Problem number 17. The rooms in Kenga's house are numbered. Baby Yuru enters the main door, passes through some rooms, and leaves the house. The numbers on the rooms that he visits are always increasing. Through which door does he leave the house? A, B, C, D, or E? The rule is that the numbers are always increasing. We are going to try different paths, applying this rule, and find out where Yuru leaves the house. He starts with the room number one, and he can go in any of the three directions. I would start with this next smallest number, which is 2, to the right here. From 2, he can go to the room number 10, but that will give us a dead end because there is no other room with a number greater than 10 around it. So he cannot go that direction. If we back up to room 2, he can go to room number 4, then number 6. We already know that if he goes into the room of the 10 now, he will not be able to go on. So we'll go on to the room with the number 9. He can continue going down to 12, but the only exits lead to rooms of number 9. So this path has to end here. So backing up, if we start in room 1 and go to the room of the number 9, from here we can go to number 10. This is the same room as before. Again, there is no exit. That path also will not work. Now let's try a room with the number 1 and go into the room number 3. We can go into the room of the 4, and from there the room of the 6. But it's surrounded by 5, so we cannot go on. We will back up. The 6 was the only place to go to from room number 4, so we're going to have to back up further. From the room number 3, we can go to number 5. We already tried the 6 to the left, so go to the 6 below here. From here, we can only go to the room of the 7. We can try actually either of the rooms of the 8. If we go to the right, we know not to go to the room with the 10 above it, but we can go to the room of the 9 here. If we had gone to the room of the 9 that's on the left, we also would have ended having to go down and end our path there. And just so we see this, if we had gone to the other room of the 8 from the 7, we also end up having to go to this room of the 9. From here, we could go to 12, but then we would be stuck, so let's not do that. We can, however, also go to the room of the 10, which will lead us to exit D. So if Roo took this path that we just marked, going to the rooms of the 1, 3, 5, 6, 7, either of the 8s, 9, and 10, he would leave the house at the door marked D. So the answer is D. Problem number 18. Four balls each weigh 10 grams, 20 grams, 30 grams, and 40 grams. Which ball weighs 30 grams? A, B, C, D, or answer is it could be either A or B. Start by looking at the second balance and figuring out how B and D together can weigh as much as ball C. There are two ways that this could happen, given that the weights of the balls are 10 grams, 20 grams, 30 grams, and 40 grams. Either C weighs 30 grams and B and D are, one of them is 10 and the other is 20, we don't know which one is which, or C could weigh 40 and then B and D are 10 and 30. Again, we don't know which one is which, but now we do know that C weighs either 30 grams or 40 grams and that one of B and D weighs 10 grams. This also tells us that ball A must weigh either 20 grams or 40 grams because on the other scale where A does not occur, we know that one of the balls, either B or D, has to weigh 10 and there must be a ball that weighs 30. Looking at what we know from the second balance, we are going to take the ball C off from the first balance and substitute for the two balls B and D because we know that will keep the weight exactly the same. So B and D. Notice that at this point, ball B occurs on both sides of this balance, so we could take both of them off. Now we see that ball A weighs more than two balls D. That means that D cannot weigh 20, 30, or 40 because A is at most 40. And even if D was 20, then the weights would be equal. So D has to be 10 and A then is 40. If A weighs 40, then ball C, which we said must be either 30 or 40, must be 30. So now we know that the ball that weighs 30 is C. So the answer is C. Problem number 19. The band shown in the drawing can be fastened in five ways. How much longer is the band fastened in one hole than the band fastened in all five holes? A, 4 centimeters. B, 8 centimeters. C, 10 centimeters. D, 16 centimeters. E, 20 centimeters. Let's first look at an animation of the band. It shows it being closed in one hole, then two holes, then all five holes. As we saw in the video, no matter how we fastened the band, the back did not move. We can see the back ending right here on the band. When it was fastened in one hole, this was the area of our overlap. Now let's look at the band that was fastened in two holes. We see that the band starts here. So this is the area of our overlap. Before, the overlap was here. This is the same as saying