Problem number 21 states. From the following numbers, 3, 5, 2, 6, 1, 4, and 7, Masha chose 3 numbers with a sum of 8. From the same list, Dasha chose 3 numbers with a sum of 7. How many of the numbers chosen by both girls are the same? Again, just to repeat, the numbers are 3, 5, 2, 6, 1, 4, and 7. So Dasha chose 3 numbers to make a sum of 7. The only way to do this is if she chose the numbers 4, 1, and 2. And Masha chose 3 numbers to make a sum of 8. The only way to do this is with 4, 1, and 3, or with the numbers 5, 1, and 2. If Masha chose the numbers 4, 1, and 3, and Dasha chose the numbers 4, 1, and 2, there would be two similarities, 4 and 1. If Masha, however, chose numbers 5, 1, and 2, and Dasha chose numbers 4, 1, and 2, there would also be two similarities, 1 and 2. Regardless of how Masha chose to make 8, either with 4, 1, 3, or with 5, 1, 2, there would be two similarities. The question asked us, how many of the numbers chosen by both girls are the same? The answer is 2, letter C. Problem number 22 states, 5 balls, A, B, C, D, and E, weigh 30 grams, 50 grams, 50 grams, 50 grams, and 80 grams each, not necessarily in that order. Which ball weighs 30 grams? First, let's take a look at the scale on the right. From the scale on the right, we know that the weight of ball D plus the weight of ball A equals the weight of ball B plus the weight of ball E plus the weight of ball C. Here are the possibilities for the weights of the balls. The only way to make this equation true is 80 plus 50 equals 30 plus 50 plus 50. There is no other way. This means that D and A either weigh 80 or 50, B, E, and C either weigh 30 or 50. It is important to keep in mind that only one of D or A can weigh 80, and only one can weigh 50, and only one ball from B, E, and C can weigh 30, and the remaining two have to weigh 50. Let's look at the other two scales, and let's see what information we can gather from them. First, let's assume that A equals 80 grams. This means that at least one of these two sides has to weigh 80 grams plus 50 grams, or 130 grams. We get the 80 grams since there is an A present on both sides, and we get 50 grams from the fact that we know that two of the three balls of B, E, and C have to weigh 50 grams. This might be a little bit confusing, so let me show you some specific examples. If B weighs 30 grams, and both E and C weigh 50 grams, that means that this tray weighs 130 grams. If B and C weigh 50 grams, and E weighs 30 grams, that means that both of these trays weigh 130 grams. If C weighs 30 grams, and both B and E weigh 50 grams, that means that this tray weighs 130 grams. So, no matter which ball weighs 30 grams, at least one of the trays will weigh 130 grams. If the tray on the right weighs 130 grams, that means that the maximum weight of the tray on the same scale is 100 grams. If the scale on the left weighs 130 grams, the maximum weight of the tray on the same scale is only 80 grams. Obviously 100 grams and 80 grams are not greater than 130 grams. This means that our original assumption that A equals 80 grams is wrong. This of course means that A is equal to 50 grams, and if A is equal to 50 grams, that means that D equals 80 grams. The scale on the left is telling us that the weights of balls E and B together is greater than the weights of balls C and E. We already know the weight of ball A, so let's plug in 50 grams for A. The only way to make this inequality true is to say that E and B are 50 grams, and C is 30 grams. There is no other way. This means of course that B is 50 grams, E is 50 grams, and C is 30 grams. We have solved for all the weights of our balls. The question asked us, which ball weighs 30 grams? The answer is ball C. Letter C. Problem number 23 states, A, B, and C represent three different digits. The greatest possible six digit number that uses digit A three times, digit B two times, and digit C just once, will never be written in the form. In order to solve this problem, let's go over all the possibilities of the digits. Which digits are greater than others? So for example, one possibility is that digit A is greater than digit B, which is greater than digit C. Another possibility is digit A is greater than digit C, which is greater than digit B. Another possibility is digit B is greater than digit A, which is greater than digit C. The next possibility is digit B is greater than digit C, which is greater than digit A. The next possibility is digit C is greater than digit A, which is greater than digit B. And finally, the last possibility is digit C is greater than digit B, which is greater than digit A. If digit A is greater than digit B, which is greater than digit C, then the largest possible number we can make with using digit A three times, digit B twice, and digit C only once, is A, A, A, B, B, C. This may be hard to grasp, so let's see a concretion. If digit A is 9, digit B is 4, and digit C is 2, then we can use 9 three times, 4 twice, and 2 once. The largest possible number we can make with these numbers is 999442. If A is greater than C, which is greater than B, the largest number we can make is A, A, A, C, B, B. If B is greater than A, which is greater than C, then the largest number we can make is B, B, A, A, A, C. If B is greater than C, which is greater than A, then the largest number we can make is B, B, C, A, A, A. If C is greater than A, which is greater than B, then the largest number we can make is C, A, A, A, B, B. If C is greater than B, which is greater than A, then the largest number we can make is C, B, B, A, A, A. The question asked us to find which of these forms