💡Okay, so I have this problem where there are several zeros, ones, and twos written on a board. The allowed operation is to erase two different digits and write one digit instead, which is different from the erased ones. I need to prove that if after several such operations only a single digit remains on the board, then this digit does not depend on the order in which the erasures are made.Hmm, let me try to understand the problem step by step. First, let's restate the problem in my own words. We have a board with some number of 0s, 1s, and 2s. We can perform operations where we take two different digits, erase them, and replace them with a third digit that's different from the two we erased. The goal is to show that no matter the sequence of operations we perform, if we end up with a single digit, that digit is always the same, regardless of the order of operations.So, for example, if I start with some combination of 0s, 1s, and 2s, and I keep replacing two different digits with the third one, eventually I'll be left with just one digit. The claim is that this final digit is uniquely determined by the initial counts of 0s, 1s, and 2s, and not by the choices I make during the operations.I think to approach this, I need to find some invariant—something that doesn't change regardless of the operations performed. Invariants are often key in problems like this because they can help us determine the final state without worrying about the specific steps taken.Let me think about what could be invariant here. The operations involve replacing two different digits with the third one. So, for example, replacing a 0 and a 1 with a 2, replacing a 0 and a 2 with a 1, or replacing a 1 and a 2 with a 0.Maybe I can look at the counts of each digit modulo some number. Since we're dealing with three digits, perhaps modulo 2 or modulo 3 could be useful.Let me try modulo 2 first. Let's denote the number of 0s as A, the number of 1s as B, and the number of 2s as C. So, initially, we have A, B, C.When we perform an operation, say replacing a 0 and a 1 with a 2, then A decreases by 1, B decreases by 1, and C increases by 1. So, in terms of modulo 2, A becomes A - 1 mod 2, B becomes B - 1 mod 2, and C becomes C + 1 mod 2.Similarly, replacing a 0 and a 2 with a 1 would result in A - 1, C - 1, and B + 1 mod 2.Replacing a 1 and a 2 with a 0 would result in B - 1, C - 1, and A + 1 mod 2.Hmm, so each operation affects two counts by decreasing them by 1 and increases the third by 1. So, in modulo 2, each operation flips the parity of all three counts.Wait, because subtracting 1 mod 2 is the same as adding 1, and adding 1 mod 2 is also the same as subtracting 1. So, each operation flips the parity of all three counts.But if all three parities are flipped each time, then the overall parity configuration changes. So, maybe modulo 2 isn't the right approach because the parities are all changing each time.Alternatively, maybe I should consider the differences between the counts. For instance, A - B, B - C, or A - C modulo something.Wait, another thought: since each operation replaces two different digits with the third, maybe the differences between the counts of each pair of digits are being altered in a specific way.Let me think about the total number of digits. Each operation reduces the total number of digits by 1 because we're replacing two digits with one. So, starting with N digits, we'll end up with N - k digits after k operations. Eventually, we'll have 1 digit left.But the total number of digits isn't an invariant; it's decreasing by 1 each time. So, that's not helpful.Wait, perhaps considering the counts modulo 3? Since we have three digits, maybe modulo 3 could be useful.Let me see. Suppose I define some function of A, B, C that is invariant modulo 3. For example, maybe A + 2B + C modulo 3 or something like that.But I need to see how each operation affects this function.Let's take an example. Suppose I replace a 0 and a 1 with a 2. Then A decreases by 1, B decreases by 1, and C increases by 1.So, if my function is f = A + 2B + C, then after the operation, f becomes (A - 1) + 2(B - 1) + (C + 1) = A + 2B + C - 1 - 2 + 1 = f - 2.So, f decreases by 2 modulo 3.Similarly, if I replace a 0 and a 2 with a 1, then A decreases by 1, C decreases by 1, and