that the overlap before was the same as this area here and this area here. Or, the only part that's added is the two centimeters that we know are between each hole. So when we move from the band being fastened in one hole to the band being fastened in two holes, we are only decreasing the length by two centimeters. Now let's look at the band that is fastened in all five holes. Here is the end of the back of the band. So this area and this front were the original overlap. And we have one, two, three, four intervals of two centimeters each between the holes. Two centimeters times four is eight centimeters. Looking back at our original question, we see that eight centimeters is answer B. So B is our answer, eight centimeters. Problem number 20. In an ancient language, the symbols, I'm going to call them I, sun, orbit, because it reminds me of the way the planet orbits the sun, comb, and fish, represent the following numbers, one, two, three, four, and five. Nobody knows which symbol represents which number. We know that orbit plus orbit equals fish, sun plus sun equals orbit, sun plus fish equals comb. Which symbol represents the number three? A, the I, B, the sun, C, the orbit, D, the comb, or E, the fish? One thing to remember is that when we add two of the same number, we will get an even number. So when we add two orbits together, we get the fish, so that has to be an even number. And when we add two suns together, the orbit has to be an even number. Now, we only have two even numbers among those listed, numbers two and four. Now, the other thing that we see is that two orbits added together equals the fish, which means that we have to use the numbers two and four to make this first equation true. This works if we say that the orbit is two and the fish is four, because two plus two is four. Now, we see that in the second equation, if the orbit is two, then the sun has to equal one, because one plus one is two. We are left with the third equation. We can put in one for the sun and four for the fish, which tells us that the comb is five. We have not found which symbol is equal to three yet, but we have found a symbol that corresponds to each of the other numbers. One is the sun, two is the orbit, four is the fish, and five is the comb. This leaves the I to represent number three, so the answer is A. Problem number 21, the hexagonal stained glass tile is flipped. One of the flips is shown. What does the stained glass tile look like at the far right? Let me pop a close-up of the picture. We can use our imagination to think about the flipping, or we can draw the pictures to help us. The first flip is shown. Each picture looks like its mirror image. When we flip the next time, we are flipping around this line. It's fairly easy to see that the dolphin will be here. The head is on this end, the tail here, with the fin up here. The tail here, with the fin up here. Looking at the tiles next to it, we see the ant on this tile. The butterfly will be at a tile next to the ant, so it's over here. Now it's starting from the other direction. We have the little cat, which was to the left of the dolphin. It would be here, with the tail closest to the dolphin. The ladybug will be on the next tile. Its head is on the side closer to the cat. The ladybug will be on the next tile. Its head is on the side closer to the cat. The last tile is the little dog. Head pointing this way. Body. If we use the images as given, they would look something like this after the flip. I colored it in pink to make it stand out. Now, we see that when we are flipping the tile around this line, the butterfly will be right next to the place where we flip. The ant will be above the butterfly, and the little dog will be to the right of it. Here's a picture of the tile after it's been flipped. We see that the dolphin is on the top, and the little dog is on the bottom. We see that the dolphin is on the top, and the little dog is on the bottom. Let's go back to the original problem and find the answer that corresponds to it. This is answer B. Problem number 22. The large rectangle is made up of a number of squares of various sizes. The three small squares each have an area of 1. What is the area of the large rectangle? A, 165, B, 176, C, 187, D, 198, or E, 200. We calculate the area of a rectangle by multiplying the length of the shorter side by the length of the longer side. We can find the length of the sides of the large rectangle by keeping in mind that a square has sides of equal length. The problem tells us that the area of each of the smallest squares is 1. No matter what units we use, the only way a square can have an area of 1 is if each of its sides is 1, because 1 times 1 is 1. That means that this distance is 1, this distance is 1, and this distance is 1. This will make the side length of this square 3. The length is 3 here, and the length is 3 here. We can use this information to find the side length of the next bigger square right here. This part of the length is 3, and this part is 1, meaning that the side length is 4. That's 4, and this is 4. Now we can move on to the next bigger square, the one right here. Part of the length is 4, the other part is 3, given this square, so this is 7, which means that this side is also 7. Now we can find the length of the short side. We have 7 plus 3 plus the side length of the small square, which is 1. 