Question number 20. Anne wrote the following equation. 2012 equal to m to the power m times quantity m to the power k minus k for some positive integer values of m and k. What is the value of k? In such a question about factorization, it's always helpful to know the prime factorization of 2012, which is 2 to the power 2 times 503, and 503 is a prime number. So comparing that to the form in Anne's equation, we have the factor m to the power m matching 2 to the power 2, so therefore we will use 2 is equal to m in our calculation. And now we must solve the following equation. 503 is equal to 2 to the power k minus k for k in some set of integers. Okay, so we recall that 2 to the power 10 is 1024, 2 to the power 9 is 512, which is the closest power of 2 in the neighborhood of 512, and so we should probably work with k is equal to 9, and checking that, 2 to the power k minus k is 512 minus 9, which is indeed 503. Our answer to number 20 is k is equal to 9, or d. Question number 21. A jeweler has 12 pieces of chain, each with two links, so they're shown here on the left. He wants to make one big closed necklace out of them, as shown on the right, by opening some of the two link chains and closing them afterwards. What is the smallest number of links he has to open? So two strategies are apparent here, taking two link chains at a time. One link can be opened and then closed to create a chain with four links, or both links in one chain can be opened and then used to create a chain with six links, that is by joining three two-link pieces together. And that second strategy is the more efficient one, so we will perform that on our chain and see how many links we must open and close to perform our operation. And let's see, let's take a link here, open both, and use them to create a longer chain like that, and I can keep going like that, each time creating a longer and longer chain. So now I'm up to six, and finally, removing the fourth chain, I'm up to eight links that I opened and closed. This last one allows me to create one long chain, but it's still open, and so I use the remaining eighth link to close up the chain and make a necklace. So the answer here is A takes eight links to perform this operation. Question number 22. The rectangular piece of paper, A, B, C, D, measuring 4 by 16 centimeters, is folded along the line M, N, so that vertex C coincides with vertex A. What is the area of the pentagon A, B, N, M, D prime? Note that this area consists of the shaded region, which is composed of two triangles, triangle A, D prime, M is a right triangle, and triangle A, N, M, and also we need to find the area of the triangle A, B, N, and together the three triangles will give us the area of the pentagon we are asked to compute. So first, let's label the sides with the information that's given to us. This is a 4 by 16 sheet of paper, so side A, B here measures 4 centimeters. Side C, D also measures 4 centimeters, but now that is the side A, D prime, like so. We don't know the length of side M, N. Let's skip that for now. The length here M, C is also unknown, but that became the hypotenuse of triangle A, B, N, and now its leg we can say measures 16 minus x centimeters. Similarly, the length M, D is not known, but that is now the same as the length of segment M, D prime, and the hypotenuse of triangle A, D prime, M, we can call 16 minus y centimeters. Now it should be clear which angles are right angles. Angle D prime here corresponds to angle D, so both of those measure 90 degrees, and angle B is also a right angle. We don't have information regarding any of the angles in triangle A, M, N, although the angle here at vertex A that originally corresponded to angle C is a right angle, but that is not useful to us. What we do have is that the height of triangle A and M, which I just drew in here, is also equal to 4 centimeters, so that would be enough for us to compute the area. So let's go ahead and do that. Let's get started with solving for x. By the Pythagorean theorem, we have that 4 squared plus quantity 16 minus x squared must be equal to x squared, and simplifying this, we see that 16 plus 16 squared minus 32 x plus x squared is equal to x squared, so the powers of x squared cancel nicely. We obtain that 16 plus 16 squared is 32 x, and so we can simplify that and obtain that 8.5 is equal to x. Okay, so now moving on to y, the same calculation gives us 4 squared plus y squared is equal to quantity 16 minus y squared. Simplifying again, the powers of y squared cancel, and we obtain that 4 squared is equal to 16 squared minus 32 y, or after simplifying, 7.5 is equal to y. And now we have all the information needed to compute our areas. First, let's do the area of triangle A, B, N. Okay, that would be equal to 1.5 times its base, which is 16 minus x, 16 minus 8.5, so 7.5 times its height, which is 4. Then we have the area of triangle A, D, prime, M, which would be 1.5 times its base, so 7.5 times its height, so 4. And finally, the last area of triangle A, N, M, which is 1.5 times its base, and now that's 16 minus y, so 8.5 times its height, which we calculated to be 4. And adding all those up, we obtain that the area of the pentagon A, B, N, M, D prime comes out to 47 centimeters squared, so the answer to question 22 is D, 47. Question number 23. We are asked to consider two trains moving at a constant speed. Train G takes 8 seconds to pass a milestone. Train H takes 12 seconds to pass a milestone. The two trains then pass each other in 9 seconds. Which of the following statements about the lengths of the trains is true? What we observe is that the length of a train is related to its velocity, so let me denote by Lg the length of a train by Vg the velocity, and we are told that it takes 8 seconds for the train to pass a milestone. That means from the tip to the very back of the train, as it passes at a stationary point, it takes 8 seconds for that to happen, so velocity times 8 seconds would be the length. Then the length of train H is given by this relationship, 12 times its velocity is its length, and together as the trains pass one another, we have the following relationship between velocity and length. And as the trains are passing, we expect in our calculation to have velocity of one be the negative of the velocity of the other, or some multiple thereof, and we interpret the minus sign as moving in opposite directions. So now with that, let's solve for one of the lengths. We can add the first two equations, and therefore obtain another relationship between the combined length, like so, and then after all, this is supposed to be equal to 9 times the individual velocities, like this, and so we can subtract these equations and obtain the following relationship. 