B increases by 1.So, f becomes (A - 1) + 2(B + 1) + (C - 1) = A + 2B + C - 1 + 2 - 1 = f.So, f remains the same.Wait, that's interesting. So, replacing 0 and 2 with 1 doesn't change f, but replacing 0 and 1 with 2 decreases f by 2 modulo 3.Similarly, replacing 1 and 2 with 0: B decreases by 1, C decreases by 1, A increases by 1.So, f becomes (A + 1) + 2(B - 1) + (C - 1) = A + 2B + C + 1 - 2 - 1 = f - 2.So, replacing 1 and 2 with 0 also decreases f by 2 modulo 3.Hmm, so depending on which operation we perform, f either stays the same or decreases by 2 modulo 3.Wait, but in the case of replacing 0 and 2 with 1, f remains the same. So, f is not entirely invariant, but it can either stay the same or change by -2 modulo 3.Is there a way to adjust the function so that it's invariant?Alternatively, maybe f modulo 3 is not the right approach. Maybe I need a different function.Wait, another idea: consider the counts modulo 2 for each digit. So, A mod 2, B mod 2, C mod 2.Each operation flips two of these parities and leaves the third one the same. For example, replacing 0 and 1 with 2 flips A and B, and flips C (since C increases by 1, which is equivalent to flipping its parity).Wait, replacing 0 and 1 with 2: A becomes A - 1, B becomes B - 1, C becomes C + 1.So, modulo 2, A flips, B flips, and C flips.Similarly, replacing 0 and 2 with 1: A flips, C flips, and B flips.Replacing 1 and 2 with 0: B flips, C flips, and A flips.So, each operation flips all three parities.Therefore, the parities of A, B, and C all flip with each operation.But if we start with some parities, after an even number of operations, the parities would return to their original state, and after an odd number of operations, they would be flipped.But since we're starting with some number of digits and reducing by one each time until we have one digit left, the number of operations is fixed: it's equal to the initial total number of digits minus one.Wait, so if the initial total number of digits is N, then the number of operations is N - 1.Therefore, the number of operations is fixed, so the number of parity flips is fixed.Therefore, the final parities of A, B, and C are determined by the initial parities and the number of operations.But since we end up with one digit, two of the counts will be zero, and one will be one.So, the final parities are: two counts are 0 mod 2, and one is 1 mod 2.But depending on which digit remains, the parities will be different.Wait, maybe this can help us determine which digit remains.Let me think.Suppose we denote the initial counts as A, B, C.Each operation flips the parities of A, B, and C.After k operations, the parities of A, B, and C will be:A_initial + k mod 2,B_initial + k mod 2,C_initial + k mod 2.But k is equal to N - 1, where N = A + B + C.So, k = A + B + C - 1.Therefore, the final parities are:A + (A + B + C - 1) mod 2,B + (A + B + C - 1) mod 2,C + (A + B + C - 1) mod 2.Simplify:For A: A + A + B + C - 1 mod 2 = (2A + B + C - 1) mod 2 = (B + C - 1) mod 2,Similarly for B: (A + 2B + C - 1) mod 2 = (A + C - 1) mod 2,For C: (A + B + 2C - 1) mod 2 = (A + B - 1) mod 2.So, the final parities are:A_final = (B + C - 1) mod 2,B_final = (A + C - 1) mod 2,C_final = (A + B - 1) mod 2.But in the end, only one of A, B, C is 1, and the others are 0.So, let's suppose that the final digit is 0. Then, A_final = 1, B_final = 0, C_final = 0.So, we have:1 = (B + C - 1) mod 2,0 = (A + C - 1) mod 2,0 = (A + B - 1) mod 2.But since A_final = 1, and B_final = C_final = 0, we can write:1 = (B + C - 1) mod 2,0 = (1 + C - 1) mod 2 = C mod 2,0 = (1 + B - 1) mod 2 = B mod 2.So, from the second and third equations, we get that C ≡ 0 mod 2 and B ≡ 0 mod 2.Substituting into the first equation: 1 = (0 + 0 - 1) mod 2 = (-1) mod 2 = 1 mod 2. So, that's consistent.Similarly, if the final digit is 1, then B_final = 1, A_final = 0, C_final = 0.So,0 = (B + C - 1) mod 2,1 = (A + C - 1) mod 2,0 = (A + B - 1) mod 2.From the third equation: 0 = (A + B - 1) mod 2 => A + B ≡ 1 mod 2.From the second equation: 1 = (A + C - 1) mod 2 => A + C ≡ 0 mod 2.From the first equation: 0 = (B + C - 1) mod 2 => B + C ≡ 1 mod 2.But since A + B ≡ 1 and A + C ≡ 0, subtracting these gives B - C ≡ 1 mod 2 => B ≡ C + 1 mod 2.From B + C ≡ 1, substituting B = C + 1: (C + 1) + C ≡ 1 => 2C + 1 ≡ 