7 plus 3 plus 1 is 11. 11 is the length of the shorter side. We have only one square left with side lengths that we have not determined yet. But we know that one part of it is 7, and the other part is 4. This is the same as the length we already found, which was 11, which means that the other side is 11. We can add this 11 to the 7, which is the length of the side of the other square making up this side. 11 plus 7 is 18. So we have 18 as the other side length. All that is left for us is to multiply 18 by 11. Let's multiply the ones first. 1 times 8 is 8. 1 times 1 is 1. And then the tenths. 1 times 8 is 8, but we need to write it here for the tenths place value. 1 times 1 is 1. Add those two together. 8. 1 plus 8 makes 9. And 1. The answer is 198, which is answer D. Problem number 23. Lian wants to write the numbers from 1 to 7 in the grid shown. Two consecutive numbers cannot be written in two neighboring cells. The neighboring cells are those that meet at the edge or at the corner. What numbers can he write in the cell marked with a question mark? A. All 7 numbers. B. All of the odd numbers. C. All of the even numbers. D. Only the number 4. E. Only the numbers 1 or 7. First, let's figure out how many neighbors the cell marked with a question mark has. The problem tells us that neighboring cells are those that meet at the edge or at a corner. That means that any cell that shares an edge or a corner with the cell marked with a question mark is its neighbor. So, this top cell will be a neighbor because it shares a corner. The cell under it will also be a neighbor. It shares an edge. The cell underneath shares an edge, so it's a neighbor. The other top cell is a neighbor because it shares a corner. And the cell under that also shares an edge, so it is a neighbor. So, the cell has five neighbors. There's actually only one cell which is not a neighbor of the cell with the question mark. Now, let's look at the numbers that Leon needs to write into the grid. These are the numbers 1, 2, 3, 4, 5, 6, and 7. Leon needs to include all the numbers in the grid, but there's only one spot on the grid, right on the bottom, that a number that is consecutive to the number with the question mark can be written. Because this is a series of consecutive numbers, each number except the first and the last will have two numbers that cannot be next to it. For example, if we choose the number 3, the number 4, and the number 2 cannot be next to it. But if we look at the number 1, this sequence has only one number that cannot be next to it in the grid. That's the number 2. Likewise, the number 7 has only one number that cannot be next to it, and that's the number 6. This means that if Leon chooses any number that has two neighbors in the list of consecutive numbers, he will need to break the rules by putting one of the neighbors next to the cell with the question mark. This means that the only numbers he can put in the cell with the question mark are the numbers 1 or 7. This is answer E. Problem number 24. To defeat a dragon, Matthias has to cut off all the dragon's heads. If he can cut off three of the dragon's heads, one new head immediately grows. Matthias defeats the dragon by cutting off 13 heads in total. How many heads did the dragon have at the beginning? A, 8. B, 9. C, 10. D, 11. E, 12. To solve this problem, let's work backwards. Matthias had to cut off the dragon's final head in order to kill the dragon. That's one head. That head would have grown when he had cut off three other heads. That gives us four heads so far that he has cut. So we need to keep going further back. Each of the three heads in this row grew when he had cut off three other heads. So this head grew from three heads. This head grew from three heads. This head grew from three heads. And this head grew from three heads. So there were, at that point, nine heads that he had cut off. Now let's add the numbers together so far. Now let's add the numbers together so far. 1 plus 3 plus 9 is... 9 plus 1 is 10 plus 3, that's 13. So we have found all the heads that Matthias had cut, which were 13. Which means that at the beginning, the dragon had nine heads. So our answer as to how many heads the dragon had at the beginning is B, 9.

logical reasoning

step-by-step calculations

problem-solving

puzzles

deduction

mathematics