LinkBack URL
About LinkBacksYou have 15 tiny gold bars and a balance scale. One of the bars is lighter than the others, but you can't tell the difference by feel. What is least number of times you could use the scale to determine which one is the light one?
You have 15 tiny gold bars and a balance scale. One of the bars is lighter than the others, but you can't tell the difference by feel. What is least number of times you could use the scale to determine which one is the light one?
1)Take any 7, and take any 7.Originally Posted by qbkr21
2)Weigh them.
a)If equal then the remaining coin is lighter.
b)If unequal take the lighter one.
3)Spilt 3 with 3.
a)If equal then remaining coin is lighter.
b)If unequal take the lighter one.
4)Split 1 with 1
a)If equal the remaining coin is lighter.
b)If unequal take the lighter coin as the solution.
The worst case is 3 weighings.
----
Does anybody know how to do this problem with math?
Looks like a graph theory problem, but I do not know.
The reason why I move so many threads is because of obsessive-compulsive disorder. I just cannot keep them if the wrong place.Is there an award for the most moved post?
I moved this from Advanced Topics to Gerneral HS Maths Help, now
moderator or moderators unknown have moved it here
RonL
Now that's going to mess with his head! He'll create 100 forum bots, each to carry out a separate, unique algorithm on every single post for deviations from the normal posting habits.He hasn't noticed the prank we have been playing on him since last April first yet
RonL
..This is the sort of thing that he'll cover his bedroom walls with...in blood!
Here is one way.
So the odd bar is lighter than any of the rest.
Divide the 15 bars into 3. We now have 3 groups of 5 bars each, say, A, B and C.
Weigh A and B. -------1 weighing.
a) If the scale is balanced, then the lighter bar is in group C.
a.1) Separate the 5 bars in C into 2-2-1
Weigh the 2 and 2. -------another weighing.
a.11) If they are balanced, then the 1 is the lighter bar. ------hence, 2 weighings only altogether.
a.12) If not balanced, the 2 that is lifted has the lighter bar.
a.121) Weigh the those 2. -----another weighing.
The bar that is lifted is the lighter bar.
Hence, 3 weighings altogether.
--------------------------
b) If in weighing groups A and B, they are not balanced, say group B is lifted, then the lighter bar is in B.
Same procedure,
b.1) Separate the 5 bars in B into 2-2-1
Weigh the 2 and 2. -------another weighing.
b.11) If they are balanced, then the 1 is the lighter bar. ------hence, 2 weighings only altogether.
b.12) If not balanced, the 2 that is lifted has the lighter bar.
b.121) Weigh the those 2. -----another weighing.
The bar that is lifted is the lighter bar.
Hence, 3 weighings altogether.
----------------
Therefore, the "maximum" number of times to use the scale to determine the lighter bar is 3 times. The "minimum" number is 2 times only.
But which of those is the " least number of times you could use the scale to determine which one is the light one?"
I'd say three times. -------------answer.
I would say once...
1) Pick 2 at random (In this situation you pick the light one and a normal one so you are very lucky!)
2) Put them each on a balance
3) They unbalance and woohoo you find the lightest one.
Not mathematically sound but there is a probability that this could happen therefore this would count as the least weighings you could have to determine the lighter one.
Games? I think the laws of probability allow this to happen giving one as the least number of weighings! The riddle says nothing about the person doing the weighing being constantly mathematically unlucky
EDIT: And doesn't your model base itself on the person weighing being lucky too?
So you are told to find the lighter bar from among the 15 bars. At least how many times do you have to use the balance scale to find it.What? No one dare to challenge me on this?
If once, then you are lucky.
Suppose you can not do it in one time? Then you can't solve the problem.
If you can do it in two times, then you are lucky again.
Suppose two times using the balance scale does not do it?
You stop?
Then 3 times solves the problem.
Therefore, you need at least 3 times to solve the problem.
---------------
If that doesn't challenge your lucky guess, then at least it explains why you need at least the "highest" possible times to use the scale to solve the problem.
  |