1 => 0 + 1 ≡ 1, which is true.So, consistent.Similarly, if the final digit is 2, then C_final = 1, A_final = 0, B_final = 0.So,0 = (B + C - 1) mod 2,0 = (A + C - 1) mod 2,1 = (A + B - 1) mod 2.From the third equation: A + B ≡ 0 mod 2.From the second equation: A + C ≡ 1 mod 2.From the first equation: B + C ≡ 1 mod 2.From A + B ≡ 0, we have A ≡ B mod 2.From A + C ≡ 1, and B + C ≡ 1, subtracting these gives A - B ≡ 0 mod 2, which is consistent with A ≡ B.So, all consistent.So, depending on the initial parities, the final digit is determined.But how does this help us? Because the final digit is determined by the initial counts modulo 2, and the number of operations is fixed, so the final digit is uniquely determined.Therefore, regardless of the order of operations, the final digit is the same.Wait, but I need to formalize this.Alternatively, maybe I can think in terms of the counts modulo 3.Wait, another approach: consider the counts of each digit modulo 2.But earlier, I saw that each operation flips all three parities. So, the parity vector (A mod 2, B mod 2, C mod 2) is being flipped each time.But the number of operations is fixed, so the final parity vector is determined by the initial one and the number of operations.Since the number of operations is fixed, the final parity vector is uniquely determined, which in turn determines the final digit.Therefore, regardless of the order, the final digit is the same.Alternatively, maybe I can think about the problem in terms of equivalence classes.Each operation is essentially a transformation on the triplet (A, B, C). The key is to find an invariant that remains the same regardless of the operations.Wait, another idea: consider the counts modulo 2 for each digit, but also consider the total number of digits modulo something.Wait, but earlier, I saw that the parities flip each time, so the total number of operations affects the final parities.But since the number of operations is fixed, the final parities are fixed, so the final digit is fixed.Therefore, regardless of the order, the final digit is determined by the initial counts and the number of operations, which is fixed.Therefore, the final digit doesn't depend on the order.Alternatively, maybe I can think about the problem in terms of linear algebra.Each operation can be represented as a vector that subtracts 1 from two counts and adds 1 to the third.So, for example, replacing 0 and 1 with 2 can be represented as the vector (-1, -1, +1).Similarly, replacing 0 and 2 with 1 is (-1, +1, -1), and replacing 1 and 2 with 0 is (+1, -1, -1).So, these are the three possible operations.Now, the problem is to show that regardless of the sequence of these operations, the final state is uniquely determined.This is equivalent to showing that the set of operations generates a certain equivalence class, and the final state is the same regardless of the path taken.In linear algebra terms, if we can show that the operations span a certain subspace, and the final state is determined by the initial state modulo this subspace, then the final digit is uniquely determined.But maybe this is getting too abstract.Alternatively, let's consider specific examples.Suppose we start with one 0, one 1, and one 2. So, A=1, B=1, C=1.What happens if we perform operations?First, replace 0 and 1 with 2: Now we have A=0, B=0, C=2.Then, replace the two 2s: but wait, we can't replace two of the same digits. The operation requires replacing two different digits.Wait, so in this case, after replacing 0 and 1 with 2, we have two 2s. But we can't replace two 2s because the operation requires two different digits.So, we're stuck with two 2s. But the problem states that we can perform operations until only one digit remains. So, in this case, maybe we can't perform any more operations because we only have two 2s, and we can't replace them.Wait, but the problem says "several zeros, ones, and twos," so maybe the initial counts are such that it's possible to reduce to one digit.Wait, in my example, starting with one of each, after one operation, we have two 2s, and we can't proceed further. So, that suggests that the initial counts must satisfy certain conditions for it to be possible to reduce to one digit.Wait, maybe the initial counts must satisfy that the total number of