## October 8, 2015

### 12 men on an island riddle

There is an island with 12 islanders. All of the islanders individually weigh exactly the same amount, except for one, who either weighs more or less than the other 11.
You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. There are no scales or other weighing device on the island.

How can you find out which islander is the one that has a different weight?

N.S.Sivasankaran said...

FIRST DO IT WITH 12..SIX ELEMINATED

SECOND WITH SIX....THREE ELEMINATED

THIRD TIME DO IT WITH TWO PERSON..YOU CAN FIND.IF IT IS EQUAL...

THE OTHER MAN IS LESS WEIGHT

David and Anne lymn said...

"FIRST DO IT WITH 12..SIX ELEMINATED" how do you know which six to eliminate if you don't know if the odd person is heavier or lighter

Jerry Critter said...

What if the seesaw is NOT even with two people? Which of the two people on the seesaw is it?

Unknown said...

If the odd person is less weight or over weight is not confirmed the answer is not correct

If the odd person is less weight or over weight is not confirmed the answer is not correct

revolver said...

Seems impossible with just 3 weighings.

Worst case scenario:

1. weigh 3 vs. 3, narrows it to 6 balanced, 6 unbalanced.

2. weigh 3 unbalanced vs. 3 balanced narrows it to 3 unbalanced.

3. weigh 2 unbalanced vs. 2 balanced narrows it to 2 unbalanced.

4. weigh 1 unbalanced vs. 1 balanced gives answer.

Cijo Paul said...

Divide the 12 into 3 groups of 4 people each.
Let's call them A, B and C.

Step 1: Now lets compare A and B on the seesaw.
if the seesaw is balanced we know the person with the unbalanced weight is in C.
Step 2: lets take C and divide it into two groups having two people each. lets call this group C1 and C2.
Step 3: lets compare C1 with 2 members from A(which has members of equal weight) on the seesaw
if the seesaw is not balanced we know the person with the unbalanced weight is in C1 since A is balanced. else the person with the unbalanced weight is in C2. perform Step 4 with the group with the unbalanced weight.

Step 4: lets compare 1 person from the unbalanced C2 with 1 person in A.
If the seesaw is not balanced we have found our person with unbalanced weight(total # of comparisons = 3)
Else: the other person from C2 is the person with unbalanced weight. (C2 written for convenience perform with the appropriate group)

From Step 1...
If the Seesaw wasn't balanced it means the balanced group is C and either A or B are unbalanced.

We'll just compare A with C. If the seesaw isn't balanced. A is our unbalanced group. Otherwise B.

Now with the unbalanced group...
repeat steps 3 and 4...

This portion will take 4 steps on the seesaw...
can anybody workaround this to do it in 3?

Jerry Critter said...

The question does NOT ask you to determine if the person is lighter or heavier, only which person it is.

Chris Boensel said...

The only problem I have with this is that we still do not know if the person weighs more or less. If you take the scenario of weighing 6 on one side and 6 on another that still does not tell us which group it is. Of course if we know he weighs more then it will be the side that goes down. But, if he weighs less the side that goes up will be the one with the person of different weight. Since we have no idea that solution does not really help us. If you use the A,B, C solution if is hit or miss if you get it right. Certainly you would hope the first time you measure the two groups of four you get an even balance and know it is the other group but you then still run into a problem. I could be totally looking at this wrong. Please let me know if that makes sense. Thanks!

Anonymous said...

Yes