Question number 21. The side lengths of two cubes are whole numbers which differ by one. The difference between the volumes of these cubes is equal to 217. The volume of the larger cube is equal to which of the following? Let's let X be the side length of the larger cube. then the volume of that cube is X to the third power and the volume of the smaller cube would be X minus 1 to the third power and we know that this expression here is equal to exactly 217 so on the left-hand side we have a difference of cubes which we can factor we have X minus X minus 1 times X squared plus X times X minus 1 plus X minus 1 quantity squared so here I'm using the following formula for the factorization if a is equal to X and B is equal to X minus 1 then we have that a cubed minus B cubed is equal to a minus B times the quantity a squared plus a B plus B squared and so that's what I have here in black so let's continue the parentheses here simply becomes a 1 and then we continue working in the second parenthesis we can multiply out all the terms X squared plus X squared minus X plus X squared minus 2x plus 1 and that is still equal to 217 now we have 3x squared minus 3x plus 1 and we can solve this quadratic equation we subtract 217 and divide by 3 so we obtain 0 is 3x squared minus 3x minus 216 dividing by 3 X squared minus X minus 72 3 times 72 is 216 and then we can factor 9 times 8 is 72 so X minus 9 is one factor X plus 8 is another factor and this equation has two solutions X is equal to positive 9 which makes sense in our setting and X is equal to negative 8 which we discard so here we have that X is equal to 9 positive 9 makes sense as the length of a side of the cube and so then X to the third power is equal to 729 and that is the volume of the larger cube and our answer here is B question number 22 pictured are two intersecting circles of different radii the center of the larger circle with radius R lies on the smaller circle and the points at which the two circles intersect are the endpoints of the diameter of the smaller circle what is the area of the shaded region our job here is to find the area of this loon that's what the name for the shaded region obtained in such a way and we will do that by first computing the area of a sector here that I will draw in red and the sector is made up of the following triangle and then in addition to that this shaded region here so that's what we will do first we have the sector the formula for its area is 1 half R squared times the angle of aperture and here R squared is the square of the radius of the larger circle which is right here that's the length of two sides of this triangle making this an isosceles triangle and the angle theta here is that angle right there which happens to be a 90 degree angle theta is equal to pi over 2 as is always the case whenever we have an angle constructed on a circle such that its hypotenuse is the diameter and it's two remaining sides meet on the circle so we have a 90 degree angle here and we can compute now the area of the sector as 1 half R squared pi over 2 which gives us pi R squared over 4 for the area of the sector now we can also compute the area of the triangle we know that this is a right isosceles triangle I will use the formula 1 half a b sine of theta where a b are two sides of the triangle and theta is the angle included between them so in our setup here again theta is equal to pi over 2 and a is equal to b and both are equal to the radius of the larger circle so what we have here is the area of the triangle is 1 half R squared sine of pi over 2 but that is 1 so we have 1 half R squared for the area of the triangle and now let's look at the area of the smaller circle has radius equal to let's mark that somewhere I will mark it in in blue it's half the diameter which is the hypotenuse here of the right isosceles triangle so right here I will label that also in blue as capital R and by for example taking a cosine of half of the angle theta we see that the radius capital R is equal to square root of 2 over 2 times little r which is the radius of the larger circle and we can do this for example by looking at an equation like this cosine of theta over 2 is equal to the radius R over R capital R over R like that so cosine of pi over 4 is equal to capital R over little r and another name for cosine of pi over 4 is exactly square root of 2 over 2 and that gives us the relationship that we need so the smaller circle has area equal to pi and in terms of little r root 2 over 2 times little r quantity squared which comes out to pi R squared over 2 and we take half of that number so finally for the loon as area equal to we take the area of the of half the circle of half the smaller circle 1 half pi R squared over 2 subtract from that the region here shaded in red but to obtain this region we have to first take the area of the sector and subtract from it the area of that triangle so we have here minus the area of the region