digits is congruent to 1 modulo 2, because each operation reduces the count by 1, so to end up with 1, the initial count must be odd.But in my example, starting with 3 digits (1+1+1=3), which is odd, but after one operation, we have 2 digits, which is even, and we can't proceed further.Wait, so maybe the initial counts must satisfy that the number of digits is congruent to 1 modulo 2, but also that the counts of each digit satisfy certain parity conditions.Wait, perhaps I need to think differently.Let me try another example.Suppose we start with two 0s, one 1, and one 2. So, A=2, B=1, C=1.Total digits: 4.We need to perform 3 operations to get down to 1 digit.First operation: replace 0 and 1 with 2. Now, A=1, B=0, C=2.Second operation: replace 0 and 2 with 1. Now, A=0, B=1, C=1.Third operation: replace 1 and 2 with 0. Now, A=1, B=0, C=0.So, final digit is 0.Alternatively, let's try a different sequence.First operation: replace 0 and 2 with 1. Now, A=1, B=2, C=0.Second operation: replace 1 and 1 with 0. Wait, but we can't replace two 1s because the operation requires two different digits.So, instead, replace 1 and 0 with 2. Now, A=0, B=1, C=1.Third operation: replace 1 and 2 with 0. Now, A=1, B=0, C=0.Same result: 0.Another sequence:First operation: replace 1 and 2 with 0. Now, A=3, B=0, C=0.Wait, but we started with A=2, B=1, C=1. After replacing 1 and 2 with 0, we have A=3, B=0, C=0.But now, we have three 0s. We can't perform any more operations because we need two different digits to replace.Wait, but we needed to perform three operations to get down to one digit. But in this case, after one operation, we have three 0s, and we can't proceed further. So, this suggests that the order of operations can affect whether we can reach one digit or not.But the problem states that "if after several such operations only a single digit remains on the board," so perhaps the initial configuration must allow for such a reduction, regardless of the operations.Wait, so in my first example, starting with A=2, B=1, C=1, I was able to reach one digit by choosing operations that didn't leave me with two of the same digits. But in the second sequence, I made a bad choice that left me with three of the same digit, preventing further operations.So, perhaps the problem assumes that the operations are chosen such that we can reduce to one digit, and in that case, the final digit is uniquely determined.Therefore, regardless of the sequence, as long as we can reduce to one digit, that digit is uniquely determined.So, to formalize this, perhaps we can define an invariant that determines the final digit.Wait, earlier, I considered the parities of A, B, C, and saw that each operation flips all three parities. Since the number of operations is fixed (N - 1), the final parities are determined by the initial parities and the number of operations.Therefore, the final digit is uniquely determined by the initial counts.Alternatively, another approach is to consider the counts modulo 2 for each digit, but also considering the total number of operations.Wait, maybe I can think about the problem in terms of the counts modulo 2.Let me define:Let’s denote the counts as A, B, C.Each operation replaces two different digits with the third, so:- Replacing 0 and 1 with 2: A -= 1, B -= 1, C += 1- Replacing 0 and 2 with 1: A -= 1, C -= 1, B += 1- Replacing 1 and 2 with 0: B -= 1, C -= 1, A += 1Now, let's consider the counts modulo 2.Each operation affects the counts as follows:- Replacing 0 and 1 with 2: A flips, B flips, C flips- Replacing 0 and 2 with 1: A flips, C flips, B flips- Replacing 1 and 2 with 0: B flips, C flips, A flipsSo, in all cases, all three counts flip their parity.Therefore, the parity vector (A mod 2, B mod 2, C mod 2) flips each time an operation is performed.Since the number of operations is fixed (N - 1), the final parity vector is determined by the initial one and the number of operations.Therefore, the final parities are uniquely determined, which in turn determines the final digit.Because in the end, only one digit remains, so two of the counts are 0, and one is 1.Thus, the final digit is determined by the final parities, which are uniquely determined by the initial counts and the number of operations.Therefore, regardless of the order of