2. Weighing Created by Inna Shapiro ©2007

3. Problem 1 Judy has 3 old coins. She knows that one coin is fake and is heavier than the other two. How can she find out which coin is fake by using the scales only once?

4. Answer Judy can put two coins on the opposite pans. If these coins are equal, the third one is fake. If one coin on the scales is heavier than the other one, it must be fake.

5. Problem 2 There are 9 similar rings. A jeweler knows that one ring is heavier than the others. The jeweler can use the scales twice. How can he find the heavy ring?

6. Answer The jeweler can put three rings on one pan of the scales and three other rings on the other pan. If the triples are equal, then the heavy ring is among the remaining three. Otherwise it is one of the three rings in the heavier triple. To find the heavy ring in a triple, use the same method as in Problem 1.

7. Problem 3 An archeologist has 27 ancient coins. He knows that there is one fake coin among them. The fake coin is heavier than others. Can he find the fake coin if he can only use the scales three times?

8. Answer The archeologist can make three 9-coin sets and put two of them on the opposite pans of the scales. This way he will find which set contains the fake coin. Now he can follow the steps from Problem 2.

9. Problem 4 Four pearls have the same shape and size. Jane knows that one fake pearl differs in weight from others, but she is not sure if it is heavier or lighter than the rest. Please help Jane to find the fake pearl. You can use scales twice.

10. Answer Let us compare two pearls on the scales. 1.If they are equal – both are real. So we can place the third pearl with a real pearl on scales. If they are equal -> the fourth pearl is fake. If not -> the third pearl is fake. 2.If one is heavier, and the other is lighter, then the third and the fourth pearls are real. So we can place a real pearl and the heavier pearl on scales. If it is still heavier -> it s fake. If both are equal-> the lighter pearl is fake.

11. Problem 5 There are 75 similar rings. Mike knows that 74 rings have the same weight, but one fake ring differs from the others. Mike can use the scales twice. Can Mike find out if the fake ring is heavier or lighter than the rest? /Mike does not need to find the fake ring./

12. Answer Mike can divide the rings into three equal groups. Then he can compare the 1 st group /of 25 rings/ and the 2 nd group /of 25 rings/. 1.If these groups are equal, both have no false rings, and he can compare one group with the 3 rd to see if the false ring is heavier or lighter. 2.Assume the 1 st group is heavier. Then the third group is good. And he can compare the 1 st and the 3 d group. If they are equal – the fake ring is in the 2 d group and it is lighter. If not – the fake ring is heavier.

13. Problem 6 There is a set of weights: 1 g, 2 g, 4 g, 8 g, 16 g. Mary has 31 candies each weighing 1 g. Prove that she can use this set of weight to counterbalance any number of candies from 1 to 31.

14. Answer We can easily counterbalance 1, 2 and 3 candies. Adding the 4g weight, we can deal with 4, 5,6, or 7 candies. Now we add the 8g weight to the previous ones, and with it we can counterbalance from 8 to 15 candies. And so on with adding the 16g weight.

15. Problem 7 There is a set of weights : 1 g, 3 g, 9 g, 27 g, 81 g. Peter has 121 cookies each weighing 1 g. He can put cookies only on the left side of scales and add weights on both sides of scales. Prove that he can use this set of weight to counterbalance any number of cookies from 1 to 121.

16. Answer You can put weights on both sides of scales to counterbalance 1, 2, 3, or 4 cookies with 1g and 3g weights. The 9g weight lets you counterbalance up to 13 cookies, as follows. If you have between 5 and 13 cookies in the left pan, you can put 9g in the right pan, and now you effectively have to counterbalance between 1 and 4 cookies in either the right or the left pan with 1g and 3g weights. The 27g weight lets you to continue to 40 in the same way, and the 81g weight, to 121 cookies.

17. Problem 8 David has a package with 9 kg of grass seed. He needs 2 kg for his lawn. He also has a 200 g weight. How can he get 2 kg of seed in a pile if he has a scales that he can only use 3 times?

