Question number 11. Three regular-sided dice, that is dice on which the sum of the dots on opposite sides, always equals 7, are stacked one on top of another in such a way that the sum of the dots on two touching sides is always equal to 5. One of the visible sides of the bottom die has two dots. How many dots are there on the top side of the top die? So in the small diagram, that's the side labeled with question mark. On my enlarged diagram here, I will label the same side x, and we will have equations relating the number of dots on opposite sides. So the side that is directly opposite, the top side, has to have value equal to 7 minus x, so that together these add up to 7. I will call the side it's touching here z, and z is such that z plus 7 minus x is equal to 5. So now we have a relationship between x and z, and that is z plus 2 is equal to x. The side below z, or opposite of z, has value 7 minus z, so that we can write equivalently 7 minus x minus 2, like so, and that gives us 9 minus x. And I'll call the last here side that's touching w, and the value here of w has to be such that w plus 9 minus x is equal to 5. And so finally, we know that x is going to be w plus 4. Now what are the possible values for w? Well, it has to be a number small enough so that x does not become bigger than 6, the largest possible number we can have. So x could be 1 plus 4, could be 5 if w is equal to 1, that's certainly a possibility, or x could be 2 plus 4, which is 6, and it cannot get any bigger. But we also know that 2 here is already present on the bottom die, so we cannot have the possibility of w being equal to 2. Therefore, the only remaining option is that w is 1, and consequently x is 5. So that's our answer here, d. Question number 12. Some of the cells in the diagram to the right need to be shaded. The numbers on the bottom and on the right indicate how many cells are to be shaded in each column and each row of the diagram, respectively. And how many different ways can this be done? So looking at our diagram here, we notice that the second column and the second row do not contain any shaded cells. And so if we were to cross that out or somehow reduce this diagram, we would have the equivalent diagram here with cells that do get shaded at some point. Then starting from the bottom right corner, we can shade, for example, this cell, and the last column in this reduced diagram and the last row are completely determined now. So these four cells can only contain one shaded cell at a time. And so if we begin shading like this with the bottom right cell, then I have to finish the shading by putting two dots in here in the first column like that, so there are two shaded cells. And now the middle column is missing shading, so it has to go in here like that, so the top row also has two shaded cells. And if we continue thus, we see that, okay, now let's shade a different cell in this 2 by 2 bottom right square, for example, here. Then the shading in this 2 by 2 square is completely determined. Now I have to finish shading by putting dots here like so. And again, the middle is unshaded, so we finish like this, so the green dots are the second shading. And if we continue like so, there are four different ways here to shade the cells based on the four different shadings of the 2 by 2 bottom right-hand corner. Now, of course, we can just choose not to shade it at all. In that case, in the larger diagram, what we would see is no shadings in here, but there must still be one shaded cell in this column and one also in this column. And likewise here, the two rows have to have one shaded cell, so this is the last possibility. And so together we have five different colorings or five different ways in which this diagram can be shaded just the right way. So the answer here is D. Question number 13. If 9 to the power n plus 9 to the power n plus 9 to the power n is equal to 3 to the power 2011, then n is equal to which of the following numbers? We have to solve our equation for n, so let's begin by simplifying. We have 3 to the power 2011 equal to 9 to the power n plus itself three times, so we can write 3 times 9 to the power n. And to obtain the same base on the left-hand side as on the right-hand side, we recall that another name for 9 is 3 squared. So we have 3 to the power 1 times 3 to the power 2n with the same base that is 3 to the power 2n plus 1, and all the while still equal to 3 to the power 2011. If the bases are the same, so must be the exponents, so we solve now 2011 is equal to 2 times n plus 1, or 2010 is 2 times n. We're finally dividing by 2. We see that n has to be equal to 1005, and so that is our answer A. Question number 14. Three drivers, Michael, Fernando, and Sebastian, competed in a Formula One race. They started in the following order, Michael in front, followed by Fernando, and then Sebastian. During the race, Michael and Fernando passed each other nine times, Fernando and Sebastian passed each other 10 times, and Michael and Sebastian passed each other 11 times. Thus, the total number of times one driver passed another is 30. In what order did these three competitors finish the race? Let's begin by making a little diagram. We can have here the start line, and on the left-hand side, the right-hand side, the finish line, and we have the following order of competitors. Michael was in front, so behind him is Fernando, and then Sebastian was last. Now, if we just have two