operations, the final digit is the same.So, to summarize, the key idea is that each operation flips the parities of all three counts. Since the number of operations is fixed (equal to the initial total number of digits minus one), the final parities are uniquely determined. Therefore, the final digit is uniquely determined by the initial configuration, and thus does not depend on the order of operations.I think this makes sense. Each operation affects the parities in a predictable way, and since the number of operations is fixed, the final state is uniquely determined. Therefore, the final digit doesn't depend on the order in which the operations are performed.

💡To solve the problem, we start by using the given information that ( x = 1 ) is a solution to the quadratic equation ( ax^2 + bx + 1 = 0 ). 1. **Substitute ( x = 1 ) into the equation:** [ a(1)^2 + b(1) + 1 = 0 ] Simplifying this, we get: [ a + b + 1 = 0 ]2. **Solve for ( a + b ):** [ a + b = -1 ]3. **Calculate ( 3 - a - b ) using the value of ( a + b ):** [ 3 - a - b = 3 - (-1) = 3 + 1 = 4 ]Therefore, the value of ( 3 - a - b ) is ( 4 ).

💡Okay, so I have this problem about a boarding school dormitory where students exchange greeting cards during New Year's Day. I need to figure out how many students are in the dormitory based on the total number of greeting cards used, which is 51. Let me try to break this down step by step.First, let's understand the problem. There are several students in a dormitory, and one of them is the dorm leader. Each student exchanges greeting cards with every other student. Additionally, each student gives a greeting card to every dormitory administrator. Furthermore, each administrator gives a greeting card back to the dorm leader. The total number of greeting cards used in all these exchanges is 51. I need to find out how many students are in the dormitory.Alright, let's start by defining some variables to represent the unknowns in the problem. Let me denote the number of students in the dormitory as ( n ). This includes the dorm leader. Let me also denote the number of dormitory administrators as ( m ). These are the people who receive greeting cards from the students and also give greeting cards back to the dorm leader.Now, let's break down the different types of greeting card exchanges:1. **Student-to-Student Exchanges:** Each student exchanges greeting cards with every other student. Since there are ( n ) students, each student will exchange cards with ( n - 1 ) other students. However, this counts each exchange twice (once for each participant), so the total number of unique exchanges is ( frac{n(n - 1)}{2} ). Each exchange involves two cards, so the total number of cards exchanged among students is ( n(n - 1) ).2. **Student-to-Administrator Exchanges:** Each student gives a greeting card to every administrator. There are ( n ) students and ( m ) administrators, so the total number of cards given from students to administrators is ( n times m ).3. **Administrator-to-Dorm Leader Exchanges:** Each administrator gives a greeting card back to the dorm leader. Since there are ( m ) administrators, the total number of cards given from administrators to the dorm leader is ( m ).Now, let's sum up all these greeting cards to get the total number used:- Cards exchanged among students: ( n(n - 1) )- Cards from students to administrators: ( n times m )- Cards from administrators to the dorm leader: ( m )So, the total number of greeting cards used is:[n(n - 1) + n times m + m = 51]This simplifies to:[n^2 - n + nm + m = 51]Hmm, this equation has two variables: ( n ) (number of students) and ( m ) (number of administrators). I need another equation or some way to relate ( n ) and ( m ) to solve for ( n ). But the problem doesn't provide any additional information about the number of administrators. So, I need to make an assumption or find a way to express ( m ) in terms of ( n ).Wait a minute, maybe I can factor the equation to make it easier to solve. Let's see:[n^2 - n + nm + m = 51][n^2 - n + m(n + 1) = 51]Hmm, that doesn't seem immediately helpful. Maybe I can rearrange the terms:[n^2 + (m - 1)n + (m - 51) = 0]This is a quadratic equation in terms of ( n ). For this equation to have integer solutions for ( n ), the discriminant must be a