Hello, and welcome to the Math Kangaroo Media Library. You are about to view interactive solutions to levels 11 and 12 of the 2014 competition. You will likely notice that some of the solutions are slightly different from the suggested solutions you may have already reviewed. So as you follow along, compare your own solutions and the suggested solutions with the current presentation and please make sure you understand the differences. If you have any questions or comments, feel free to contact me at the address provided on the bottom of the screen. My name is Luke, and I'm a past Math Kangaroo participant, and I hope you will find this presentation useful in preparing for your next Math Kangaroo competition. Question number one. If you take a certain number of 1x1x1 cubes out of a 5x5x5 cube, you end up with a solid figure consisting of columns of the same height, which stand on the same ground plate, as shown here. How many small cubes were taken out? We see that each of the remaining columns is four cubes stacked on top of one another, and the spaces between remaining columns are exactly of that same capacity. You could stack four cubes here to fill up one of those empty spaces. So we see that there are two kinds of rows here in this remaining figure. We have the rows that contain three standing columns. For example, these three contain three standing columns, and so the spaces, the missing spaces in between these rows contain two sort of missing columns each. So here we would have, let me write that down as three times two missing columns. And then in between those, we have just empty, empty rows. We have exactly here one and two of those. So two rows contain no standing columns, so they therefore can be filled in with five times two missing columns. So we have ten plus six columns have been removed, and then we multiply that by four cubes per column. And so that gives us a total of sixteen times four, or sixty-four cubes have been removed to make the figure here in the diagram. So that answer is choice C, sixty-four. Question number two. Today is Carla's, Emily's, and Lilia's birthday. The sum of their ages is now forty-four. What will the sum of their ages be the next time it is a two-digit number with two equal digits? Let's begin by writing down the information we know as of today. So today we see that the sum of their ages is, let me use C for Carla's age, E for Emily's age, and L for Lilia's age, and that is forty-four years. And then sometime in the future, so let's say X years from now, we see that each woman will be X years older, so we have C plus X as Carla's age, E plus X as Emily's age, and L plus X as Lilia's age, and that is C plus E plus L plus three times X, and that gives us forty-four from above plus three times X, and we want this to be equal to a multiple of eleven, so that the sum of their ages will be a multiple of eleven, in other words, a two-digit number with two equal digits. So then we have this equation here, eleven K is equal to forty-four plus three X, and this tells us that, so three X is divisible by eleven. In other words, we can say, so X is a multiple of eleven, X would have to be divisible by eleven, let's call it eleven times Y. And then our equation reduces to eleven K is equal to forty-four plus three times eleven Y, or we can divide by eleven everywhere and look at four plus three Y. And then we can summarize this information in a table. We can have here the values of K versus the values of four plus three Y, and these have to be integers. So on the left column I will have the value of K, and then in the right column the value of four plus three Y, and then let's make one more column and answer the question here is Y is an integer? And that's going to be a question over here. So let's say, well, okay, the smallest value of K we can have is five, so that would be answer A, fifty-five. So if K is five, then Y is definitely not an integer. So we would cross that out and say no, that's not an integer, or let me just undo that and write in no. And if K is equal to six, that would be answer B, and again Y would not be an integer. If K is equal to seven, then Y would be one. So that's good. If K is eight or nine, again, Y would not be an integer. And Y has to be an integer. So we choose the only remaining possibility, which is when K is seven, the sum of their ages is going to be seventy-seven years, and that is here answer C, seventy-seven. Question number three. If A raised to the power of B is equal to one-half, what is the value of A raised to the power of negative three B? First, let's review some rules of exponents. If we have a base M and we raise that to the power B, then raising again to a power K, we have to multiply the exponents. So the exponent is now K times B. So in our problem, first we let M be equal to A, and we set K equal to negative three. So then we have the calculation here, A to the power negative three B, and we work from right to left, is going to be equal to A to the power of B, and then raised to the negative three. Now we know that A to the power of B is one-half, so we raise one-half to the power of negative three, and then we can rewrite this again as one-half first raised to the power of negative one, and then raised to the power of three. So that would be saying, in our example, that M is equal to one-half, that K is equal to three, and B is equal to negative one. So we continue, and one-half raised to the power of negative one is two, and then the three remains. So finally, the answer comes out to positive eight, so that is choice B. Question number four. There are 48 balls placed into three baskets of different sizes. The smallest and the largest basket together contain twice as many balls as the middle basket contains. Okay, so before going on, before the information gets too confusing, let's