shaded in red which is the area of the sector and that came out to be pi R squared over 4 from that we again subtract the area of the triangle which is 1 half R squared and we see that there is some cancellation here the first term is exactly pi R squared over 4 and then we subtract from that pi R squared over 4 then we have a minus negative 1 half R squared so the answer is indeed 1 half R squared where little r is the radius of the large circle and so that is our answer here C question number 23 a five digits natural number is called good provided that in decimal notation no two of its digits are the same and that the first digit from the left is equal to the sum of the remaining four digits how many good numbers are there we see that beginning with just trying to make an example already provides us with important information about how good numbers look like if we just have five spots here to fill with digits we see that respecting the rule that no digit appears more than once we make a minimal example 1 2 3 & 4 and run into trouble 1 plus 2 plus 3 plus 4 is equal to 10 which forces us to have a 0 always in fact so always there will be a 0 and so now we can begin counting how many good numbers there are let's make a model here for a good number again I have five spots to fill the first digit from the left will have to be the sum of the remaining digits and we know for sure that there is a zero so let's put a zero in here and let's name the remaining three digits a b and c so that as here the first digit from the left will be the sum of the other four so it will be a plus b plus c and now what we have to do is count how many different ways we in how many different ways we can create this digit s so let's begin again with a minimal example we have here 6 is equal to 1 plus 2 plus 3 and then we see that to go on we cannot increment one that would cause us to have two twice we cannot increment two but we can increment three so seven is the next possibility that's 1 plus 2 plus 4 and then continuing on 1 plus 3 plus 4 would be 8 but there is another way to do this we can also say 1 plus 2 plus 5 and that is the second and last way in which we can construct an 8 and then 9 we can do in three different ways we can say 1 plus 2 plus 6 or 1 plus 3 plus 5 or finally 2 plus 3 plus 4 so we have all together seven ways of constructing s the first digit from the left okay so s can be constructed in seven different ways so we multiply 7 by and then we permute we take all the possibilities of a b c and 0 here appearing in any order so that is going to be 7 times 4 factorial which is 7 times 24 and that comes out to 168 so that is exactly how many good numbers exist question number 24 each of the numbers X and Y is greater than 1 which quotient has the greatest value to tell these apart and in particular to order them by size we will have to rewrite the denominators and I will do that in such a way that the denominator will always begin with the number one or in other words I will factor out the first term from the denominator and when I'm finished with that we'll see why it's a nice idea to do that and the first expression if we divide out by Y and the denominator we have 1 plus 1 over Y remaining and B we obtain 1 minus 1 over Y and we see that these are easy to compare and C factoring out to Y produces here an extra factor of 2 but we have cancellation and C looks very much like the expressions in A and B so let's keep going and D we would obtain 2 X 2 Y 1 minus 1 over 2 Y and again cancel the common factors and in E 3 X 3 Y times quantity 1 plus 1 over 3 Y and again there is cancellation so now let's look at these denominators and we always have a factor of X over Y so our job here is to find the smallest possible denominator so that the quotient will have the greatest possible value so we immediately discount the answers here or the possibilities where we are adding to one because one plus a positive quantity will automatically be a smaller value of the quotient and if we subtract from one so we focus in on the expressions where we have a smaller value than one now we note that since Y is positive and in fact greater than 1 2 Y is going to be greater than 2 and reversing this expression by for example taking reciprocals we see that 1 over 2 Y is less than 1 over Y and that is then less than 1 this tells us that 1 minus 1 over Y is going to be a smaller value than 1 minus 1 over 2 Y and therefore we conclude that the largest value of the quotient has to be the value where we take X over Y and we divide it by the smaller denominator so that the value of the quotient is then increased as much as possible so B here is the answer question number 25 choose four edges of a cube in such a way that no two edges will share a vertex and how many ways can this be done so let's look at our example of a cube here and choose edges in such a way that no two will assure a vertex if I choose this edge then I must choose