Question number 17. The chessboard shown in the picture is damaged. How many black squares are missing on the right side of the line? So here we have a chessboard, some black and some white squares are shown, there is a line dividing the board in two, and pieces are missing, squares are missing. Our job is to figure out how many black squares are missing from the right-hand side of the line. So one way to do this is to try to draw in the correct number of black squares. I can do that here with the computer, so let me draw lines across like that in black, and then across the other way, like so, and once I have all of them, we will color in the squares which have to be colored black, and then count how many we have on the right-hand side of the line. OK, so here is the grid, like so, and then we can continue coloring the squares. So here would be the first black square over here, so let me just put a 1 here, then we skip a white square, and there is a 2 here, second black square, then here would be a third one, this one is white, 4, and then we continue over here with 5, 6, then 7, 8, 9, 10, 11, and in the corner here, 12. So those have to be the black squares, we can check our work here by looking at diagonals, the diagonal here would match up with the diagonal of the black squares across the board, and likewise over here, where the numbers are filled in, that matches up the diagonal black squares, likewise over here. So we have filled in a total of 12 squares in black, and so that's the number missing on the right-hand side of the board, and the answer is B, 12. Question 18. Peter Rabbit eats cabbages and carrots. Each day he eats either 10 carrots or 2 cabbages. Last week Peter ate 6 cabbages. How many carrots did he eat last week? We can keep track of what Peter eats in a week by writing the information down in a table where I will have a column here for the day of the week, a column for the number of carrots he eats, and a column for the number of cabbages. So the first column is day, and I'll just label that with the number, so from 1 to 7, and then I'll have a column for the number of carrots eaten on each day, and a column for the number of cabbages eaten on each day. And so let's say that on day 1 Peter Rabbit eats 2 cabbages, then we know a total of 6 have been eaten during the week, so on day 2 he must have eaten some more, so let's say 2 more, because he eats them 2 at a time, or he doesn't eat any, and that makes a total of 4, so we need 2 more. Let's say on day 3 he ate 2 more, and now that gives us a total of 6 cabbages, which is exactly the number that we require, so he does not eat any cabbages for the remaining days. Now we can fill that information in and say on day 4 no cabbages are eaten, so that's 0, no cabbages on day 5, on day 6, and on day 7 he does not eat any, so there is a total of 6. Now he either eats cabbages or carrots, so when a cabbage is eaten, he does not eat carrots, so on days 1, 2, and 3, while he's eating cabbages, he does not eat any carrots, and when he is eating carrots, he is not eating cabbages, so on day 4 he needs to eat 10 carrots, on day 5 he needs to eat 10 more, on days 6 and 7, while he's not eating cabbages, he's eating carrots, 10 at a time. And now we can add up these numbers, we have 10 plus 10, and 2 more 10s, that gives us a total of 40, so 40 carrots must have been eaten during the week, that's also 6 cabbages were eaten, and so that's choice in D here, 40. Number 19, what should you put in the square on the bottom to get a correct diagram? And the diagram is filled in by following the direction of the arrow from one piece to the next, and performing the operation next to the arrow, so from 7 for example, if we follow the arrow, the arrow says add 2, and we add 2 to 7 and obtain 9. So moving on, we have to divide by 3, 9 divided by 3 is 3, next we move on clockwise from 3 in the direction of the arrow, that says add 9, 3 plus 9 is 12, and next we have to multiply by 4, so 12 times 4 is 48. To move on to the next piece, we have to know what's in the square. So we cannot move on for now, but what we can do is move backwards, fill in the pieces here to the left of 7, and the way we do that is we say, imagine we're in the piece here where I'm pointing, there's some number here that's missing, but I know that when I divide it by 2, the answer is 7, so that number is 14, 14 divided by 2 is 7, and continuing on this way, there is a number over here that when we subtract 2 from it, we get back 14, so that number has to be 16, and finally we can see that 8 is the last missing number because 8 times 2 is 16. So we have filled in all the numbers in the diagram, but we're not done yet, we have to find the operation here that is the relationship between 48 and 8. So let's check the possibilities. Is it A? Is it minus 38? So 48 minus 38 is not 8, it's 10, so it's not A, we can cross that out. Is it divide by 8? 48 divided by 8 is 6, it is not 8. Is it subtract 45? 48 minus 45 is 3, not 8, so that is not correct. Is it multiply by 6? No, because 48 times 6 is much larger than 8, and finally we have E, divide by 6, so that's the only remaining choice, let's write that in, divide by 6, and then we can check that 48 divided by 6 is indeed 8, 8 times 6 is 48. So that is the correct choice, and we mark E as the correct answer. Question number 20. Put the digits 2, 3, 4, and 5 in the squares and calculate the sum to get the largest possible value. What is that value? So here in the four squares, we have to use each of the numbers exactly once, and create the largest possible value in the first term, and the largest possible value in the second term. So let's begin by using the largest number we have, 5, and putting it in the tenths place to create a number that will be 50-something. Now 5 is used, and the next largest value we have is 4, and we should put it here in the tenths place of the next term, so that we will have something like 50 plus 40, rather than, for example, 54 and 32. And now we have 2 and 3 remaining. We can put it here, and then the 2 over here. So here are some possibilities. If we start with a 53 and a 42, we obtain a 95, or if we start with a 52 and add to that a 43, we also get a 95. So 95 is the largest possible sum here, and we obtain that by making sure the tenths digits are as large as possible, and then the ones digits can be interchanged, like in these two calculations, to obtain, in either way, a 95. So the answer is D. Question number 21. The central cell of the square was removed. So we have here a 5 by 5 square with the center square removed. We cut the rest of the square into identical pieces. Which type of piece is not possible to get? We can see that by stacking pieces A in a column, we can obtain exactly a copy of the square here. So if I make a copy of piece A several times, and stack it on top of itself, for example, here is one, here is the second one, it is pretty clear that with two more of these, I