You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Hello everyone please help me
- Thread starter abtia
- Start date
D
Deleted member 4993
Guest
Hello everyone please help me
we have 25 coins that the weigh of 24 of them is equal and the weigh of the last one is less than others.At least how many times should we weigh them to find that coin(which is not equal to others)????
Please share your thoughts/work with us, so that we may know where to begin to help you.
Hello, abtia!
Please give us the original wording of the problem.
You've omitted some important details.
If this is the classic "Counterfeit Coin" problem, we are using a balance-scale.
Number the coins from 1 to 25.
Weighing 1: Weigh {1,2,3,4,5,6,7,8,9} vs.{10,11,12,13,14,15,16,17,18}
[1] If they are equal, the light coin is among the other 7 coins.
. . Weighing 2: weigh (19,20,21} vs. {22,23,24}
. . If they are equal, the light coin is #25.
. . If they are unequal, the light coin is on the lighter side of scale.
. . Suppose the light side is {19,20,21}.
. . . . Weighing 3: weigh {19} vs. {20}.
. . . . If they are equal, the light coin is #21.
. . . . If they are unequal, the light coin is on the lighter side.
[2] If they are unequal, the light coin is on the lighter side of the scale.
. . .Suppose the lighter side is {1,2,3,4,5,6,7,8,9}.
. . Weighing 2: weigh {1,2,3} vs. {4,5,6}.
. . If they are equal, the light coin is in {7,8,9}.
. . . . Weighing 3: weigh {7} vs. {8}.
. . . . If they are equal, the light coin is #9.
. . . . If they are unequal, the light coin is on the lighter side of the scale.
. . If they are unequal, the light coin is on the lighter side of the scale.
. . Suppose the lighter side is {1,2,3}.
. . . . Weighing 3: weigh {1} vs. {2}.
. . . . If they are equal, the light coin is #3.
. . . . If they are unequal, the light coin is on the lighter side of the scale.
The light coin can be detected in three weighings.
Please give us the original wording of the problem.
You've omitted some important details.
We have 25 coins that are identical except one of them is lighter than the others.
At least how many times should we weigh them to find that light coin?
If this is the classic "Counterfeit Coin" problem, we are using a balance-scale.
Number the coins from 1 to 25.
Weighing 1: Weigh {1,2,3,4,5,6,7,8,9} vs.{10,11,12,13,14,15,16,17,18}
[1] If they are equal, the light coin is among the other 7 coins.
. . Weighing 2: weigh (19,20,21} vs. {22,23,24}
. . If they are equal, the light coin is #25.
. . If they are unequal, the light coin is on the lighter side of scale.
. . Suppose the light side is {19,20,21}.
. . . . Weighing 3: weigh {19} vs. {20}.
. . . . If they are equal, the light coin is #21.
. . . . If they are unequal, the light coin is on the lighter side.
[2] If they are unequal, the light coin is on the lighter side of the scale.
. . .Suppose the lighter side is {1,2,3,4,5,6,7,8,9}.
. . Weighing 2: weigh {1,2,3} vs. {4,5,6}.
. . If they are equal, the light coin is in {7,8,9}.
. . . . Weighing 3: weigh {7} vs. {8}.
. . . . If they are equal, the light coin is #9.
. . . . If they are unequal, the light coin is on the lighter side of the scale.
. . If they are unequal, the light coin is on the lighter side of the scale.
. . Suppose the lighter side is {1,2,3}.
. . . . Weighing 3: weigh {1} vs. {2}.
. . . . If they are equal, the light coin is #3.
. . . . If they are unequal, the light coin is on the lighter side of the scale.
The light coin can be detected in three weighings.
http://www.freemathhelp.com/forum/members/70-sorobanThanks to soroban,your answer was true and my problem was solved
D
Deleted member 4993
Guest
Thanks to soroban,your answer was true and my problem was solved by others without producing single line of original work - what a system
.
abtia,
this is another complaint (by me) about how you were allegedly "helped," especially
as Subhotosh Khan prompted you first to share all your thoughts/work with us.
Last edited: