Question number 21. There are 2015 marbles in the box. The marbles are numbered from 1 to 2015. Marbles with equal sums of their digits are the same color, and marbles with different sums of their digits are different colors. How many different colors of marbles are there in the box? So with the 2015 marbles we have, with face values being the first 2015 positive integers, we are asked to consider all the different face values as separate numbers with possibly distinct digit sums. So how many digit sums are we going to have that are distinct? That's what we have to count here. Count distinct digit sums. And each such distinct sum will have to require its own color. So we can begin with a list of 1 as the face value, 2, 3, 4, 5, all the way up to 9. And at least here we'll have 9 separate categories of colors. I won't try to come up with 9 different names of colors. I'll just label them as colors 1, 2, 3, all the way up to 9, and then possibly we'll have more if required. And if we keep going and look at marble number 10, 10 has a digit sum of 1. So marble number 10 will fall into here the same category as number 1, then 11, 12, 13, 14, and so on. And marble 18 will fall here in the same category in the same color as number 9. But the next marble, 19, requires its own color. So that's the 10th color. And we can continue with marble number 20. Marble number 20 will fall into the digit sum of 2 categories, 20, 21, 22, 23, all the way up to 27, the digit sum of 9. Marble number 28 falls into the same category as marble 19, and then 29 here will require the 11th color. Like so. Okay. So let's keep going. Let's not bother with 30 through 90, but let's look at the last value here. 90 will have the same digit sum as 9. So 90 falls into here, then we have the 91, 92, all the way up to 99. And 99 has a digit sum of 18, so we need at least an 18th color. But we see that all the colors in between 11 and 18 are used up, really, on some marble. So the 12th color would, for example, be the marble with number 39, because 3 plus 9 is 12, and so on. So we use up all 18 colors, but we're not done yet. We keep going. Marble 100 has a digit sum of 1, so that falls into the first category. Then 101, 102, 103, 104 goes here. 108, and so forth, until the largest number we can obtain with 3 digits is 999, somewhere later along the way. So that has a digit sum of 27, and we need a 27th color here to mark that number on the marble. And then we continue with 1000 having a digit sum of 1, 1001, 1002, and so forth. 1003, 1004, and then 1008 here has the same digit sum as 9. And then, finally, the largest number we can make here within our range is 1999, and that requires a 28th color. Okay, so so far we have 28 colors. And are we done? Well, let's keep going. We only have a few more digits left. So 2000 falls in the second column here with a digit sum of 2. Then we have 2001, 2002, 2003, 2007 has the digit sum of 9. And then 2015 will need a 17th color, which we already use. We already have a 17th color somewhere earlier on in the list. But the 28th color is required to color number 1999 as a marble. So we do need 28 colors. We have shown here that the middle colors from 1 to 28 all get used up. So indeed, 28 different colors are required. And that, in our choices, is C. We need 28 colors for the 28 different digit sums. Question number 22. When reading the following statements from the left to the right, what is the first statement that is true? So here we have five statements. And we would begin by considering statement A on the very left, which references statement C and claims that C is a true statement. Okay, so first we need to know the truth value of statement C. Moving on to C, we see that it references statement E. So we need to know its truth value. And statement E is an equation here, which is 1 plus 1 is equal to 2. Definitely a true statement. So here we assign the truth value of true to statement E and then backtrack to all the statements that reference E. So back to C, we have a claim that statement E is false. Well, that is a false claim. So the truth value of C is false. C is referenced in A, which claims that C is a true statement. And that claim is false. So A has a truth value of false. Moving on to B, we see that the claim here is that A is true, and that is not the case. So that is another false statement. And the last one here we need to decide on is D, which claims that D is a false statement. And that is indeed the case. So this true value is true. And so looking from the left to the right, the fourth statement is the first one with the truth value of true. And that would be statement D. And it is the answer to this question. Question number 23. In how many ways can we color the edges of a cube, A, B, C, D, E, F, G, using four distinct colors, if each of the faces needs to have one edge of each of the four colors? Okay, so the colors I will use are going to be red, orange, blue, and green. And I already constructed a cube, so we don't have to spend time making one. Here is my cube. And I'll begin by coloring one of the faces. Let's look at the bottom face. And I'll use red first, just to have a starting point. Edge A, D is going to be red. For orange, I'll choose C, D. And then blue, I'll use A, B for the blue color. And B, C is going to be green. So here we have a coloring of one face. And in how many ways can we actually accomplish this? So that bottom face can be colored in 24 different ways. And that is four choices for the first edge, then three choices for the second edge. Once two edges are colored, the third edge only has two remaining colors available. And finally, the last edge is stuck with the remaining colors. So that's four times three times two times one, or 24 ways to color a face. Now let's look at edge BF over here. And BF adjoins a green edge and a blue edge, so we cannot use those two colors again. But we are allowed to use red, or we can use orange. So let's first use red for BF, like that. And then what I'll do, I guess I should make another version of my cube. So here's another copy. And the second, eventuality, I'll use orange here for the edge BF. So just to make everything the same, let me take a second here to color the same edges in the same colors, like that. Okay, and then let's keep coloring. So in the left cube, let's look at my edge CG. Now, CG cannot be orange, it cannot be green, and it cannot be red. So the only possibility here is that we use blue. So let's make CG blue, like that. And now I have just one color remaining for the edge FG. It has to be orange. I'll use orange over here. Now let's move on to the front face. On the front face, I have an edge AE. AE cannot be red, it cannot be blue. Can it be green? Yes, it can be green. And can it be orange? No, it cannot be orange. Because if it were, we would kind of run into trouble. So let's make it green. Like so. If that is green, then the remaining color I have to choose for the edge EF is orange, and here we actually do run into trouble. So I misspoke last time, we have to reverse these colors. Orange has to be on AE, and green has to be on EF. Okay, so that seems to be okay so far. Let's keep going. Let's move on to the edge BH right there in the back. It's still not colored, it cannot be red, it cannot be orange. What can it be? It cannot be blue, because it would then conflict with CG. So the only thing that it can be is green. So let's put green over here, like that. And then that probably finishes our color. So the two edges are HG, which cannot be orange. It cannot be green, so we have a choice of blue and red. While it cannot be blue, it has to be red. So we have a red over here, and blue is going to be BH, like that. So here is our cube, and it's completely colored. And that's really just the only possibility. At each point, we didn't have any ambiguities. So that's one complete color. Now we have a second cube left over here, and we'll see if we can color it in more than one way. So let's keep going. Let's look at the edge over here, CG. CG is in the back. It cannot be orange, obviously. We're going to use that. It cannot be green. It can either be blue, or it can be red. So let's try red. Let's see what happens if we put red over here. If we have red on CG, then the only remaining color is blue on FG. So that doesn't seem to cause any trouble yet. Let's keep going. Let's move on to the front face and look at edge AE. It cannot be blue. It cannot be red. It cannot be orange. So that means it has to be green. If this is green, are we making the correct choices? If that is green, then the last edge here has to be red. And we really have no choice about it. So we have red over here like that. OK, so far so good. Let's move on to DH in the back. DH here borders red and orange. It cannot be green or blue. Well, it cannot be green because that color is already included in our face so we have to be blue and if we have blue in the back there what remains for the top face we have red and we have blue already so we can choose orange or green orange cannot be hg because dc is already orange so orange would have to belong to the edge h and if i have the h being orange then the only remaining color for the top face there is green on edge hg and i have colored it and i didn't have any choices at any point so those are the only two possible colorings once we decide on a way one of 24 ways to color the bottom face okay so that is the key over here once the bottom face is colored there are only two ways to finish the coloring as we have shown so we have 24 possibilities for a coloring of one face and once we color that one face we chose bottom of this example but it could be any one then that's it we have just two more possible colorings remaining and therefore the answer there is 3 times 24 that would be choice a question number 24 how many regular polygons exist such that the measure of each one of their interior angles in degrees is an integer we are talking about measures of interior angles of regular polygons at our disposal we should have the formula for the angle sum of all the interior angles in a regular polygon so let's recall that formula and let's write that a regular polygon with n sides that polygon will have an interior angle sum given by 180 times n minus 2 and that is in degrees so here we have an important piece of information that we can glean from the formula it should also be quite obvious that here n has to be greater than 2 otherwise we don't have a polygon and that angle sum comes out to be a negative number okay so we can keep going now the question we're trying to answer is to decide on how many of those regular polygons have each one of their interior angles as an integer so what is the measure of a particular angle in one of these polygons well there are exactly n angles so each of the n angles will measure 180 times n minus 2 divided by n degrees and this number we want to be an integer so if this number so if this number is an integer we have 180 times n minus 2 divided by n we can simplify that a little bit we have 180 times n minus 360 in the numerator divided by n and n has to divide 360 why is that well if n is an integer looking at the first fraction here after splitting this up 180 times n is certainly divisible by n and the second part must also be an integer so 360 would also have to be divisible by n so here's what we have so far we have that n is a number greater than 2 the number of sides and also angles in a regular polygon has to be in this problem be a factor of 360 so let's obtain the prime factorization of 360 and count how many factors it has so we have 360 we have 360 can be factored into 2 to the third power times 3 to the second power times 5 and the way we count factors distinct factors as we look at the exponents in the prime factorization we increase each exponent by 1 if an exponent is not written it's understood that there is a 1 there in the exponent we increase each exponent by 1 and multiply those numbers together so there would be 3 plus 1 times 2 plus 1 times 1 plus 1 factors including the number 1 the number itself or 360 and the number 2 as 360 is even so out of all the 24 factors after all we have 4 times 3 times 2 of all the 24 factors 1 is a factor 2 is a factor and so 360 so of the 24 factors we have to discount 1 and 2 why is that well those are factors those are possible values of n and n being the number of sides of a regular polygon has to be greater than 2 so of all the 24 possible factors we do have to throw away two of them the numbers 1 and 2 and that leaves us with 22 possible factors and that number is also then the number of sides of a regular polygon where each interior angle is equal in measurement and that number is in fact an integer so that is the answer to our question that would be choice c 22 question number 25 how many three-digit positive integers can be represented as the sum of exactly 9 different powers of 2 and here i have to know something about powers of 2 i recall that 2 to the power 10 is 1024 but 2 to the power 9 is 512 so i will not be working with 1024 the highest power i will use in this problem is 9 and we are adding up different powers of 2 let's add up all of the ones that are less than a thousand so let's compute here 2 to the zeroth power plus 2 to the first plus 2 to the second all the way up to 2 to the power 9 which can be compactly written in summation notation like so the index running from 0 to 9 and what we have here is a geometric series which i can sum up using a formula and we'll see how large that number is that would be the first term so 2 to the zeroth power over 1 times 1 minus 2 which is the common ratio here to the power equal to the number of terms in this summation which is 10 and then we have 1 minus the common ratio here so i have 1 times 1 minus 1024 divided by negative 1 and the answer comes out to be 1023 positive which is too much so what i what i have to do is take out one of these terms as power of 2 in my summation and that number has to be at least 24 okay so what we have to do is remove one term from the summation with a value of at least 24 so the remaining nine terms will sum to a number that is less than 1000 so how many of my terms here are powers of 2 that are at least 24 well that would be 2 to the 9th power for sure 512 2 to the 8th power for sure that's half of that so 256 2 to the 7th power half of 2 to the power 8 is 128 2 to the 6th power half of that is 64 2 to the 5th power half of that is 32 and then 2 to the 4th power is 16 which is too little so we need at least 24 let me just erase this 2 to the 4th power and we have exactly one two three four five terms that can be removed one at a time from the summation of 10 terms to give us a summation of nine remaining terms which will add up to a number that is less than a thousand so it will be a three-digit integer so that is how many ways we can accomplish our task we can do that in five different ways that would be choice e over here question number 26 how many triangles abc with angle abc measuring 90 degrees and leg ab measuring 20 exists such that all sides have integer lengths if we draw a triangle and label the legs here that's always a good start to a problem so we have a 90 degree angle over here that has to be at vertex b and then it doesn't really matter how we label the remaining vertices so let's do it like this i'll call the hypotenuse here c i'll call the bottom leg a and then ab is 20 we know that now we have some information about triangles to work with here we have the pythagorean theorem which tells us that c squared is going to be equal to a squared plus 20 squared and the triangle inequality implies that the hypotenuse being the longest side here has to be at least 20 so c has to be at least 20 and we'll try to combine all this information here to recover possible values of c okay so let's do that we can work with the pythagorean theorem and factor a little bit so i will move the variables here on one side and then factor that difference of squares as c minus a times c plus a let that be equal to 400 and then we observe that both of these factors the way they're constructed is they have the same parity so these are both even or both odd well two odd numbers multiply to an odd number and two even numbers multiply to an even number so these are in fact both even so what we're going to do next is factor 400 into all possible pairs of even factors so these are in fact both even they cannot be both consulates crossed they are both even and we will factor 400 first into its prime factors so that's 2 to the fourth power times 5 to the second power and then from here we cover the possible values of our c and a so these pairs are 2 times 200 or we can have 4 times 100 or 10 times 40 or 8 times 50 and finally we can have 20 times 20 and that's it now those are not quite c and a they're c minus a and c plus a so a little closer look is warranted that this pair of factors we observe that we note that c plus a plus c minus a is equal to 2c so with that we can look at each of our possible pairs of factors and just add them up divide by 2 and that's going to be c so here 2 plus 200 divided by 2 is 101 so that's uh one value of c here we have 104 so the value of c would be 52 with 50 as the sum 25 is the value of c with 58 29 is the value and here with 220 20 is the value of c so finally we know that c is an element of the following set starting with the largest number we have 101 and then 52 and then 29 and then 25 and remember that since c is the hypotenuse it has to be at least 20 as we have noted here that's the triangle inequality at work so c equal to 20 is not a valid choice and we close the parentheses here so we have five possible numbers of which only four are correct values of c and so for each value of c in our equation here we will come up with the corresponding value of a in the pythagorean theorem and hence we'll be able to construct a triangle with those integer values for the lengths of sides and the pythagorean theorem guarantees our triangle will have a 90 degree angle so we have four choices here and the answer to the problem 26 is d four question number 27 in the rectangle a b c d shown in the figure m1 over here is the midpoint of the segment bc m2 is the midpoint of segment a m1 m3 is the midpoint of the segment b m2 and m4 is the midpoint of the segment c m3 find the ratio between the areas of the quadrilateral m1 m2 m3 and m4 contained here in the interior of the rectangle and the rectangle a b c d so to begin with i'll enlarge this figure here i'll be labeling right on it so let's make it a little bigger and i'll solve this problem by imagining my rectangle is sitting on a cartesian coordinate plane with a the point at the origin and with c somewhere on the plane so the coordinates of c are x y that makes the coordinates of b x comma zero and the coordinates of b zero comma y so the length of side a b here is y the length of side a b is x and the area of the rectangle is x times y now from the area of the rectangle we'll subtract the areas of the four triangles and to obtain their areas i need coordinates of the midpoints so we'll average the coordinates as necessary to obtain the coordinates of the midpoints m1 here is with coordinates x over 2 comma y after averaging the x coordinates and the y coordinates of b and c then i'll average the coordinates of a and m1 to obtain the coordinates so that would be x over 4 comma y over 2 the coordinates of m3 x average with x over 4 is 5 x over 8 y over 2 average with 0 is y over 4 and finally m4 we average 5 x over 8 with x that's 13 x over 16 and y with y over 4 is 5 y over 8 and let's move on here to label the triangles as areas one two three four and then from x y subtract the areas one two three four so triangle labeled one the area would be one half times the base along the segment ad is y in length and the height along segment bc would be x over 2 in triangle 2 i have the base along the segment ab with length x and the height is the length of the following segment over here i'll label that orange so that is y over 2 then in blue we have one half the base is exactly the distance from m1 to c so that's x over 2 and the height is over here and that is not quite y from y we have to subtract 5 5 y over 8 so that this times 3 y over 8 and finally green the base is along the segment bc so that is y and the height here is the length of the following segment in green and that is the x coordinate of m3 taken away from x so times we're going to have 3 x over 8 and that gives us x over y x times y rather minus the first product in red is x y over 4 minus in orange we also have x y over 4 minus in blue that is going to be 3 x y over 32 and in green 3 x y over 16 so that gives me the first three terms are x y over 2 then 3 x y over 32 and let's write 6 x y over 32 here with the common denominator of 32 i end up looking at 32 x y minus 3 x y minus 6 x y all over 32 that comes out to um well excuse me small error over here 16 x y instead of 32 is the first term so that comes out to 7 x y over 32 16 minus 3 minus 6 is 7 and that is the area of the quadrilateral m1 m2 m3 m4 so that i can kind of fill in over here that is 7 x y over 32 and the area of the rectangle a b c d is just x y so taking the quotient here of those two leaves me with no x and no y i just remain with the fraction 7 over 32 so that is the ratio that we're looking for and that is in this problem sir c 7 over 32 question number 28 a student drew blue and red rectangles on a blackboard exactly seven of the rectangles are squares there are three more red rectangles than blue squares there are two more red squares than blue rectangles how many blue rectangles are there on the blackboard so let's begin by naming some variables