drivers, let's just call them A and B, what happens if they pass each other once? Well, in that case, A has to pass B because A is behind B, so A ends up over here. This would be one pass. If we have another pass, if A passes B, and then also after that change in position, B then goes ahead and passes A, we have two passes, and we see that the position of the drivers, the competitors, remain in the same order as initially. So, we know that Michael and Fernando, for example, passed each other nine times. So, that's an odd number of passes, and they would, in fact, switch positions. So, we have Sebastian, then Michael, and then Fernando. Then we know that Fernando and Sebastian passed each other 10 times. So, we can say that these two passed each other 10 times. So, in fact, they do not switch positions. So, we have, again, after they're done passing each other 10 times, the same order, and finally, we know that Michael and Sebastian passed each other 11 times. So, these two have 11 passes. That means they do switch positions, and we have Michael followed by Sebastian, and Fernando is closest to the finish. So, Fernando finishes first, then Sebastian, and then Michael, and that is exactly the situation here described in B. Question number 15. A cube with side length equal to 10 centimeters was covered with a certain number of identical yellow squares in such a way that all the sides look identical. So, we have this illustration here. The center yellow square seems to be at a 45-degree rotation with respect to the face that contains it, so that the corners where the other yellow squares begin cause the squares straddling the edge to do so in such a way that exactly their diagonal is on the edge. And our question here is, what area of the cube in square centimeters is covered by the yellow squares? So, since we know that the diagonal here has a special property like that, let's express the length of one of the edges being 10 centimeters in terms of diagonals, and so how many of these red lines here make up the length of the edge? And by drawing one more red line like that here through the center, we can kind of see that that's one diagonal here, and then half of one and half of one. So, two diagonals make up the length of the edge, so that's exactly 10 centimeters. So, let d be the length of the diagonal of one of these yellow squares. So, then we know that 10 centimeters is exactly two of these, which tells us that a diagonal measures 5 centimeters, and in turn, d squared over 2 equals to 25 over 2. That's the area of one yellow square. And then we should count how many of the yellow squares we actually have, and we have six faces of the cube. Each face contains exactly three yellow squares, three squares per face and then we multiply that by 25 over 2 centimeters squared per square per yellow square we should say and that comes out to 6 times 3 times 25 over 2 centimeters squared which is 9 times 25 centimeters squared and that comes out to 200 plus 25 225 centimeters squared and so the answer here is C question number 16 a sector of a disc with radius equal to 15 centimeters was cut out and rolled into a cone viewed from the side the cone has the shape of an equilateral triangle how deep is the cone in centimeters to illustrate this situation i have here a disc with radius equal to 15 centimeters we cut out a sector here out of that disc in such a way that the edge that is missing here that i have filled in on the right makes the triangle an equilateral triangle so all sides have length 15 and what we have to find is the length of this perpendicular bisector here that i have labeled d so the angle we are using is 30 degrees it's a 30 90 60 triangle with the hypotenuse equal to 15 centimeters we can use several methods of obtaining the length v i'm going to take the cosine of 30 degrees which is equal to root 3 over 2 and also as a ratio of sides that's equal to the adjacent side divided by the hypotenuse so 15 halves times root 3 centimeters is the length of that perpendicular bisector and so our answer here has to be c question number 17 in a certain month there were five mondays five tuesdays and five wednesdays in the month directly before it there were only four sundays which of the sentences below is true before evaluating the truth value of each of these five sentences let's mark exactly the days we have to consider in our month we know that to include five mondays as i have included here five tuesdays and five wednesdays we also cannot have more than 31 or less than 28 days in the month and so just with five mondays tuesdays and wednesdays here we do have already in red and closed 31 days so that is in a sense a maximal number of days and exactly enough to have the desired number of mondays tuesdays and wednesdays so we have a month here an unknown month with 31 days several months have 31 days and we're not sure which one we're looking at but before that if we enclose exactly four sundays we have exactly 28 days and going back in time one more day to enclose one more sunday in this month would put us well at 29 days which is allowed but we would have five sundays and we're only allowed four therefore in blue here we have the prior month containing exactly 28 days that has to be february and then we know that our 31 day month has to be march and what we're looking for is the beginning of april here which always has 30 days so let's enclose here in orange 30 days and