Imagine that you have in front of you fifty coins. They all look exactly alike except one of them is a fake. Because it’s a fake, it weighs a couple of grams more than a real coin. So, if you had a balance scale, and you knew which coin was the bogus coin, you would put it on one side of the scale, a good coin on the other side and it would be immediately obvious from this imbalance which was the phony coin, because it’s heavier than a real coin.
The question is, what is the fewest number of weighings that you need to perform, to determine which coin is bogus?
Bonus Question: How can you do it in four?
June 1st, 2013 | Category: 
Warning: Use of undefined constant bfa_comments - assumed 'bfa_comments' (this will throw an Error in a future version of PHP) in /home/customer/www/riddledude.com/public_html/wp-content/themes/atahualpa/comments.php on line 132
Step 1: Split them into 25, 25 coins and place on the balance and choose the set of 25 that is heavy.
Step 2: From the above set which is heavier, split them into 12, 12 if they weigh equally then the balance is the heavier one. If either one weighs heavier then choose that set of 12 coins.
Step 3: From the above split them into 6 and 6 and choose the heavier one.
Step 4: from set of 6, split into 3, 3 and weigh.choose the heavier 3.
Step 5: from the heavier 3 select 2 & weigh. The heavier one if present should showup. If they balance exactly then the third one is heavier.
Hence 5 weighings.
In 4 Steps:
Step 1: Weigh 24 + 24, If one is greater choose the heavier 24, split into 9+9+6 and goto step 2. If found equal weigh the balance in 1:1 and decide the heavier coin.
Step 2: Weigh 9 + 9, If one is greater choose the heavier 9, split into 3+3+3 select any 2 sets of 3 coins and goto step 3. If found equal goto step 3 with the balance 6 coins.
Step 3: Weigh 3 + 3, if found equal, divide the balance 3 into 1+1+1 and goto step 4. Else chose the heavier 3, dive it into 1+1+1 and goto step 4.
Step 4: segregate one coin. Weigh the balance two to find the heavier one. If found equal then the unweighed coin is the heavier one.
Correct Chandramouli.
At first thought, you would think that you would divide the 50 coins in half.
So, you’d do 25 and 25. That’s weighing number one.
You find out that the bogus coin is on the left side. Then you weigh the coins on the left, 12 and 12 with one left over.
Now, assume the worst case scenario. One of them is heavier. So now that’s two weighings. Now you divide it six and six, making it three weighings. Then three and three, for four weighings.
And you’re done for. It takes five.
No matter how you do it using that system, it won’t work. So you had to come up with something more clever. And what you do is divide the coins into three piles–two piles of 17 and one of 16.
You take the two piles of 17 and you put those on the scale, and you keep the 16 pile aside,. Right away, you can see that you’re going to eliminate not half the coins, but two thirds of the coins.
So, let’s assume that one of the 17 is the heavier one. You throw everything else away.
You’ve only made one weighing and you’ve narrowed it down to 17 coins. Now, you could divide the 17 in half, but better still, divide it thirds so you’ve got six and six and five.
OK? And you can see very clearly that you’re going to be able to do this because the next group would be three and three, and then one and one. And bingo! And the key is, once you figure out the idea that you’re going to divide it into three piles and not two, it jumps right out at you. Four weighings.
You are today’s winner.
Dude please comment on my answer #2.
I did. You are the winner.