 Here are the answers to the brainercises. I hope you got them all right! We’ll start off with the one that was corrected.

4. Correct version:You have 100 balls divided evenly into ten boxes (ten balls in each box). One of the boxes contains balls that are 1.5 grams lighter than the balls in the other boxes. You are given a digital weigh balance. How do you figure out which of the ten boxes contains the balls that are 1.5 grams less by using the balance only once– you can only take one digital reading. (Note: The boxes are numbered 1 through 10 and so are the balls from the boxes i.e. a ball from box 2 will be labeled 2)

Answer: You can only take one measurement, so you need a way to weigh at least one ball from each box. However, since you can only take one reading, all the balls must be on the balance at the same time and need to be differentiated somehow. You know that the balls and boxes are numbered. Take the number of balls from each box that corresponds to its number and put it on the balance. For instance, take 1 ball from box 1, 2 balls from box 2, 3 balls from box 3, etc. Now, you have at least one ball from each box. If you do this for all ten boxes, you should have 55 balls on the balance at the same time. If all the balls were the same weight (let’s say 10 grams), then the digital reading would be 550 grams. However, one of the boxes has balls that are 1.5 grams less. Thus take the difference of your digital reading from 550 grams. Then divide this answer by 1.5 grams and you know which box is the odd one out! For example, let’s say your reading was 544 grams. 550 minus 544 grams is 6 grams. 6 divided by 1.5 grams is 4. Thus you know that the lighter balls are from box 4.

New Brainercises:

1. Let us take a variation of the old problem with the 4 stones. Last time, you had four stones and one was either heavier or lighter. Now, you have eight stones and you know for sure that the odd one out is heavier. How do you find the heavier stone by using the weighing scale only twice? Remember this is the judge’s scale not the digital one.

Answer: Your technique for answering this question should be similar to the one with the 4 stones. In fact, this time you have the distinct advantage of knowing that the odd one out is heavier. Let us call the eight stones A, B, C, D, E, F, G, and H. First, weigh three stones against another three stones. Weigh ABC against DEF. Scenario 1: ABC turns out to be heavier than DEF, so you know that the heavier stone is either A, B, or C. Weigh A against B. If one of them is heavier, then it will show on the scale. If both are equal, C must be the heavier one. Scenario 2: DEF is heavier than ABC, so D, E, or F is the odd one out. Weigh D against E. If D or E is heavier, it will show on the scale. Otherwise, F is the odd one out. Scenario 3: ABC and DEF weigh out the same. This means the heavier stone is either G or H. Just weigh G against H and the heavier one will show. As you can see, no matter what scenario you choose, you only have to weigh twice!

2. A school has 1,000 students and 1,000 lockers, all in a row.  They all start out closed.  The first student walks down the line and opens each one.  The second student closes the even numbered lockers.  The third student approaches every third locker and changes its state.  If it was open he closes it; if it was closed he opens it.  The fourth student does the same to every fourth locker, and so on through 1,000 students.  To illustrate, the tenth locker is opened by the first student, closed by the second, reopened by the fifth, and then closed by the tenth.  All the other students pass by the tenth locker, so it winds up being closed.  How many lockers are open?

Answer: This seems rather complicated doesn’t it? Let us focus on different parts of the problem to elucidate the solution. Since the second student closes each even numbered locker, he or she is changing the state of the even-numbered lockers (since they were all opened by Student 1). Student 3 changes the state of every 3rd locker. The pattern is that if the student’s number is a factor of the locker’s number, then he or she will change that locker’s state. This is well illustrated by the tenth locker’s scenario given in the problem. The factors of 10 are 1, 2, 5, and 10. As you can see, those students were the ones that changed the state of the tenth locker. Since student 1 is a factor of all the lockers, he will open all of them. To remain open, the lockers need to be handled only by an odd number of students. As you can see, locker 10 was handled by 4 students and ended up closed. Factors always come in pairs, so this will be hard to accomplish. For example, 1×10 and 2×5 are the factor pairs of the number 10. The only numbers that have an odd number of factors are square numbers. Take the number 16, for instance. The factors are 1, 2, 4, 8, and 16. The factor pairs are 1×16, 2×8, and 4×4 (which is not really a pair). So the square number lockers will be open (ie. 1, 4, 9, 16, 25, etc.) You have to go all the way to 31×31= 961. This is the last square smaller than 1,000; thus, 31 lockers are open. 3. The island of Elbonia has a rather eccentric postal system. Postage for an item can be anything from 1 dinari to 15 dinari, and you must use exact postage. Frustratingly, there is only space on the envelopes in Elbonia to attach a maximum of three stamps. What is more, they only have three different denominations of stamps, can you work out what they are?

Answer: We need three different types of stamps such that we can cover all the numbers between 1 and 15. The most obvious denomination we need is 1 dinari. The next most obvious is 5 dinari. You can only have 3 stamps on an envelope so 5+5+5= 15. That leaves only one denomination. The last denomination is 4 dinari. This becomes obvious if you try guessing and checking different numbers between 1 and 15. There is no way to get 4 dinari or 9 dinari on an envelope without the 4 dinari denomination. So the answer is 1, 4, and 5 dinari stamps.

4. What shape completes the bottom line?

triangle pentagon square

square hexagon hexagon square

pentagon hexagon hexagon hexagon square triangle

hexagon octagon octagon octagon octagon ==?==

Answer: Let us convert the shapes into numbers by their number of sides. The first line would read 3 4 5. The second line would be 4 6 6 4. The pattern is that the square of the first number in the line equals the sum of the other numbers in the line. For example, take line 1: 3×3=4+5. Thus in the last line, 6×6=8+8+8+8+? By simple arithmetic, the missing number is 4. There are 4 sides in a square, so the missing word is square!

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