perfect square. The discriminant ( D ) of a quadratic equation ( ax^2 + bx + c = 0 ) is ( D = b^2 - 4ac ). So, let's compute the discriminant:[D = (m - 1)^2 - 4 times 1 times (m - 51)][D = m^2 - 2m + 1 - 4m + 204][D = m^2 - 6m + 205]For ( n ) to be an integer, ( D ) must be a perfect square. Let's denote ( k^2 = m^2 - 6m + 205 ), where ( k ) is an integer. So,[m^2 - 6m + 205 = k^2]This is a Diophantine equation, and solving it might give us possible integer values for ( m ) and ( k ). Let's rearrange it:[m^2 - 6m + 205 - k^2 = 0]This can be rewritten as:[(m^2 - 6m + 9) + 196 - k^2 = 0][(m - 3)^2 + 196 = k^2][k^2 - (m - 3)^2 = 196]This is a difference of squares, which can be factored as:[(k - (m - 3))(k + (m - 3)) = 196]Let me denote ( a = k - (m - 3) ) and ( b = k + (m - 3) ), so that ( a times b = 196 ). Also, since ( k ) and ( m ) are positive integers, ( b > a ).Now, I need to find pairs of positive integers ( (a, b) ) such that ( a times b = 196 ) and ( b > a ). Let's list the factor pairs of 196:1. ( (1, 196) )2. ( (2, 98) )3. ( (4, 49) )4. ( (7, 28) )5. ( (14, 14) )Wait, but ( b > a ), so we can exclude the pair ( (14, 14) ) since ( b ) is not greater than ( a ) in this case.Now, for each of these factor pairs, I can solve for ( k ) and ( m ):1. For ( (a, b) = (1, 196) ): [ a = 1, quad b = 196 ] [ k - (m - 3) = 1 ] [ k + (m - 3) = 196 ] Adding these two equations: [ 2k = 197 implies k = 98.5 ] But ( k ) must be an integer, so this pair is invalid.2. For ( (a, b) = (2, 98) ): [ k - (m - 3) = 2 ] [ k + (m - 3) = 98 ] Adding: [ 2k = 100 implies k = 50 ] Subtracting: [ 2(m - 3) = 96 implies m - 3 = 48 implies m = 51 ] So, ( m = 51 ) and ( k = 50 ).3. For ( (a, b) = (4, 49) ): [ k - (m - 3) = 4 ] [ k + (m - 3) = 49 ] Adding: [ 2k = 53 implies k = 26.5 ] Not an integer, so invalid.4. For ( (a, b) = (7, 28) ): [ k - (m - 3) = 7 ] [ k + (m - 3) = 28 ] Adding: [ 2k = 35 implies k = 17.5 ] Not an integer, so invalid.So, the only valid factor pair is ( (2, 98) ), which gives ( m = 51 ) and ( k = 50 ).Now, let's plug ( m = 51 ) back into our quadratic equation for ( n ):[n^2 + (m - 1)n + (m - 51) = 0][n^2 + (51 - 1)n + (51 - 51) = 0][n^2 + 50n + 0 = 0][n(n + 50) = 0]This gives two solutions: ( n = 0 ) or ( n = -50 ). However, the number of students cannot be zero or negative, so this doesn't make sense. Did I make a mistake somewhere?Let me go back and check my steps. I think I might have made an error in setting up the equation or in the factor pairs. Let me revisit the discriminant part.We had:[D = m^2 - 6m + 205 = k^2]And then I rewrote it as:[(m - 3)^2 + 196 = k^2]Which led to:[k^2 - (m - 3)^2 = 196]And then factoring as:[(k - (m - 3))(k + (m - 3)) = 196]Wait, let me double-check that step. The equation ( k^2 - (m - 3)^2 = 196 ) is indeed a difference of squares, so it factors correctly. Then, setting ( a = k - (m - 3) ) and ( b = k + (m - 3) ), with ( a times b = 196 ), is correct.But when I solved for ( m ) and ( k ) using the factor pair ( (2, 98) ), I got ( m = 51 ), which led to an invalid number of students. Maybe I need to consider negative factors as well? Because ( a ) and ( b ) could be negative integers as well, since ( k ) and ( m ) are positive, but ( a ) and ( b ) could be negative.Let me try that. Let's consider negative factor pairs of 196:1. ( (-1, -196) )2. ( (-2, -98) )3. ( (-4, -49) )4. ( (-7, -28) )5. ( (-14, -14) )Again, ( b > a ), so we need ( b ) to be greater than ( a ). For negative factors, this would mean ( b ) is less negative than ( a ).Let's try ( (a, b) = (-2, -98) ):[k - (m - 3) = -2][k + (m - 3) = -98]Adding these:[2k = -100 implies k = -50]Subtracting:[2(m - 3) = -96 implies m - 3 = -48 implies m = -45]But ( m ) cannot be negative, so this is invalid.Next, ( (a, b) = (-4, -49) ):[k - (m - 3) = -4][k + (m - 3) = -49]Adding:[2k = -53 implies k = -26.5]Not an integer, invalid.Similarly, other negative factor pairs will lead to negative or non-integer values for ( m ) and ( k ), which don't make sense in this context.Hmm, so it seems like the only valid factor pair leads to an invalid number of students. Maybe my initial setup is wrong.Let me go back to the beginning and re-examine the problem statement."During New Year's Day, each student in the dormitory exchanges greeting cards with each