that would be starting from from our thursday the following collection of days here that i am enclosing in orange that's exactly 30 so now let's evaluate our statements we have statement a claiming that in our april there will be exactly four fridays and we see that we have exactly five fridays that is a false statement statement b claims that our april will have four saturdays which is what happens so statement b is true statement c claims we have five sundays which is false and statement d claims we have five wednesdays which is also false so the only possibility here is that we have exactly four saturdays so statement b here is our answer question number 18 consider sequences of consecutive positive numbers in which each term is a three-digit number with at least one odd digit how many terms does the longest such sequence contain the matter of choosing the digits in the 100s place will not be too hard so long as we have considered the proper two-digit sequence so let's look at those two-digit numbers the proper two-digit sequence so let's look at those first let's look at two-digit sequences and if we begin with an odd number for example 10 then we can have several terms in our sequence right away all the way up to 20 so here we have 10 terms and if i start with any other odd number like 70 then the same thing happens i continue my sequence all the way up to 80 and i still have here exactly 10 terms now is it possible to extend such sequences to the right i cannot as i have already shown but i can take one step back step back and before 1 0 there is going to be a 0 9 and then a 0 8 and here i will have a 69 and a 68 so it is possible to extend such a sequence to 11 terms okay so we have 11 terms now let's look at three-digit sequences if i begin with a even digit then i have to have odd digits next so if two is first i can't have 200 i must begin with 201 so that's that's losing terms here and uh and then i pretty much have to stop after that like that so we can try several numbers here and we pretty quickly discover that it's a good idea to start with an odd number first if i start with three then the odd number i need is always assured to be there and i can continue like this for a long time until 399 and the next higher number in the hundreds place will end my sequence so i have exactly here 100 terms and so now it's just a matter of appending these the right way if i want 300 here to start my three digit sequence then i should add uh to the beginning of it an 11 term sequence for example a 279 271 270 and and so on like that except i can't really have consecutive terms in this construction so let's let's do this let's let's here put a 299 and then a 298 and have the same two digit sequence that we have considered already and that will stop 11 terms away at 289 and so that is 11 terms i have appended and altogether we have 111 terms and that is the longest sequence i can construct by appending two of the longest sequences i have considered up to this point so the answer is d 111 question number 19 paul wrote numbers into the empty cells of a 3x3 table so here we have a table some of the cells are filled in already and paul wrote numbers into the empty cells he did this in such a way that the sum of the numbers in any 2 by 2 square always equals to 10 what is the sum of the numbers which paul wrote into the table so he filled in the empty cells of the table let's label them in no particular order let's say a b c and d like so and then what we want is to evaluate the sum a plus b plus c plus d and so we will group these terms two at a time like that because these do come from two by two cells that we know have to sum to 10 so let's take a plus b from the upper left two by two square that also contains one plus two that has to be equal to 10 and we know that a plus b is therefore going to be seven now in the lower right we have c and d together with two plus three two plus three and that also sums to 10 so solving for c plus d we have c plus d is equal to five and finally adding these together we have 12 as the sum we needed to evaluate and so that is the answer here as the sum of the numbers written by paul into the table question number 20 if the ratio between positive integers a to b is equal to three to two then the ratio between the least common multiple of a and b and the greatest common divisor of a and b is equal to which of the following let's express our ratio in fraction form which is a little more our ratio in fraction form which is a little more wielding so we have a to b is equal to three to two and that's in lowest terms and so that tells us that for some number x we have a is equal to three x and b is equal to two x so that dividing three x by two x we obtain exactly three halves a fraction already on lowest terms hence we can find the least common multiple hence six x is the least common multiple of two x which is b and three x which is another name for a now what is the greatest common divisor since the ratio is in lowest terms we have exactly x here as the greatest common divisor x is the greatest common divisor of two x and three x since we cannot really simplify the fractions since three halves is in lowest terms hence the ratio we're looking for is what is six x two x and that is six or we can say one to six but the answer either way is six e

math problems

stacked dice

sequence shading

geometrical shapes

algebraic manipulation

problem-solving