18. Answer David can place a 200g weight on the left side of the scales and then place seeds on both sides to get a balance. That means he will get 4.4 kg on the left and 4.6 kg on the right side. 4.4 kg could be easily divided with scales by two equal parts 2.2 kg each. Finally, David can subtract 0.2 kg from 2.2 kg with a 200 g weight.

19. Problem 9 Judy has four similar melons. She can compare melons on the scales five times. How she can arrange the melons in order of increasing weight?

20. Answer Suppose the melons have weights a, b,c, and d. Judy can compare the 1 st melon with the 2 nd, and the 3 rd with the 4 th. Suppose a>b and c>d. The third time Judy has to compare the 1 st and 3 rd melons. If a>c, then a>c>d and she can compare b and c. If b>c, then a>b>c>d. On the other hand, if c>b, then Judy has to use the scales a 5 th time and compare b and d to find the lightest of the four melons. The case c>a is similar.

21. Problem 10 A package contains 24 lb of granulated sugar. Can you measure 9 lb of sugar using the scales 3 times? /You cant use weights./

22. Answer You can divide 24 lb into two equal piles that weigh 12 lb each. After this you can divide 12 lb into two equal parts again. And then 6 lb can be divided into 3 lb and 3 lb by using the scales. You can now put a 6 lb pile and and a 3 lb pile together to get 9 lb of granulated sugar.

23. Problem 11 An archeologist has 101 ancient coins. He knows that there is one fake coin among them, and its weight is different from the weight of the others. Can he find if the false coin is heavier or lighter than the others, if he can only use the scales twice?

24. Answer Yes, he can. He can divide the coins into three groups of 33, 33 and 35 coins (group a,b and c). Then he must compare groups a and b on the scales. If they are equal – al 66 coins are genuine. After that he can place group c on the left part of the scales an 35 genuine coins on the right part. If the left part is heavier – the fake coin is heavier, otherwise it is lighter. If group a is heavier, then all coins in group c are genuine, and he can use the scales to compare 33 coins from group c with group a. If they are equal – the fake coin is in group b and it is lighter than others. If group a is heavier, the fake coin is in group a and is heavier than others.

25. Problem 12 A wizard shows Dorothy 12 coins which look the same, but one of them is fake and is either heavier or lighter than the others. How can Dorothy find the false coin if she can use scales three times?

26. Answer (1) Dorothy can follow these steps: Divide coins into tree groups: group A contains coins 1, 2, 3, and 4; group B has coins 5, 6, 7, and 8; and group C, 9,10,11, and 12. Weighing 1. Compare groups A and B. If they are equal – all 8 coins are real. Weighing 2. Compare 2 real coins with 9 and 10. If they are equal – the fake coin is 11 or 12, if not – 9 or 10. Weighing 3. Compare one of the two coins that can be fake with a real coin to find out whether it is fake. If A is heavier than B, then coins in group C are real. The fake coin can either be in group A and be heavier than the others, or it can be in group B and lighter than the rest. Weighing 2. Put coins 1, 2, 3, and 5 on one pan of the scales, and coins 4, 9, 10, and 11 on the other pan. If they are equal, the fake coin is lighter and is either 6, 7, or 8 /Problem 1/. If (1,2,3,5) is heavier, the fake coin is heavier and is either 1, 2, or 3 /Problem 1/. If (1,2,3,5) is lighter, the fake coin is either 5 or 4 and we can find it by Weighing 3. Compare coin 5 with any good coin /say, 12/. If they are equal, 4 is the fake coin. Otherwise 5 is the fake coin.

27. Answer (2). Hint There is another way to find the fake coin. Weighing 1. Compare the coins (1,2,3,10) against (4,5,6,11). Weighing 2. Compare the coins (1,2,3,11) against (7,8,9,10). Weighing 3. Compare the coins (1,4,7,10) against (2,5,8,12). After that Dorothy can think and point to false coin. Prompt. What will she see if coin 1 is fake? What will she see if coin 2 is fake? And so on.