is not a possible largest number. Letter A is a possible number if A is greater than B, which is greater than C, so A cannot be the answer. B is also a possible largest number if C is greater than A, which is greater than B, so B can also not be the correct answer. Letter C can also be the largest number if B is greater than A, which is greater than C, so it can also not be the correct answer. And finally, letter E can also be the largest number if A is greater than C, which is greater than B, which means that E also cannot be the correct answer. This just leaves us with letter D. So the question asked us, the greatest possible six digit number that uses digit A three times, digit B two times, and digit C just once will never be written in the form A, A, A, B, C, B. Letter D. Problem number 24 states, the sum of the ages of Kate and her mother is 36, and the sum of the ages of her mother and her granny is 81. How old was her granny when Kate was born? In order to solve this problem, we need to figure out the difference in the ages between Kate's granny and Kate. So in other words, we need to find Kate's granny's age minus Kate's age. In order to do this, let's first rearrange the equations. The equations are still the same, Kate's granny plus Kate's mom equals 81, and Kate's mom plus Kate equals 36, they're just rearranged. Now, in order to get Kate's granny's age minus Kate's age, we need to multiply the bottom equation by negative one. So if we distribute the negative, we get negative Kate's mom's age minus Kate's age equals negative 36. Now we just need to combine the equations. Kate's granny's age plus zero is still Kate's granny's age, positive Kate's mom's age minus Kate's mom's age cancels each other out to make zero. Zero plus negative Kate's age is still negative Kate's age, and 81 minus 36 is 45. This means that the difference between Kate's granny's age and Kate's age is 45 years. This means that when Kate was born, Kate's granny was 45 years old. The question asked us, how old was her granny when Kate was born? The answer is 45 years old, letter C. Problem number 25 states, Nick wants to arrange the numbers 2, 3, 4, 5, 6, 7, 8, 9, and 10 into several groups so that the sum of the numbers in each group is the same. What is the largest number of groups he can make? First, let's determine what the sum is going to be of all the groups. The sum of all the numbers is 54. 54 is divisible by 2, 3, 6, 9, 18, and 27. The numbers that 54 is divisible by are possible sums of the groups. It is impossible for the sum to be smaller than the largest number in the numbers. For example, it would be impossible for us to make a 9 with a 10, so therefore 2, 3, 6, and 9 cannot be our sums. For our sum, let's choose 18, because if we choose 18 over 27, then we can have more groups. So, our first group is going to have 10 and 8, which adds up to 18. Our second group can have 9, 7, and 2, which also adds up to 18. And our third group will have the remaining four numbers, 3, 4, 5, and 6, which also add up to 18. So the question asked us, what is the largest number of groups he can make? The answer is three groups. Letter B. Problem number 26 states, Peter cut a long board that was 8 cm wide into 9 pieces. One piece was a square and the rest were rectangles. Then he put all the pieces together as shown in the picture. How long was the board? First, let's say that one width of the board is considered one unit. This means that the whole arrangement of the boards is 5 units by 5 units. Which means that the total area is 25 units squared. The only way to make a board 25 units square with whole numbers is either a 5 by 5 board or a 1 by 25 board. Since the picture shows a 5 unit by 5 unit arrangement, this is impossible, which means that the original board was 1 unit by 25 units. If we arrange all the pieces end to end, we get that the board is 25 units long and 1 unit wide. The question told us the long board was 8 cm wide, which means that 1 unit equals 8 cm. In order to determine how long the board is, we just need to convert our units. We start off with 25 units, and we know that in each unit there is 8 cm, so we multiply 25 units by 8 cm over 1 unit to get 200 cm. This means that our board is 200 cm long. So the question asked us, how long was the board? The answer is 200 cm, letter D. Problem number 27 states, write 0 or 1 in each cell of the 5 by 5 table so that each 2 by 2 square of the 5 by 5 table contains exactly 3 equal numbers. What is the largest possible sum of all the numbers in the table? Before we tackle on this problem, let's solve a simpler problem. Let's only fill in a 3 by 3 table. Since we want to maximize the sum of the values, let's initially set all of the values to 1. The question stated that each 2 by 2 square has to contain exactly 3 equal numbers, and as you can see, in this 2 by 2 square, there are 4 equal numbers. In order to solve this problem, let's set the middle value to 0. This way, wherever we place our 2 by 2 square, there will always be 3 exactly the same numbers. Now let's tackle on a bigger grid. Let's solve a 3 by 5 grid. Again, let's fill in all the empty grid cells with 1s, because we want to maximize the sum of all the numbers in the table. And we can see here that we have a similar situation, a 3 by 3 square of all 1s. In order to solve this, let's set the middle value equal to 0. Now, all the 2 by 2 cells in the grid have exactly 3 equal numbers. Now, let's solve the actual problem, and let's solve a 5 by 5 table. Again, we will fill in all the empty squares with 1s, since we want to maximize