just write down what we know. We see that 48 is the total number of balls in all three baskets of three sizes. So the sizes will be S for small, M for middle, and L for large. And the second sentence then tells us that the smallest and the largest basket, so if we say S plus L, that contains twice as many balls as the middle basket contains. So let's keep going. Now from the third sentence, we see that the smallest basket contains half as many balls as the middle basket contains. So we can keep writing down equations. is exactly equal to half the number of balls as in the middle basket. And then our job here is to find out how many balls are there in the largest basket, so we have to solve the system of equations for L. So I will begin here by working with these two equations, and if we subtract from the top equation the bottom equation, we see that S plus L minus S is going to be equal to 2M minus one half M, or in other words, L is equal to three halves M. Now we have a relationship between the largest and the middle basket, and we have a relationship here between the smallest and the medium basket. So we can use this together to rewrite the top equation. We see that 48 is going to be equal to S, which is one half of M, plus M, plus L, which is three halves of M. And that is all together equal to four halves, so 2 plus M, 3M. And so we have 3M is equal to 48, or dividing by three, the middle basket contains exactly 16 balls. And if the middle basket contains 16 balls, we see that we can use that information here in our bottom equation. L is equal to three halves times 16, which comes out to be equal to exactly 24. So the largest basket contains exactly 24 balls. And that is our answer here. That comes out to be answer C, 24. Question number five. What is this expression equal to? We have in the numerator 2 raised to the power 2014, subtract 2 raised to the power 2013. And in the denominator, we have 2 raised to the power 2013 minus 2 raised to the power 2012. So let's begin by taking out some common factors. We can rewrite this expression in the following way. We can say that 2 to the power 2014 minus 2 to the power 2013 is equal to 2 to the power 2013 times 2, that would be decreasing the exponent by 1, minus 2 to the power 2013. And then we can factor out the 2 to the power 2013. What remains is 2 minus 1. OK? And so that was the numerator. Now in the denominator, we follow the same procedure. We have 2 to the power 2013 minus 2 to the power 2012. So I will decrease here the exponent by 1 and write, instead of 2 to the power 2013, 2 times 2 to the power 2012, subtract 2 to the 2012, and then factor. What remains is 2 minus 1 again. So then if we divide, what we have is the expression equals 2 to the 2013 in the numerator and divide that by 2 to the power 2012 that's the denominator and again we can factor out a power of 2 so that's 2 times 2 to the 2012 up top on the bottom 2 to the 2012 which cancel and so the answer is 2 that would be answer e question number 6 which of these expressions does not contain b plus 1 as a factor well let's just begin by factoring each of these expressions and see what happens the first expression we see is to be plus 2 each term has a common factor of 2 so let's factor that out and we have a b plus 1 so a does not work it does have a factor of b plus 1 let's keep going b squared minus 1 factors as a difference of squares b minus 1 times b plus 1 and so again we see a factor of b plus 1 so b does not work and see we have b squared plus b each term has a common factor of b and if we take it out we have b plus 1 times b and again a factor of b appears in d we have negative 1 minus b each expression has a where each term has a common factor of minus 1 so if we remove that we do end up looking at a factor of 1 plus b or b plus 1 and so by the process of elimination we should be looking at answer e and just to make sure b squared plus 1 factors as b plus the square root of negative 1 times b minus the square root of negative 1 and that requires the knowledge of non-real numbers and does not factor with b plus 1 as a factor so there we go that is the choice we need to mark as our answer e question number seven how many digits long is the result of the multiplication of 2 to the power 22nd raised to the power 5 times 5 to the power 55 raised to the power 2 let's see if we can simplify this expression so we have first of all 2 to the 22nd power raised to the power 5 that's the first factor and we can multiply the exponents 22 times 5 is 110 then we have a second factor here of 5 to the 55th power that being raised to the second power so we multiply our exponents and we have 5 to the power 110 now we recall that if we have two different bases but they're raised to the same exponent then we can distribute the exponents to each of the bases like this and that's what we're going to do but backwards so here m is going to be equal to 110 a is 2 b is 5 so the product the product that we have equals 2 times 5 a times b raised to the power 110 which gives us 10 raised to the power 110 and now we can count how many digits we will have we have here a 10 so there is going to be so this is a 1 followed by we're going to have many zeros and how many are we going to have there are going to be exactly as many zeros as the integer and the exponent so followed by 110 zeros so exactly if we add up this one one with 110 zeros counting the digits we have that there are going to be 111 digits in the product so answer e question number eight handsome Harry has a secret email account that only four friends know today he received eight emails in that account which of the following is certainly true since we know nothing about how the emails were received we can suppose or we can ask the question what if what if just one friend sent all eight what then then we could