20 Oct 05
Originally posted by ViAThree times. Take 8 coins, balance them. If they are equal, the
Two times.
odd man out is it. If not, the heavier side eliminates 4. Second
time, take the lighter side of the 8, you have eliminated 5 coins
so far, put the 4 left two by two in the balance. Heavy side,
eliminates two, you are up to 7 eliminated.
Third time, one one each balance, bingo, lighter side gives it away.
20 Oct 05
Originally posted by sonhousenope....ViA is rite. It is 2 times
Three times. Take 8 coins, balance them. If they are equal, the
odd man out is it. If not, the heavier side eliminates 4. Second
time, take the lighter side of the 8, you have eliminated 5 coins
so far, put the 4 left two by two in the balance. Heavy side,
eliminates two, you are up to 7 eliminated.
Third time, one one each balance, bingo, lighter side gives it away.
20 Oct 05
Originally posted by gaurav2711Well if you guess right, its one time! Don't see how you do it in two,
nope....ViA is rite. It is 2 times
you have to have equal number of coins on each side to start.
How can you eliminate more than 5 in the first weighing?
That leaves you with 4 to figure out. You have to have two on each
side so you would only know which two of the four are heavier, leaving
you with two coins which you don't know which is heavier till you
weigh them. So whats the trick I am missing?
20 Oct 05
Originally posted by sonhousehere it is.....
Well if you guess right, its one time! Don't see how you do it in two,
you have to have equal number of coins on each side to start.
How can you eliminate more than 5 in the first weighing?
That leaves you with 4 to figure out. You have to have two on each
side so you would only know which two of the four are heavier, leaving
you with two coins which you don't know which is heavier till you
weigh them. So whats the trick I am missing?
u divide the 9 coins into 3 groups of 3 each. Put 2 groups on either side of the balance. If either is lighter, the fake coin is in that one. If not, the fake coin is in the 3rd bunch. Now u have eliminated 6 coins. out of the remaining 3 coins, put 1 on either side. If either is lighter it is the fake coin. Else, the last coin is fake.
cheers!!
20 Oct 05
Originally posted by gaurav2711Interesting. I was on to that last tact after I thought about separating
here it is.....
u divide the 9 coins into 3 groups of 3 each. Put 2 groups on either side of the balance. If either is lighter, the fake coin is in that one. If not, the fake coin is in the 3rd bunch. Now u have eliminated 6 coins. out of the remaining 3 coins, put 1 on either side. If either is lighter it is the fake coin. Else, the last coin is fake.
cheers!!
into 4 and 4 wouldn't do it. I was working out three sets when you
answered it! Good one.
21 Oct 05
Originally posted by sonhouseHe's right. It's all in the way the question is worded. Yes, I can see how you could do it in two. On the other hand, suppose I broke the coins up into three groups...4, 4 & 1, and on the first try, I put 4 on one side and four on the other. If the scale balanced perfectly, I'd know that the other coin was the fake. You asked what was the MINIMUM number of times I could use the scale to make this determination. Yes, there may be some luck involved, but the answer is 1.
Well if you guess right, its one time! Don't see how you do it in two,
you have to have equal number of coins on each side to start.
How can you eliminate more than 5 in the first weighing?
That leaves you with 4 to figure out. You have to have two on each
side so you would only know which two of the four are heavier, leaving
you with two coins which you don't know which is heavier till you
weigh them. So whats the trick I am missing?
Here's a question...what's the minimum number of times I need to buy a lottery ticket to win? ANSWER: 1.
21 Oct 05
1 edit
Originally posted by ddebenedwell i suggest we go over the last line of the question again. It says: "What is the minimum number of times i need to use the balance to be able to unmistakably detect the fake coin?"
I think perhaps the introduction of the word "unmistakably" into the question is technically intended to eliminate the options based on luck. But you are right, it is ambiguous.
Plz look carefully at the words...."NEED" & "UNMISTAKABLY DETECT". Its like, before i start the operation i have to tell how many times will i need to use the balance to be sure of getting the fake coin.
Its not "in minimum how many attempts MIGHT you detect the fake coin"
If i had to put it that way i wud never put it on this forum.
21 Oct 05
Originally posted by gaurav2711It's still ambiguous. You could interpret it either way. Considering the fact that it's a logic question, the set up can't be open to interpretation. Otherwise, it becomes an exercise in creativity or something like that. If you want to be REALLY technical, I don't "need" to use the scale at all. There are many other methods I could use to determine which coin is the fake. I like the problem. I just think it should have been worded differently.
well i suggest we go over the last line of the question again. It says: "What is the minimum number of times i need to use the balance to be able to unmistakably detect the fake coin?"
Plz look carefully at the words...."NEED" & "UNMISTAKABLY DETECT". Its like, before i start the operation i have to tell how many times will i need to use the balanc ...[text shortened]... MIGHT you detect the fake coin"
If i had to put it that way i wud never put it on this forum.
21 Oct 05
Originally posted by ddebenedAnd with twelve coins 😉
This problem is often stated without the knowledge of whether the odd coin is heavier or lighter, in which case the answer is three. As stated we can do it in only two because we know that the odd coin is lighter.