the parallel edge on the same face if I use a perpendicular edge belonging to the same face then I will have an edge sharing a vertex with the other edge and then we cannot use any edge belonging to any of the faces here that are adjacent to this face so we must go to the opposite face and we have two choices we either again choose the parallel edges of the opposite face like this or what we can do is choose these edges here which are again parallel but on the opposite face also and then we think about rotations of this cube so if I were to rotate the cube 90 degrees I can do that along one or two of the possible axes and imagine that the cube here is spinning about a line I will draw here roughly through the center like that so imagine first that the cube is being rotated around this axis in the following direction and so we have one configuration that works and I'm just here looking at the four edges marked in blue now this can be rotated 90 degrees and create a second configuration so for blue we have one rotation and another one which is represented as it is and what we can then do is also rotate the cube 90 degrees through the following axis like that and that will be a rotation for example in this direction and that will give us the blue parallel edges moving to the bottom face and then that's another configuration so we have three now we can sort of do the same thing if we choose to use our two blue edges here and the two green ones we can rotate about either of the pictured axes and we would obtain in that way the green edges here they can move to to this edge that's one configuration or they can move to this edge here that's a second or to this edge that's a third so in addition to the one pictured we would have four possibilities and then we can also rotate around the other axis so the two green edges here would be here and here that's another configuration or they can then switch positions and be here and here and that gives us some more possibilities in fact we would have two more and that is all together nine possibilities so the answer here after careful counting is C question number 26 how many ordered pairs of natural numbers X comma Y satisfy the equation 1 over X plus 1 over Y is equal to 1 third let's rewrite our equation so that we have Y as a function of X and then and then we'll discuss some conditions on X that are required we can say 1 over Y is 1 third minus 1 over X and then obtain a common denominator here X minus 3 divided by 3 X is equal to Y so that Y is the reciprocal 3 X divided by X minus 3 and as a natural number since Y belongs to the set of natural numbers Y has to be positive and so we need X to be at least 4 otherwise we have a negative denominator so we try what happens when X is equal to 4 then the calculation shows that that Y is equal to 12 we found therefore 4 comma 12 and also 12 comma for us clearly in our equation here X can be interchanged with Y the addition here is of course commutative so we have 412 and 12 for now the question is are there any more can we find any more pairs we have to can we find another one so let's let n be some natural number and we set n is equal to our value here 3 X over X minus 3 this equation has a solution in the natural numbers provided that n equal to X over X minus 3 has a solution so in other words what we're saying here is that after solving for Y if we divide by 3 we must also have a natural number otherwise 3 times a natural number does not produce a natural number Y if X over X minus 3 is not a natural number so now we have a simplified expression here and let's solve for X now so we solve for X and we see that n times X minus 3 is equal to X that's n X minus 3 n is equal to X and we will solve for X by bringing the terms with X to one side everything else to the other side then factoring out we have X n minus 1 is equal to 3 n so finally X has to be 3 n divided by n minus 1 and this is a natural number provided that n over n minus 1 is also a natural number so n is a natural number and n divided by n minus 1 is a natural number well let's think when that can happen this happens only this is only possible if n is equal to well we cannot use 1 how about 2 if n is equal to 2 so if n is equal to 2 what do we have X would be 3 times 2 divided by 2 minus 1 that gives us 6 and if X is 6 so is Y and that's the only solution we have been able to obtain rigorously and interchanging X with Y gives us nothing new so we have here two solutions already 412 12 comma 4 and we add our last one to the list 6 comma 6 so that gives us a total of 3 and the answer is D question number 27 for each integer n greater than or equal to 2 let the symbol here which I will pronounce brackets n denote the largest prime number not greater than 2 how many positive integers satisfy the equation brackets quantity k plus 1 plus brackets quantity k plus 2 is equal to brackets quantity 2k plus 3 so we're looking for positive integers K that