Weigh One coin first and find the weight. Then Seperate the 200 coins into 4 groups
200/4 = 50 coins in 1 group
Now weigh 1 group at a time. Each time you weigh check if each group matches the weight of one coin *50. The group that doesn't match it has the lighter coin.
Edit:
I was answering the question to the best of my knowledge. Your answer, I must admit is confusing to read, but im not even sure what your saying. I answered it because first he put it in the math section. Second it is a grade 8 problem. So im guessing they needed an answer something to do with division and thats what i just posted. I am going to read your post and try to figure out what you have just said and see how insane my answer is.
Second Edit:
"To find a solution to the problem we first consider the maximum number of items from which one can find the lighter one in just one weighing. The maximum number possible is three." How did you get three? You have written the maximum number of items is 3. Are you saying 3 coins? 3 groups of coins. Please explain. There are 200 coins and dividing it into 3 groups would give 66 coins in one group each approximately.
" To find the lighter one we can compare any two coins, leaving the third out. If the two coins tested weigh the same, then the lighter coin must be one of those not on the balance - otherwise it is the one indicated as lighter by the balance. "
This is the part that confused me the most. You weigh any two coins and the last one is not in the balance? Please explain this. It is very confusing to read.
"Now, assume we have three coins wrapped in a bigger coin-shaped box. In one move, we can find which of the three boxes is lighter (this box would contain the lighter coin) and, in the second weighing, as was shown above, we can find which of the three coins within the box is lighter. So in two weighings we can find a single light coin from a set of ."
Set of what?
And bigger coin shaped box? What is a coin shaped box. I am being honest here. i don't understand some of the things you have just said. You want to find the lighter of three box and that would contain the lighter coin. Please explain your logic. I am trying to make sense out of your answer but it is hard to. Is this the exact same answer you gave for your solution? And you got it correct. This is very confusing. If you can explain some of the parts it would be helpful to understand what your trying to say. You have written quite a long answer and i didn't really understand your solution. So please tell me what exactly you mean by all this. And i still don't see how mine is insane.
I had a problem very similar to this, I believe some of the math is different (we had three tries), But, here was my answer and it was correct.
To find a solution to the problem we first consider the maximum number of items from which one can find the lighter one in just one weighing. The maximum number possible is three. To find the lighter one we can compare any two coins, leaving the third out. If the two coins tested weigh the same, then the lighter coin must be one of those not on the balance - otherwise it is the one indicated as lighter by the balance.
Now, assume we have three coins wrapped in a bigger coin-shaped box. In one move, we can find which of the three boxes is lighter (this box would contain the lighter coin) and, in the second weighing, as was shown above, we can find which of the three coins within the box is lighter. So in two weighings we can find a single light coin from a set of .
Note that we could reason along the same line, further, to see that in three weighings one can find the odd-lighter coin among 27 coins and in 4 weighings, from 81 coins.
Comments
Weigh One coin first and find the weight. Then Seperate the 200 coins into 4 groups
200/4 = 50 coins in 1 group
Now weigh 1 group at a time. Each time you weigh check if each group matches the weight of one coin *50. The group that doesn't match it has the lighter coin.
Edit:
I was answering the question to the best of my knowledge. Your answer, I must admit is confusing to read, but im not even sure what your saying. I answered it because first he put it in the math section. Second it is a grade 8 problem. So im guessing they needed an answer something to do with division and thats what i just posted. I am going to read your post and try to figure out what you have just said and see how insane my answer is.
Second Edit:
"To find a solution to the problem we first consider the maximum number of items from which one can find the lighter one in just one weighing. The maximum number possible is three." How did you get three? You have written the maximum number of items is 3. Are you saying 3 coins? 3 groups of coins. Please explain. There are 200 coins and dividing it into 3 groups would give 66 coins in one group each approximately.
" To find the lighter one we can compare any two coins, leaving the third out. If the two coins tested weigh the same, then the lighter coin must be one of those not on the balance - otherwise it is the one indicated as lighter by the balance. "
This is the part that confused me the most. You weigh any two coins and the last one is not in the balance? Please explain this. It is very confusing to read.
"Now, assume we have three coins wrapped in a bigger coin-shaped box. In one move, we can find which of the three boxes is lighter (this box would contain the lighter coin) and, in the second weighing, as was shown above, we can find which of the three coins within the box is lighter. So in two weighings we can find a single light coin from a set of ."
Set of what?
And bigger coin shaped box? What is a coin shaped box. I am being honest here. i don't understand some of the things you have just said. You want to find the lighter of three box and that would contain the lighter coin. Please explain your logic. I am trying to make sense out of your answer but it is hard to. Is this the exact same answer you gave for your solution? And you got it correct. This is very confusing. If you can explain some of the parts it would be helpful to understand what your trying to say. You have written quite a long answer and i didn't really understand your solution. So please tell me what exactly you mean by all this. And i still don't see how mine is insane.
Thankyou!
I had a problem very similar to this, I believe some of the math is different (we had three tries), But, here was my answer and it was correct.
To find a solution to the problem we first consider the maximum number of items from which one can find the lighter one in just one weighing. The maximum number possible is three. To find the lighter one we can compare any two coins, leaving the third out. If the two coins tested weigh the same, then the lighter coin must be one of those not on the balance - otherwise it is the one indicated as lighter by the balance.
Now, assume we have three coins wrapped in a bigger coin-shaped box. In one move, we can find which of the three boxes is lighter (this box would contain the lighter coin) and, in the second weighing, as was shown above, we can find which of the three coins within the box is lighter. So in two weighings we can find a single light coin from a set of .
Note that we could reason along the same line, further, to see that in three weighings one can find the odd-lighter coin among 27 coins and in 4 weighings, from 81 coins.
try to find it 5 times or less