9 times the velocity of train G minus 8 times the velocity of train G is equal to 9 times the velocity of train H minus 12 times the velocity of train H, so the velocity of train G is equal to negative 3 times the velocity of train H, where again the minus sign we interpret as movement in the opposite direction. And then let's go and substitute this information into our original equation. We obtain the length of train G is equal to the velocity of train G, which is now negative 3 times the velocity of train H times 8, which comes out to negative 24, the velocity of train H, and that is an absolute value equal to twice the speed of train H, like that, and that also tells us that train G is twice as long as train H, so the answer here is A. Question number 24. Consider the following number given by 2 to the power 59 times 3 to the power 4 times 5 to the power 53. What is the last non-zero digit of this number? To answer that question, I will begin by rewriting the prime factorization of K into a factorization I'm more comfortable working with, and I notice that there are some factors of 2 together with some factors of 5, and I can rewrite this by factoring out some powers of 2 as a product of 10 to some power and some other numbers, so there are two factors here to consider, 2 times some power 5 to some power, that power must be 53 if I hope to combine my factors in any way, so now I have 2 times 5 to the power 53, and so I obtain a 10 to the power 53, and so multiplying by 1 with lots of zeros introduces lots of zeros at the end of the product, so we now have to focus in on the last digit of this number. We know that 2 times 6 is 64, 3 times 4 is 81, so the product 2 to the 6th times 3 to the 4th will be some number that ends in a 4. Now multiplying that by 10 to the power 53 will leave this 4 and then introduce lots of zeros at the end, so the last non-zero digit of the number k is equal to 4, and the answer is question number 25. Peter creates a kangaroo game, and here the diagram shows the possible moves. The kangaroo starts at position s, which is school, and he can move from that position into any other position. In fact, from any starting point except from home, which is marked here by H, the kangaroo can jump to either of the two neighboring positions. When the kangaroo lands at home, the game is over. The question is to determine in how many ways can the kangaroo move from position S to position H in exactly 13 jumps. So what we have to do is come up with sequences of moves that begin at school, and then in 13 moves, end up at home. So we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 moves, with the last 13th move putting us back at home. OK, now let's look at the move that's before the last move, and what are the possibilities here. If the kangaroo is at school, then he can move to home or the playground. If the kangaroo is at the library, he can jump to home or the playground. But if he happens to be at the playground, he cannot move to home, because that is not a neighboring position. So there are two possibilities for the previous move before ending up at home and finishing the game. That is either S or L. Now, the move prior to that has to be the playground. From the playground, the kangaroo can jump to either school or the library, but he cannot move to home. And we repeat this process like so. These positions are reserved for the playground. And in every other position we have two possibilities, school or library. And depending on the choice of each, we have here all the possible moves, like so. The capital P's are determined, but in each of the places where we have a choice, we have two choices at each step. So that's 2 to the power of 1, 2, 3, 4, 5, 6, or 64 different combinations. So the answer to number 25 is C, 64 possibilities. Question number 26. We are looking at five lamps, each which can be switched to the position on or off. Each time we switch any lamp, or in other words, change its status from on to off, the status of exactly one other randomly chosen lamp is changed. At the beginning, all the lamps are off, and then we make ten such switch operations. Which of the following statements is true? So, in a diagram, we can describe our lamps as digits in a five-digit binary number with zeros and ones. So here, one would indicate on, and zero would indicate the off position. Initially, we are told that all the lamps are off, so we have five zeros like that. After making one move, what happens is one lamp will change state, and then one other randomly chosen lamp will change state. So let's say something like this. And what happens is, from having zero on after move one, we have two that are on. From here, making another move can create three situations. Either we add two lamps in the on position, or we happen to turn off two lamps from the on position. Or, we can have a third situation where we increase this number exactly by one. For example, if in the second move, the first lamp is switched from off to on, and the second lamp from on to off, like that. Now, in either of the three cases, we observe that the number of lamps that are on, the number of ones in our binary representation, remains even. And so, after no matter how many moves of each type, we are left with an even number of lamps that are on. Since there are five lamps, it's not possible for all of them to be on at the same time after performing any number of switching operations. Therefore, the true statement here among the list is C. It is impossible for all the lamps to be on at the same