You own a goldmine with 50 employees...

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exiledforcefreak

RIP Jacob KT 2/16/06
...every day they are given a 16oz bag and told to fill it with gold. Someone tips you off that an employee has been stealing gold from you for a year by filling their bags with only 15 oz of gold. The bags your employees hand in are numbered so that you know which bags belongs to which employee but you must first weigh them in order to figure out which bags are missing 1oz from them. You have a digital scale with no weight limit.

What is the least number of measurements you can make garenteeing you will find the culprit? (you may not say one measurement with the reasoning you got lucky and weighed the right bag first)

Edit: explain your reasoning.
 
You split the 50 into two groups of 25. Weighing one of them tells you which group contains the ligh sack. Split that into a group of 13 and a group of 12. Weighing one of them tells you which it is in. Assume that it is in the 13 group. Split it into a group of 6 and a group of 5. Weigh again and assume it is in the 6. Split that into two groups of 3. Weigh again. Now split the group of 3 into 1 and 2. Weigh again. Assume it is in the 2. Weigh again to determine the short bag.
 
Are we guessing best case or worst. If we break it up into 2 piles, 1 of 12 and 1 of 13 do we have to weigh it twice or can we weigh them once to figure out which pile is light?
 
deepdark said:
Are we guessing best case or worst. If we break it up into 2 piles, 1 of 12 and 1 of 13 do we have to weigh it twice or can we weigh them once to figure out which pile is light?
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We're assuming WORST case.

If we break into a pile of 12 and a pile of 13, we only need to weigh once to figure out which is the light pile.
 
Jason_C said:
You split the 50 into two groups of 25. Weighing one of them tells you which group contains the ligh sack. Split that into a group of 13 and a group of 12. Weighing one of them tells you which it is in. Assume that it is in the 13 group. Split it into a group of 6 and a group of 5. Weigh again and assume it is in the 6. Split that into two groups of 3. Weigh again. Now split the group of 3 into 1 and 2. Weigh again. Assume it is in the 2. Weigh again to determine the short bag.
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you mean 7 and 6????? Last i checked 5+6=11
 
okay we have
one pile of 12 and one of 13, 1 weighs
one pile of 6 and one of 7, 2 weighs
one pile of 3 and one of 4, 3 weighs
two pile of 2, 4 weighs
1 more weigh and the answer is 5
 
You could weigh one time. Place all the bags on the scale, then remove them 1 at a time until the difference between the weight after removing and the difference before removing is 15oz.
 
exiledforcefreak said:
you mean 7 and 6????? Last i checked 5+6=11
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Right. Darnit.

From 50, two groups of 25.
From 25, a group of 12 and a group of 13.
From 13, a group of 6 and a group of 7.
From 7, a group of 3 and a group of 4.
From 4, two of two.
From that, two of one.
 
if you're lucky 3. But more likely 4.

Step 1:
weight 17 and 17 on the scale, if they are equal, then its in the remaining 16.

Step 2
Weigh 6/5 on each scale then if they are equal its in the remaining 5/6.

Step 3: 2 on each side, if they are equal it is the remaining 2 or one.

Step 4: weight the remaining 2 if necessary.
 
So....what happens if the empty bags don't weigh the same to begin with? How do you determine the answer then?
 
Step 1: Divide the 50 bags into three piles. Two of those piles contain 16 bags and the other containing 18. You weight each of them. The two piles that contains 16 bags should weigh 256 oz each. The other pile should weigh 288 oz. The pile that doesn't match either of those weight numbers contains the bag of the culprit. Lets say it's a pile that has 16 bags in it.

Note: Total of 3 weighs

Step 2: Now divide the pile of 16 bags into two equal halves. Meaning two piles of 8 bags. Each pile of 8 bags should way 128 oz. The pile that weighs less has the bag of the culprit.

Note: Total of 5 weighs

Step 3: Divide that pile of 8 bags into two equal piles. Meaning 4 bags in each pile. Each of these 4 bag piles should weigh 64 oz. The pile that ways the less has the bag of the culprit.

Note: Total of 7 weighs

Step 4: Divide that pile of 4 bags into two equal piles. Meaning 2 bags in each pile. Each of these 2 bag piles should weigh 32 oz. The pile that ways the less has the bag of the culprit.

Note: Total of 9 weighs

Step 5: All you need to do is weigh one of the remaining bags. If the bag you weigh matches 16 oz. then the other bag is the culprit's bag by default and there is no reason to weigh it. If the bag you weigh happens to be less than 16 oz. then the other bag isn't the culprit's bag by default.

Note: Total of 10 weighs.

My answer is 10.
 
Place one of the 'other' bags underneath the scale, then place 17 on one side and 17 on the other side ;). DOn't bother reading what the digital scale says.
 
Jason_C said:
Tkwiget, I already got it down to six.
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And he said your answer is wrong. I really don't care if you got it down to six or not. An incorrect answer means nothing to me.
 
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