we are looking for blue rectangles here so let x be that number let x be the number of blue rectangles we then have a sentence here relating the number of blue rectangles the number of red squares so we have two plus x red squares and going back to the previous sentence we know that of the rectangles exactly seven are squares and of the seven squares two plus x are red as we have discovered so seven minus two plus x are blue squares okay and then we can compare this number of blue squares to the number of blue rectangles after all blue squares are included in the number of two rectangles so since squares are rectangles we have the number of blue squares that would be seven minus two plus x that is less than or equal to x number of blue rectangles well that gives us five minus x is less than x or five is less than or equal to two x so that's going to be one inequality now moving on we have some information about red squares uh red rectangles rather and blue squares so we know that there are three more red rectangles than blue squares and we have already an expression for blue squares so let's write that down we have here five minus x blue squares so three more would be three plus five minus x red rectangles okay and the number of red rectangles has to be greater than the number of red squares and we know exactly how many of those we have since again red squares are rectangles also we have two plus x red squares and that has to be less than the number of red rectangles which is eight minus x as we have just discovered and that gives us a second inequality over here which comes out to be that two x is less than or equal to six and we can solve this and say that x is less than or equal to three now x is exactly three then our two inequalities are satisfied at the same time so if x is equal to three then x is definitely less than or equal to three and two times x is at least five and that is in fact the only value of x that will work over here so we could have solved the first inequality and said that x is at least 2.5 and then that makes things clear that has to be a whole number so x is at least three the second inequality says it is at most three well then we conclude it has to be exactly three and that is answer b over here question number 29 96 members of accounting club are standing in a large circle they start saying the numbers 1 2 3 etc in turn going around the circle every member that says an even number steps out of the circle and the rest continue starting the second round with 97 they continue in this way until only one member is left which number did this member say in the first round okay so we are going to recover the number that the member said in the last round and then work backwards to see what was the number he spoke in the first round so we need to know the number of rounds and with 96 members we can say that half of them leave after each round because half of the numbers between 1 and 96 are even so half the members leave after round one then 96 since that number can be factored as 2 to the 5th power times 3 we will have 2 to the 4th power times 3 members remaining after the first round and so forth dividing by 2 each time until we have 3 members remaining and then one more round is necessary more round is necessary because two of the numbers here that will be left in the last set of three will be even and one of them will be odd okay so since 96 is factored as 2 to the 5th power times 3 we know that there will be 6 rounds we have 6 rounds during which first of all half of 96 so 2 to the 4th power times 3 then 2 to the 3rd power times 3 then 2 to the 2nd power times 3 then 2 to the 1st power times 3 then 3 half of that and then 2 members leave Okay, so that is six rounds. Now, continuing this thought, the last remaining person counts to which number? Well, 96 was the highest number reached in the first round, and then people continued the counting with 97, and after the first round, half the members have to be counting, so that's 2 to the 4th power times 3 has to be added, and then they split in half, half of them leave, and the remaining half of those members keep counting, so we add here the number of members remaining after each round to obtain the highest number reached by the last person remaining, like that, and that gives us 191 as the number spoken by the last person remaining. Okay, so now working backwards, now working in reverse, start with 191 and recover the number from round 1, and how are we going to do that? Well, this number has to be always a power of 2 apart from the number spoken in the next round, because that's the only way this last standing member can remain in the game, he is not allowed to speak even numbers, all the numbers he spoke ending with 191 are going to be even, and so from 191, we subtract powers of 2 until we reach the number we're interested in, so that's going to be 191 minus 2 to the 1st power, and then subtract 2 to the 2nd power, 2 to the 3rd power, 2 to the 4th power, 2 to the 5th power, and finally 2 to the 6th power, because that's how many rounds we have, and that gives us 191, we're subtracting here a total of 2 plus 4 plus 8 plus 16 plus 32 plus 64, which gives us, in the end, the odd number, 65, so that is the number our last standing member in this contest must have spoken each time the numbers he reaches are a power of 2 apart, and this guarantees that they will always remain odd throughout this counting process, so 