IS THE DEFECTIVE COIN
3) PAN 1 < PAN 2...
JUST LIKE THE ABOVE CASE,,,, THIS TIME THE DEFECTIVE COIN WOULD BE B3 ( BELONGS TO B GROUP.. WHICH WAS LIGHTER ETC).......
NOW GOING BACK TO THE INTIAL PAN 1 AND PAN 2
IF PAN 1 NOT EQUAL TO PAN 2
IE DEFECTIVE COIN IS AMONG A1 , B1 , B2 , A2 , A3
B) PAN 1 > PAN 2
SINCE THIS IS THE PREV RECORDING ONLY ( FOR A > B)... THIS MEANS MOVING A2 AND A3 TO THE OTHER SIDE HAD NO EFFECT ... THE DEFECTIVE COIN REMAINS IN ITS ORIGINAL POSITION... SO IT HAS TO BE AMONG A1 , B1 , B2 OUT OF WHICH WE KNOW ( THROUGH WEIGHIN 2 ) THAT EITHER A1 IS HEAVY OR ONE OF B1 AND B2 IS LIGHT AS PAN 1 > PAN 2..
NOW THE DEFTIVE COIN CAN BE OBTAINED BY THE SAME PROCEDURE AS MENTIONED BEFORE ( FOR PAN 1 = PAN 2)
C) PAN 1 < PAN 2
THIS MEANS MOVING A2 AND A3 BROUGHT ABOUT A CHANGE IN THE POSITION OF THE DEFCTIVE COIN SO THE DEFCTIVE IS EITHER A2 OR A3.... WEIGH ANY OF THEM AGAINST A NORMAL ONE ... IF EQUAL THEN THE OTHER ONE IS THE DEFECTIVE COIN ( THE DEFECTIVE COIN WOULD BE HEAVIER PLS NOTE AS PAN 2 > PAN 1) IF NOT EQUAL THEN OBVIOUSLY THAT VERY COIN IS DEFECTIVE
SO I HOPE THE SOLUTION IS FULL PROOF ... ANY QUERIES OR LOOPHOLES PLS DO COMMENT
alrite here goes the soln
let the 12 coins be divided into 3 groups of 4 coins each such that
group A - A1, A2, A3, A4 COINS
group B - B1, B2, B3, B4 COINS AND SO ON FOR GROUP C
WEIGHING (1)
MEASURE A AGAINST B
1) A = B
WELL THIS ONE IS SIMPLE...
THE WRONG ONE LIES IN GROUP C...
TAKE C1 , C2..
WEIGH AGAINST ANY 2 COINS OF GROUP A AND B ( SINCE THEY ARE ALL NORMAL) { WEIGHING 2}
3 RD WEIGHIN :
IF UNEQUAL THAT MEANS EITHER C1 OR C2 IS DEFECTIVE..... AGAIN WEIGH OF THEM AGAINST 1 NORMAL COIN AND YOULL GET THE DEFECTIVE COIN.....
IF EQUAL THEN DEFCTIVE COIN LIES IN OTHER 2 C COINS IE C3 OR C4... FOLLOW THE SAME PROCEDURE ... WIEGHING AGANIST 1 RIGHT COIN ETC ETC
2) A > B ...( CAN BE THE OTHER WAY ALSO... BUT JUST FOR INSTANCE)....
WELL THIS MEANS C1- C4 ARE ALL NORMAL COINS
WEIGHING (2)
PAN 1 CONSISTS OF - A1 , C1 , C2 , C3
PAN 2 CONSISTS OF - B1 , B2 , A2 , A3
SO WE 'RE LEFT WITH ( THE OTHER POSSIBLE DEFCTIVE COINS) - A4 , B3 , B4
AFTER THIS THERE ARE 3 CASES
A) PAN 1 = PAN 2
THAT MEANS DEFCTIVE IS AMONG A4 , B3 , B4 ... OUT OF WHICH EITHER A4 IS HEAVY OR ONE OF B3 AND B4 IS LIGHT SINCE WE ALREADY KNOW ( BY 1 WEIGHING ) THAT A > B
SO 3RD WEIGHING WILL BE AS FOLLOWS...
PAN 1 - A4 , B3
PAN 2 - C1, C2 ( IE ANY OF THE NORMAL ONES)
AGAIN 3 POSSIBLITIES,,,,
1) PAN 1 = PAN 2..
SO A4 AND B3 ARE NORMAL... IE B4 IS DEFCTIVE
2) PAN 1 > PAN 2
SINCE WE KNOW THAT OUT OF A4 AND B3 ... A4 CAN POSSIBLY BE THE HEAVY ONE ( AS BELONGS TO A... WHICH WAS HEAVIER)... HENCE A4 IS THE DEFECTIVE COIN
3) PAN 1 < PAN 2...
JUST LIKE THE ABOVE CASE,,,, THIS TIME THE DEFECTIVE COIN WOULD BE B3 ( BELONGS TO B GROUP.. WHICH WAS LIGHTER ETC).......
NOW GOING BACK TO THE INTIAL PAN 1 AND PAN 2
IF PAN 1 NOT EQUAL TO PAN 2
IE DEFECTIVE COIN IS AMONG A1 , B1 , B2 , A2 , A3
B) PAN 1 > PAN 2
SINCE THIS IS THE PREV RECORDING ONLY ( FOR A > B)... THIS MEANS MOVING A2 AND A3 TO THE OTHER SIDE HAD NO EFFECT ... THE DEFECTIVE COIN REMAINS IN ITS ORIGINAL POSITION... SO IT HAS TO BE AMONG A1 , B1 , B2 OUT OF WHICH WE KNOW ( THROUGH WEIGHIN 2 ) THAT EITHER A1 IS HEAVY OR ONE OF B1 AND B2 IS LIGHT AS PAN 1 > PAN 2..
NOW THE DEFTIVE COIN CAN BE OBTAINED BY THE SAME PROCEDURE AS MENTIONED BEFORE ( FOR PAN 1 = PAN 2)
C) PAN 1 < PAN 2
THIS MEANS MOVING A2 AND A3 BROUGHT ABOUT A CHANGE IN THE POSITION OF THE DEFCTIVE COIN SO THE DEFCTIVE IS EITHER A2 OR A3.... WEIGH ANY OF THEM AGAINST A NORMAL ONE ... IF EQUAL THEN THE OTHER ONE IS THE DEFECTIVE COIN ( THE DEFECTIVE COIN WOULD BE HEAVIER PLS NOTE AS PAN 2 > PAN 1) IF NOT EQUAL THEN OBVIOUSLY THAT VERY COIN IS DEFECTIVE
SO I HOPE THE SOLUTION IS FULL PROOF ... ANY QUERIES OR LOOPHOLES PLS DO COMMENT
posted by Abhishek Jain at 2:27 PM
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