fit this equation here and unless we're talking about prime number 2 each term on the left-hand side has to be an odd number and as a consequence the sum of two odd numbers has to be an even number on the right-hand side which will not be prime unless it is 2 so either the number 2 appears on the right-hand side as the sum of two odd numbers or it appears on the left-hand side well let's make a note of that 2 has to be either one of these expressions brackets quantity k plus 1 brackets quantity k plus 2 or brackets quantity 2k plus 3 but it's pretty easy to see that unless k is equal to negative 1 2k plus 3 is going to be bigger than or equal to 3 if k is equal to 0 that's what happens if k is equal to 1 we get 5 and then and then it grows faster but K has to be a positive integer so the smallest value of K here is 1 so we conclude that 2k plus 3 is not 2 in particular we know that it has to be odd and that forces one of the two remaining factors brackets quantity k plus 1 or brackets quantity k plus 2 to be the number 2 now we also note that if we consider any value of K for any positive integer K and then we ask for the greatest prime number not greater than K plus 1 adding or incrementing K by 1 will possibly increment the value of our prime number here and that greater than K plus 2 so we have this following relationship so if 2 is one of these elements it's got to be the smaller one so this tells us that 2 has to be the value of brackets when applied to K plus 1 and that forces K to be 1 if K is equal to 2 then we no longer have equality here so that is the only solution in fact K has to be 1 there is just one solution so the answer is B question number 28 in the triangle ABC displayed here and which I have enlarged we construct three smaller triangles in the following way we choose a point D belonging to the side BC and we do so at random and then connect that point to the vertex A we obtain the segment AD on which we at random choose a point E and connect it to the vertex B now in this way we also obtain nine angles labeled here 1 2 3 4 5 6 7 8 and 9 now the question is what is the smallest number of different angle measures for these nine angles that we can obtain in this way if we in addition may begin with any triangle we like so just about the only thing we can begin looking at is a relationship between these angles that add up to 180 degrees so inside of a triangle we have the sum of the angle measures equal to 180 and also over here these are supplementary angle measures so exploiting that relationship we can say that the sum of the angles 1 and 2 is indeed equal to the measure of angle 4 as these are supplementary similarly adding the measures of angles 4 and 5 gives us the supplement of angle 6 which is exactly the measure of angle 7 and then looking at the measure of angle 1 you conclude that it is strictly less than the sum of the angles the angle measures 1 and 2 which is exactly equal to the measure of angle 4 likewise that measure is strictly less than the measure of angle 7 which tells us that there are at least three different numbers here and now by means of an example we will illustrate that three is possible so in the larger diagram here I will fill in angle measures I will let angle 1 be 36 degrees I will let angle 8 also be 36 degrees angle 2 will be 36 degrees and angle 5 will be 36 degrees so therefore angle number 3 is 180 minus 72 or 108 degrees angle 4 is 72 degrees and so is angle 6 we can check that pretty easily that 72 plus 72 plus 36 is 180 so the supplement of angle 6 measures also 108 degrees and consequently angle number 9 has to measure 36 degrees so I have used three different numbers here to label these angles and we check that the sum of the angles inside of each triangle does indeed come out to 180 degrees likewise overall here we have the vertex angle C measuring 36 degrees and here we have 72 degrees 72 degrees so overall we also respect our angle measures so we have indeed shown that the best possible number here the smallest possible number of different angle measures is 3 and the answer is B question number 29 three large cube boxes were brought to a warehouse and placed on the floor as illustrated here so I have the same diagram enlarged in which we see that box C is placed against the wall but we need all boxes to be placed against the wall the problem is that they are too heavy to lift and can only be moved through a series of 90 degree rotations about any vertex as illustrated here which of the diagrams below represent possible arrangement of these boxes against the wall we note that after moving a box doesn't matter which vertex we choose as the pivoting vertex the box will be exactly in the cell adjacent to it so we have four possibilities either the box moves here here here or here a motion that would place the box B for example in this cell requires two rotations so any one rotation will place the box in a cell that essentially shares