65 here is the answer to the problem, that is choice B. Question number 30, Bill and Bob replace the letters in the word kangaroo with digits, they get different numbers, but each resulting number is a multiple of 11, they each replace different letters by different digits, and the same letters by the same digits, here they avoid a is equal to 0, Bill obtains the largest possible number that one can in this way, and Bob the smallest, in both cases one of the letters is replaced by the same digit, which digit is this? So let's look at the word kangaroo and decide when it would be divisible by 11 after making replacements here of letters by digits, what we need to do is spell the word out, so we have A-A-N-G-A-R-O-O, and we need to take the alternating sum, so the first operation is a minus, then a plus, and they alternate, like so, and that number has to be divisible by 11. Now we notice right away that since we have a positive and a negative value here for O, a negative and a positive value under A, this formula simplifies to just looking at A plus N, and then minus G minus R, like that, and that number has to be divisible by 11, so let's say it's equal to 11 times some integer K, like that, that's what we need. Okay, so let's begin by filling in here the largest possible values, starting with 9 for K, 8 for A, so that's 8 here again, 7 for N, 6 for G, 5 for R, and 4 for the O's. That alternating sum here comes out to be 5. So A here is 9, N is 7, minus G, minus 6, minus 5, and that gives us 5. So we do not have divisibility by 11. What we'll do then is keep these three first numbers and try to work with the remaining numbers. So G we need to keep as a variable, that's unknown, and then R we'll have to keep as a variable, that's unknown, and the O's don't really contribute, so we'll erase them here, pick the largest values that remain. Okay, and so we have this equation over here, where we have a variable G and a variable R, and this number has to be divisible by 11, and what we have is 16 minus G minus R, that has to be equal to 11 times some integer K, and what we can gather from this right now is that the letters G and R, they have to be among certain values. So the smallest value is probably 0, and we can go up to 1, 2, 3, if one value is 3, the other one is 2, we're still above 11, and then we can probably have a 4 and a 5, but if one value is 6, then we are less than 11 at that sum, and if we keep going all the way up to 9 and 8, we will have negative numbers, but they will still not be divisible by 11. So here is our range of values, and out of this set of possibilities, the largest number we would obtain by using G is equal to 5 over here, and if G is equal to 5, then R would have to be equal to 0, like that, so that 16 minus G minus R is 16 minus 5 minus 0, that is equal to 11, which is definitely divisible by 11, and to make our number as large as possible, we fill in the O's with the largest remaining digits, and that would be a 4 and a 4, so that is Bill's number. Okay, now let's follow the same line of thought for the smallest number, so Bob's number, I will replace here 1 for K, then I'll use a 0 for A, a 2 for N, and then we have a minus G, a plus A, so plus 0, A minus R, and then the O's we will leave till the very end, so that digit sum here comes out to 1 plus N, which is 2 minus G minus R, and that has to be divisible by 11 also, and so far what we have is 3 minus G minus R, so if that's going to be 11 times some integer, it will not be a positive integer, it would have to be a negative integer, and so what we have is G plus R here, like that, has to be 11 plus some integer, like that. What integer would that be to make this work? Well, if we have G plus R adding up to negative 14, let's say, then plus 3 would give us negative 11, so let's see if that works. Say, G plus R is equal to negative 14, and then plus positive 3 would give us negative 11, so that works, and what numbers can we use? Here, G and R, we have used up 0, 1, and 2 already, so maybe we could use 3, 4, 5, 6, 7, 8, and 9, like that, those are the remaining numbers. The smallest number would be obtained by replacing here a G with a 5, because we want the highest value to be as small as possible, and if we replace that with a 5, then R has to be replaced with a 9, so we make R the smallest place in this number, as large as possible, the largest we can make it is a 9, and then that forces 5 to be equal to G, like that, so we have our smallest number over here. Notice that we're not quite done yet, because the O's have not been replaced yet by anything, and those cannot be zeros, we have used up 1, so let's take that off the list, we have used up 0 and 2, we have used up a 5 and 9, so the remaining values here have to be 3s, so that's a 3 and a 3, and on the bottom here, we have Bill's number. Okay, so let's compare the digits, A is either a 9 or a 1, so that's not the same, A is an 8 or a 0, N is a 7 or a 2, but in both cases, G is a 5, and running quickly through the other replacements, G is the only replaced letter here that gets the same digit in both Bill's and Bob's numbers, so that's the answer over here, let's just circle that, emphasize it, there we go, G is replaced by 5 every time, so the answer here to number 30 is D, 5.

mathematical questions

digit sums

logical reasoning

cube coloring

regular polygons

permutations

right triangles

coordinate geometry

word problem

number elimination