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## Math for fun and glory

### Course: Math for fun and glory>Unit 2

Lesson 1: Brain teasers

# Heavier ball

## Want to join the conversation?

• I have a good brain teaser.
One day, A king found a prisoner. In his kingdom, it was tradition for one prisoner to get out of the prison every year. The king didn't like it, so he made a rule. He said that he had two cups. He said that under one of them had a silver coin and the under the other one there was a gold coin. He said that the prisoner had to guess under which one had a gold. But the king put a silver coin under each one. The prisoner knows that the king tricked him, but he can't accuse the king of cheating. What does the prisoner do? • The prisoner reaches under the cup, takes out the coin and swallows it (or throws it in a lake, or something like that so it'll never be found). The other cup is lifted and the silver coin is shown. The people, who don't know that the king cheated, would know that the gold coin would have to be under the other.
And by the way, yes, good brain teaser.
• i'm not really sure yet but at I think it's 8 • how is it heavier if the balls are identical? • For 12 balls, we need 3 times.
Split the balls into 4,4 and 4.
First time, weigh the two 4s. Let's call them group A and B.
If they weigh same, the different ball is in other 4 balls(Let's call it group C),
meanwhile, it means the ball from A or B group is common.
Take 3 balls from A, and take 3 balls from C. We weigh them. If their weight are same, the different ball is the remaining one of C. Third time, compare it and a common ball, we will know it is heavier or lighter.
On the second weighing, if the weight of the three balls from group C is defferent from another three balls(heavier or lighter.), so one of this three ball is defferent, and also we know it is heavier or lighter.. Third time, take two balls from this three balls, compare their weight, so we will know which one is defferent.
On the first weighing, if weight of group A is defferent group B(suppose group A is heavier), then 4 balls of group C are common.
We number the 4 balls of group A----"1","2","3" and "4", number the 4 balls of group B----"5","6","7" and "8".
We gather "5" "2" "3" "4" become group M, gather "1" and three ball from C become group N.
Second time, we compare group M and N. If their weights are same, then the defferent ball is "6" or "7" or "8", and we already have known it is lighter. So third time we can find the defferent ball.
If group M("5" "2" "3" "4") is heavier than group N("1" and three common balls), then the defferent ball is "2" or "3" or "4", and we already have known it is heavier. So third time we can find the defferent ball.
If group M("5" "2" "3" "4") is lighter than group N("1" and three common balls), then "1" must be heavier or "5" must be lighter. So third time we can find the defferent ball.
(end)
• But only odd number balls can use this method,how about even numbers? • This question is actually flawed. What you wrote is that this method works for an odd number. The reality is that it only works for a number that is a power of three. When you make thirds, you cannot use any odd number. If, for instance, you used 13 balls, when you divided it in your first step you would get three groups of four and one ball left over. Lets say that you were lucky and one of the groups you put on the scale did work out. Then, you would make four groups of one (three groups of one and one group of one). That wouldn't work, so just try to keep at it in thirds until you get the answer. That might mean taking a few extra steps, but it is the fastest way.
(1 vote)
• I have a riddle for anybody who wants to solve it. Here it is. A man is working in a gold mine. His employer is giving him an equal part of a gold bar each day. But the employer can only cut the bar twice. How does the man get 1/7 more each day?

HINT: You can take gold away from the man.
(1 vote) • Here's how you do it. Cut the gold at the 1/7 mark and then at the 3/7 mark so you have 3 pieces that are 1/7 long, 2/7 long and 4/7 long. On the first day give him the 1/7 piece. On the second day give him the 2/7 piece and take back the 1/7 piece. On the third day give him back the 1/7 piece. On the fourth day take the 2/7 piece and the 1/7 piece and give him the 4/7 piece. On the fifth day give him the 1/7 piece back. On the sixth day take away the 1/7 piece again and give him the 2/7 piece again. On the last, seventh, day give him the 1/7 piece for a final time.

Personally I think it would be much easier to just pay him at the end of the week and give him the whole bar. Also isn't it unfair to pay somebody and tell them you can't spend it until the end of the week?
• If they are balanced they weigh the first group against the third group. Here you will get a couple choices. If the first group is heavier then both the first and the second group contain a bad ball. Weigh 1 vs. 1 within both groups to find the bad balls (4 weighs). If the third group is heavier then it contains both bad balls. Within that group weigh one ball vs. another then weigh that one ball vs. the third. That will tell you which two balls are bad (4 weighs). If the 1st group balances the 3rd group then the 2nd group has both bad balls. Within that group weigh one ball vs. another then weigh that one ball vs. the third. That will tell you which two balls are bad (4 weighs).

If one is heavier take that group and weigh it against the third. If they balance then both of those groups has a heavier ball. Do 1 vs. 1 within both of the groups to isolate the bad balls (4 weighs). If the original one is heavier then it contains both bad balls. Within that group weigh one ball vs. another then weigh that one ball vs. the third. That will tell you which two balls are bad (4 weighs).

So if there are two bad balls that both are heavier and weigh the same you can conclusively find both bad balls within 4 weighs. • To solve the problem split the balls into 4,4 and 1. Then weigh the two 4s. If they weigh the same then the one left is the counterfeit. If one group (let's say group A) weighs more than the other group (group B) then the counterfeit is in group A, or vice versa. Divide group A (or B) into two groups of two. The group that is heavier contains the counterfeit. Then divide the heavier group into two groups of 1 and weigh them. the heavier ball is the counterfeit. • That will give an expected number of weightings to be 2 7/9, which is not the lowest possible. However, what Sal wants is the number of weightings that you can be 100% sure will be present, never based on random chance, which means the maximum possible number of weightings using your scheme, which is 3. This scheme, with 3 weightings, however, does not give the least possible number of weightings.   