time. Problem number 27. Six different positive integers are given, the biggest of them being n. There exists exactly one pair of these integers such that the smallest number does not divide the bigger one. What is the smallest possible value of n? Since we are talking about division, let's look at the divisors of each of these numbers. And perhaps among the divisors, we can find the pair of numbers that satisfies the conditions of the problem. And then we'll have our value of n. So the divisors of 18 are 1, 2, 3, 6, 9, and 18. That's six numbers. However, looking at these, we have a pair 2 and 3 where the smaller number does not divide the larger number and several other pairs like that. For example, 3, 6, 3 and 9, 3 and 18 are not possibilities, but 2 does not divide 9. So that's something we can't have. Okay, let's look at 20. Its divisors are 1, 2, 4, 5, 10, and 20. Again, we have a pair of numbers where the smaller number does not divide the larger number and another pair that also fits this description. For example, 4 and 10. So we move on to the next choice, 24. Its divisors are 1, 2, 3, 4, 6, 8, 12, and 24. That's more than six numbers, so perhaps we can throw some out to obtain six with the properties that we desire. We have our pair 2 does not divide 3. And then we cannot have any other pair with such properties. So let's see if we can remove some of these numbers. 2 divides 4, 6, 8, and 12, so that's fine. 3 does not divide 4, and if we just get rid of 4, then we don't have that problem. 3 divides 6, does not divide 8, divides 12. So if we get rid of 8, we don't have that problem anymore. Then 6 divides 12 and 24, 12 divides 24. So now we have 1, 2, 3, 4, 5, 6 numbers where the choice 2 does not divide the larger number 3 is the only such pairing. So with 24, we have our desired division properties, and so the answer to the problem is C. 24 is the smallest such possible value. Question number 28. Nick wrote down all three-digit numbers, and for each of them he wrote down the product of its digits. After that, Nick found the sum of all these products. What total did Nick obtain? I will first solve a related question, a very simple case when we consider just one-digit numbers. So the sum of all the products are just the digits themselves, and if we sum them up, 1 through 9, we obtain 45. For two-digit numbers, we write down the answer immediately, and then let's discuss it, how the answer was obtained. The square of that quantity, or 45 squared, would be the answer. This we can see happens because the following expression for the square of a trinomial, a plus b plus c is given by a squared plus b squared plus c squared, and then plus 2ab, 2bc, 2ac. We note that digits can be arranged in any order, so 11 and 11 would be the same thing. 21 and 12, however, are not. So we obtain the same product from two different two-digit numbers, in this case with three digits a, b, and c. We account for all the possibilities by squaring this sum, and so the analogous result for nine digits, 1 through 9, gives us the following answer, 45 squared. So the answer to our problem at hand, when we consider three-digit numbers, is, by extension, the quantity 1 plus 2 plus 3 all the way up to 9 cubed, and that gives us 45 cubed, and the answer to number 28 is c. Question number 29. The numbers from 1 to 120 have been written in 15 rows in the manner indicated in the diagram. For which column, counting from the left, is the sum of the numbers the largest? So in column 1, we have 1 through 106. We could sum that, but it would take some time to repeat the calculation for all the columns. In column 2, we have 3, 5, 8, 12, in that fashion, all the way up to 107. So what we should do instead is study the change in the sum between columns. So we notice a pattern. To move from one column to the next, each cell is increased by 1, and the number that was the first entry in the previous column is deleted. So to move from column 1 to column 2, what we do is decrease by 1 the number on the top, and then there are 14 cells that are increased by 1. So the net change is plus 13 to move from column 1 to column 2. And we can repeat that analysis for the remaining columns. To move from column 2 to column 3, we subtract 3 and then add 13 once. And that gives us a net change of plus 10. To move from column 3 to column 4, we subtract 6 and add 12 once. So the net change is plus 6. To move from the fourth column to the fifth column, we subtract 10 and add 11. So the net change is plus 1. And then finally, to move from the fifth to the sixth column, we subtract 15 and add 10. So now we decrease the sum. The sum, therefore, of the numbers in column 5 over here is the largest because it's increasing as we sum from the left to the right the numbers appearing in each column. And the next step produces a sum that is less than 5 from the sum of the numbers appearing in column 5. Therefore, the answer to this question is B. The fifth column contains the largest sum. Question number 30. Let A, B, C, D, E, F, G, and H be the eight vertices of a convex octagon taken in order. Randomly choose a vertex from among C, D, E, F, G, and H and draw a line segment connecting it with vertex A. Once more, randomly choose a vertex from C, D, E, F, G, H and this time connect it to B. What is the probability that the octagon is cut into exactly three regions by these two line segments? So in each of these diagrams, we have a regular octagon. We can, without loss of generality, consider regular octagons rather than just any convex octagon. And these ten cases enumerate all the possibilities in which we obtain a cut in exactly three regions. So the probability that this can happen in general would be 10 divided by the possible number of ways we can choose our line segments. And since we are choosing at random from a set of six with replacement we have six times six ways of making a choice. And so we see that the probability for obtaining a cut in exactly three regions is 5 over 18 so the answer to this problem is D.

prime factorization

chain links

area calculation

train lengths

divisibility

digit product

octagon probability