look at the choices and we say well choice a is not possible or it could have happened some other way if just one friend sent all eight emails then we have no idea if he actually received two emails from each of his friends we also right away discredit possibility be that Harry cannot have received eight emails from one of his friends that's exactly what we supposed then C says that Harry received at least one email from each friend well if all eight came from just one friend and C also has to be discounted then we look at D and we say is it true that Harry received at least two emails from one of his friends well our possibility that one friend sent eight does not automatically discredit choice D so let's leave it alone for now let's put a question mark here and let's go on to E which states that Harry received at least two emails from two different friends so our proposition that eight emails came from one friend also discredits E and the only possibility that remains is that Harry did receive at least two emails from one of his friends that certainly has to be true so that by the process of elimination and our example here should be the right answer now we have to think a little bit more to really justify that this is the correct answer we should negate that proposition and if we negate that proposition then it should follow that today Harry did not receive eight emails so negating D we see that each friend did not send at least two emails so each friend sent exactly one email and so for a total of four emails received by Harry today and this this is a contradiction because Harry received eight emails instead of four so now we are sure that the statements here are compatible and D is indeed the correct answer number nine two identical cylinders are cut open along the dotted lines and then glued back together to form one bigger cylinder as shown here in the figure what can you say about the volume of the big cylinder compared to the volume of one small cylinder so first let's compute some volumes we see that on the small cylinder we have a height equal to the length of this dotted line so that would be H and we have a radius here for the circular base so let me call that little R and then the volume would be the area of the base I R squared for a circle and then times the height the same thing would happen on the bigger cylinder the radius is now capital R it's larger the height is exactly the same age so it's volume here is again the area of the base by capital R squared times the height and we need a relationship between lowercase and uppercase R and we will obtain that by considering the circumference here the circumference of the base so that blue circle here on the left that is equal to 2 pi times the radius so 2 pi R and if we cut one of these cylinders open glue it to the other cylinder that has been also cut open we obtain twice the circumference for the large cylinder its base is also circular and it will be exactly twice the distance around so let me try to draw that in here so that's the that's the circle and its circumference here C is equal to 2 pi R from the first cylinder plus 2 pi R from the second cylinder so that gives us 4 pi R and if we factor out the pi and divide by 2 we see that this tells us that capital R is exactly 2 times lowercase R and now we can compute what the volume is so the volume of the big cylinder is pi we have to square 2 times little R times height and after doing that we see we have a 4 we have a pi R squared and the height so the factor of 4 here appears and comparing the volume to the volume of the smaller cylinder we conclude that the big cylinder has exactly 4 times the volume as one of the small cylinders and so the answer here is D. Question number 10. In the number 2014 all the digits are different and the last digit is greater than the sum of the other three digits. How many years ago did this occur the last time? We begin by counting back in time and for each year we encounter checking the condition on the digits so first we go back to 2013 and here we notice that 3 is equal to 1 plus 2 and so the condition here on the last digit is not satisfied we have to go back further in time but 2012, 2011 will not work for the same reason 2010 and any number with two zeros here repeating in that range will break the condition on all digits being different so we go back further in time start with a 1, 1999 breaks the condition on the different digits that we need and in fact anything here starting with a 1 and a 9 breaks the condition on the last digit the sum of the digits would already be too large so we have to start with a 18 in the 1800s we would have two digits here and again the sum of the digits already is a is a 9 so like in 2013 at most we can make the last digit equal here the sum of the digits if we choose for example a 1809 and 9 is equal to 1 plus 8 so an 18 in the first position here will be too large we have to go back in time all the way up to 1700s and then the number will look like 1700 XY ok and then what we need here is with 17 followed by an X followed by a Y the Y will have to be greater than 1 plus 7 plus X so that is equal to 8 plus X and then we start listing the possibilities so if X is equal to 0 we have Y greater than 8 and if X is equal to 1 we have Y greater than 9 which is which is not possible so we have to choose X equal to 0 and in that case Y would be equal to 9 so that's exactly what we do in this case we have one seven zero nine and that is the most recent year that had the first three digits the sum smaller than the last digit and all digits are different so this occurred this was we subtract from 2014 1709 which gives us 305 years ago and that is choice C 305

Math Kangaroo

interactive solutions

mathematical concepts

exponents

factorization

volume calculations

polynomial expressions