will be able to create the left column of that square, and then I can take a piece like that and rotate it 180 or 90 degrees, and then move it over here, like that, and so I have then exactly half the square, and the other half I can make also by reflection. So A is definitely possible to obtain. Now with B, what I can do is I can copy that piece also several times and show you how to obtain the other half of this square. So here is a copy of B and then I can rotate it 90 degrees like that and I can line some of those pieces up here like so. And then I'll have a copy here of the right hand side of the square. What I will need to do is delete a little bit here, delete that 2x2 piece and then copy more of the 1x3 pieces like so. And so here I'm leaving a little bit of space but we can see that everything lines up and A and B also can be used to make our square. Now with C, what we can do is just trace it out or make copies again but tracing out is a little easier here on the computer. So I can take any 2x2 square and then just trace out a bunch of the pieces in C like so and then here in blue I would have another copy and since it's possible to recover the square from pieces shaped like what blue and red here make up together, certainly it's possible to use piece C, so that is OK. And then piece D I can similarly treat. I can trace out a copy of D a couple of times. So here like so in blue and then I can do another one in red right on top of it. And it doesn't really look like it will be possible because maybe that piece is too long but I can keep going like so. That would be the next piece. And then you see that D is also possible. We can keep going like that. But the piece in E is not possible. Let's try to find a couple of copies of E and the big square here. So I would have to trace out a T-shaped piece like this on its side. OK, so here is one and then maybe another one right next to it in blue like so. And then it looks like I can just keep going like this, tracing out these pieces. But pretty soon I run into trouble. Can I use one more here in green? Let's try to fit in a green piece. And what happens is if I line it up like this vertically, then the corner piece here cannot be covered. Or if I line up the T piece horizontally like so in the missing space, then again I have even more blocks that are missing. Then the next piece would have to go on top here like that and then I wouldn't have a square anymore. I would have to go outside of the boundary of the square. So again, here something bad happens. So E cannot be used and that is exactly what we're trying to discover, which piece is not possible to get. So the answer here is E. Question number 22. To get the product 2 times 3 times 15, Bill has to press the keys of his calculator 7 times. 2 times 3 times 1, 5 equals. Bill wants to multiply all the numbers from 3 to 21 using his calculator. At least how many times will he press the keys of his calculator? So let's write out all the numbers he has to press. So starting with 3, we have 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, then 13, 14, 15, 16, 17, 18, 19, 20, and 21. And then between each of those numbers, he will have to include a time sign. So that's a bunch of times or multiplication commands in the calculator. 12 times 13 times 14. Between each pair, we have to use a times until the very end where he wants the result and he uses an equal sign. And then we count the numbers. So maybe let's count the signs first. So we have here in this column 2 multiplications, 2 over here, 2, 2, 2 over here, like so, 2 over here, and then finally 1. So that's a total of 2, 4, 6, 8, 10, 12, 14, 16, 18, 19 operations. And then how many times do we have to press a digit? So here, 3 counts for 1, so 1, 2, 3, 4, 5, 6, 7, then 10 counts for 2, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, and 31. So 31 numbers. And so those are the numbers of keys we have to press or that Bill has to press. And if we add these up, we have 0 here, carry 1, and that gives us exactly 50 keys. And that would be our answer. Just counting each symbol individually, we have here written out exactly 50 symbols, each corresponding to one press of a calculator key. So the answer is D, 50. Question number 23. Fred has 4 red cubes, 3 blue cubes, 2 green cubes, and 1 yellow cube. He builds a tower, as shown in the picture here, in such a way that no two adjacent cubes have the same color. What color is the cube in the center here, marked with the question mark? So we can imagine that that question mark represents already a color. And let's see where else the question mark can be placed so that no two question marks are adjacent. So can I move the question mark anywhere to these cubes? The answer is no, because then we would have two adjacent question marks. But I can put a question mark here in the right position, and opposite of that, the left corner. And then, on the very top, we can have another question mark. And the 6 cubes here, in the center, around the center question mark, can be colored different colors. So we know that Fred has these 4 times the question mark, and he also has 4 red cubes. So is it true that we can just use red for the question mark? So we can fill that in here. In the picture, we have 4 red cubes. Around the center cube, there are 6 that can be colored different colors. So let's try 3 blue. Let's try 2 green. And then finally, we can do a yellow one. And we see that by rotating these colors around the center question mark, we actually obtain many different answers. But the red cubes cannot be moved into any other position. If we have 4 of them, moving one red cube will put it next to another red cube. So the center cube has to be red. That's the color hiding under the question mark. And the answer here is A. Question number 24. Cog wheel A, in the picture on the left here, turns around completely once. And in the picture on the right, we are shown it's turning clockwise. The question is, here is X, and at which place is X now, after wheel A turns around clockwise completely once? Wheel A turns the middle wheel, and the middle wheel, in turn, spins the right wheel. So if wheel A turns clockwise, wheel here in the middle will turn counterclockwise, and that will cause the last wheel to turn clockwise again. And how many times will it do that? If X is here in position H, a complete revolution of wheel A will cause X to move around as many times as there are cogs on wheel A. So A has how many cogs? Let's count them. Here is 1, 2, 3, 4, 5, 6, and 7 cogs. So X will switch position exactly 7 times. So let's count that. Here will be one switch, 2, 3, 4, 5, 6, and then one more to position A. So X goes all the way around here, and then switches positions once again, and ends up at A, with one full revolution of the first cog wheel. So the answer here is letter A, and that is position A.

geometry

arithmetic

logic puzzles

chessboard

Peter Rabbit

deductive strategies