an edge with the original position so that's what we note any notation any rotation any one rotation places a box in an adjacent position and now we see that moving the box once or in general an odd number of times corresponds to a 90 degree rotation and moving the box an even number of moves corresponds to a 180-degree rotation. Now we just don't know in which direction, so let's count the number of moves. We see that box B has to end up here in this cell that I'm going to place a blue dot in, so it can get there either by moving to the left first and then up, which is two moves, it can get there by going up first and then to the left, which is still two moves, or it can go there in sort of a roundabout way. It can go to the left once, once again, that's two moves, three moves to go up, and then four moves to return. And pretty much in any way that we consider this motion of box B, it cannot directly move across its corner unless it makes an even number of moves. So that's what we note. Box B here is in this situation where it has to have an even number of moves. So it can look like it looks now, or it can be rotated 180 degrees the other way. But it cannot look like the configuration in A, and it cannot look like the configuration in C. That is just simply not possible if it's going to end up in its final resting place here, two cells to the left of where box C is resting. Similarly, box A here has to travel to the following position right next to C that I'm going to place a red dot in. And it has to get there in an odd number of moves. So one move up, another move up, and then one to the right. Or possibly one to the right, and then two moves up. But no matter how we count, it has to make an odd number of moves. So that's what's going to happen to A. A will look like a 90-degree rotation like this, or like that. But it cannot remain in its position, and it cannot remain in a position that looks like it's been placed upside down. So choice D is not possible. And that eliminates everything except B, which is indeed the correct answer here. And we can arrange these boxes in just the right way by, for example, rotating A here so that it looks like this. And then it's upright, and then finally it comes to its correct resting place like that. And similarly, we can play with box B. First, we can rotate it to the left, and then we have to take one more move. So then rotate it upright. We have the configuration and box in arrangement B. So that is the answer. Question number 30. Which of the numbers, n, from the set 1, 2, 3, 4, 5, 6, 7, all the way up to 9, have the property that it is possible to paint the unit squares in a 5-by-5 square in such a way that in each 3-by-3 square, exactly n unit squares are painted? Let's think about this property. It should be clear that if n is equal to 9, that means all squares are painted. And so the number 9 has this property. Now, the number 1 also has this property. And here, just paint the center square. This is where all 3-by-3 squares intersect. So the numbers 1 and 9 definitely have this property. So that leaves us choices D and E. Well, what happens if we paint a square with a certain configuration and then completely reverse the order in which cells are painted? So erase all the painted cells and paint all the previously unpainted cells. So what we can see, for example, is if we paint n cells, then 9 minus n cells are unpainted. And let's make up some values here. If n is equal to 1, we know everything works. If n is equal to 2, 2 cells will be painted in each 3-by-3 squares, and 7 will be unpainted. If we paint 3, then there are 6 unpainted cells. And if we paint 4, 5 cells remain unpainted. But then if we keep going, these numbers just switch positions, which is easy to see just by considering that we can always completely reverse our configuration. So here, we only have to really check that the possibilities that remain are when n is equal to 2, 3, and 4. Those are the only ones that count. So let's see. Here, in the first configuration on the left, I have in a 3-by-3 square painted just 2 items in a 3-by-3 square, so that 7 red dots are unpainted. And so that is definitely possible. We can check easily that in the remaining all 3-by-3 squares, only 2 are always painted. So, going on to the next situation, if we have 3 in any 3-by-3 square, then 1, 2, 3, 4, 5, 6 remain unpainted. And we can see that that is also the case and everything works. And finally, in the last case, if 4 dots are painted in any 3-by-3 square, then 5 are unpainted. So 1, 2, 3, 4, 5 are red and 4 are black, and the same thing here happens for all the other 3-by-3 squares. So in fact, all numbers have this property. So choice D is incomplete. We have to conclude that all numbers 1 through 9 have this property. So the answer is E.

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combinatorics

cubic equations

circle areas

5-digit numbers

prime numbers

nested triangles