the sum of the numbers in the table. Again, we can see a 3 by 3 square of 1s, so let's put a 0 inside of there, and there is another 3 by 3 square of 1s, so let's put another 0 in the middle of that square. Now, we have solved our 5 by 5 square. Before we finish the problem, let's verify that every 2 by 2 square has exactly 3 equal numbers. It looks like it does, so let's now count up the sum of all the numbers. We have 21 1s and 4 0s, which means that our total sum is 21. The question asked us, what is the largest possible sum of all the numbers in the table? The answer is 21, letter B. Problem number 28 states, 14 people are seated at a round table. Each person is either a liar or tells the truth. Everybody says, both my neighbors are liars. What is the maximum number of liars at the table? During this problem, we will represent circles with red Ls as liars and circles with green Ts as truth tellers. So let's look at some common cases that we will run into. If we know that somebody is a truth teller, and they are saying that both my neighbors are liars, since they are telling the truth, both of their neighbors are actually liars. However, if the person is a liar, and they are saying that both of their neighbors are liars, then there are two possibilities. Either both of their neighbors are truth tellers, or one of their neighbors is a liar and the other is a truth teller. So let's recap our rules. All of our truth tellers must be surrounded by two liars, and our liars must be surrounded either by two truth tellers, or one liar and one truth teller. So let's put our rules to the side and bring up our table. We need to start from somewhere, so we'll assume that the person at the top of the table is a truth teller. Rule 1 states that a truth teller must be surrounded by two liars, so therefore, both of the truth teller's neighbors must be liars. Since we want to maximize the number of liars at the table, we're going to utilize Rule 3 and say that the next people are also liars. We cannot have three liars in a row, so the next people automatically have to be truth tellers. All truth tellers are surrounded by liars, so the next people are also liars. Using the same logic we utilized earlier, we will also say that the next people are liars. Again, we cannot have three liars in a row, so the next people are going to have to be truth tellers. And finally, we're going to utilize Rule 2 and say that the final person is going to be a liar. There are nine liars sitting at the table. The question asked, what is the maximum number of liars at the table? The answer is 9, letter C. Problem number 29 states, there are eight domino tiles on the table. One half of one tile is covered. The eight tiles can be arranged into a 4x4 square, so that the number of dots in each row and column is the same. How many dots are on the covered part? So the question told us that the eight dominoes can be arranged so that the number of dots in each row and column is the same. There are four rows and four columns, which means that the total number of dots on all the dominoes must be divisible by four. If we add up all the dots on our dominoes, we get a sum of 37 plus the unknown value. The unknown value can be any number between 0 and 6. Of all the possible sums, only one is divisible by 4, 40. So therefore, the missing number must be 3. In order to solve this problem, we do not have to put any dominoes into the grid. But in case you are curious as to how they can be put together, they can be put together in this fashion. But anyway, the question asks us, how many dots are on the covered part? The answer is 3, letter C. Problem number 30 states, write the numbers 3, 4, 5, 6, 7, 8, and 9 in the seven circles to obtain equal sums along each of the three straight lines. What is the sum of all the possible numbers that can be placed in the circle marked with the question mark? In order to simplify this problem, we can ask a different question. We can ask, which of the seven numbers can we remove and have the remaining six numbers pair into three equal sums? In order to solve this problem, let's first find the sum of all of our numbers. The sum is 42. Since 42 is divisible by 3, only removing multiples of 3 from it will yield numbers divisible by 3. So for example, if we take out the 3, our sum is now 39. If we took out the 6, our sum would be 36. And if we took out the 9, our sum would be 33. All of these numbers are divisible by 3. It is good to have numbers divisible by 3 because we have three straight lines, and we want equal sums along all of those three lines. Now all we have to do is verify that 3, 6, and 9 can be placed in the middle, and all three lines will have equal sums. So let's put 3 in the middle first. If we surround 3 with 4, 5, 6, 9, 8, and 7 in that order, we get a sum of 16 along each line. This means that 3 is a possible number in the middle. If we now put 6 in the middle, and surround 6 with 3, 4, 5, 9, 8, and 7, along each line we get a sum of 18. So 6 is also a possible number. If we put 9 in the middle, and surround it with 3, 4, 5, 8, 7, and 6, we also get an equal sum along each line, 20. This means that 9 is also a possibility. Our only possibilities are 3, 6, and 9, and the question asked us, what is the sum of all the possibilities? So 3 plus 6 is 9, plus 9 is 18. So the question asked us, what is the sum of all the possible numbers that can be placed in the circle marked with a question mark? The answer is 18, letter E.

mathematical problems

number sequences

algebraic manipulation

deductive reasoning

optimization

logic puzzles