other, and each person also gives a greeting card to every dormitory administrator. Each administrator also gives a greeting card back to the dorm leader. In total, 51 greeting cards were used."Wait, maybe I misinterpreted the "each person also gives a greeting card to every dormitory administrator." Does "each person" include the dorm leader? Yes, because the dorm leader is a person. So, the dorm leader also gives a greeting card to each administrator. But then, the administrators give a greeting card back to the dorm leader. So, the dorm leader receives ( m ) cards from the administrators, and the dorm leader also gives ( m ) cards to the administrators.But in my initial setup, I considered that each student gives ( m ) cards to administrators, which includes the dorm leader. So, the total number of cards from students to administrators is ( n times m ), which includes the dorm leader's ( m ) cards. Then, the administrators give ( m ) cards back to the dorm leader. So, the total cards from administrators are ( m ).Wait, but the problem says "each administrator also gives a greeting card back to the dorm leader." So, each administrator gives one card to the dorm leader, so total ( m ) cards. But the dorm leader also gives ( m ) cards to the administrators. So, the total number of cards between the dorm leader and administrators is ( m + m = 2m ).But in my initial equation, I had:- Cards exchanged among students: ( n(n - 1) )- Cards from students to administrators: ( n times m )- Cards from administrators to the dorm leader: ( m )But actually, the dorm leader is a student, so when I counted ( n times m ), that includes the dorm leader giving ( m ) cards to administrators. Then, the administrators give ( m ) cards back to the dorm leader. So, the total cards between students and administrators are ( n times m + m = m(n + 1) ).Wait, that's what I had before. So, the equation is:[n(n - 1) + m(n + 1) = 51]Which simplifies to:[n^2 - n + mn + m = 51]Which is the same as before. So, my initial setup was correct.But when I tried solving it, I ended up with ( m = 51 ), which led to ( n = 0 ) or ( n = -50 ), which is impossible. So, perhaps there's a mistake in my approach.Maybe instead of trying to solve for ( m ) and ( n ) simultaneously, I can assume a reasonable number of administrators and see if it leads to an integer number of students.Let me try small integer values for ( m ) and see if ( n ) comes out as an integer.Starting with ( m = 1 ):[n^2 - n + n + 1 = 51][n^2 + 1 = 51][n^2 = 50][n = sqrt{50} approx 7.07]Not an integer.Next, ( m = 2 ):[n^2 - n + 2n + 2 = 51][n^2 + n + 2 = 51][n^2 + n - 49 = 0]Discriminant:[1 + 196 = 197]Not a perfect square, so no integer solution.( m = 3 ):[n^2 - n + 3n + 3 = 51][n^2 + 2n + 3 = 51][n^2 + 2n - 48 = 0]Discriminant:[4 + 192 = 196]Which is ( 14^2 ). So,[n = frac{-2 pm 14}{2}]Positive solution:[n = frac{12}{2} = 6]So, ( n = 6 ) and ( m = 3 ). Let's check if this works.Total cards:- Student-to-student: ( 6 times 5 = 30 )- Student-to-administrator: ( 6 times 3 = 18 )- Administrator-to-dorm leader: ( 3 )Total: ( 30 + 18 + 3 = 51 ). Perfect!So, the number of students is 6.Wait, but earlier when I tried to solve it algebraically, I ended up with ( m = 51 ), which didn't make sense. But by assuming small values for ( m ), I found a valid solution with ( m = 3 ) and ( n = 6 ). So, maybe my initial algebraic approach was too convoluted, and assuming small values for ( m ) was a better strategy.Alternatively, perhaps I made a mistake in interpreting the problem. Let me double-check.The problem says:"each student in the dormitory exchanges greeting cards with each other, and each person also gives a greeting card to every dormitory administrator. Each administrator also gives a greeting card back to the dorm leader."So, each student exchanges cards with every other student: that's ( frac{n(n - 1)}{2} ) exchanges, but since each exchange involves two cards, it's ( n(n - 1) ) cards.Each person (including the dorm leader) gives a card to each administrator: ( n times m ) cards.Each administrator gives a card back to the dorm leader: ( m ) cards.So, total cards: ( n(n - 1) + nm + m = 51 ).Which is the same equation as before.So, with ( n = 6 ) and ( m = 3 ), it works.Therefore, the number of students is 6.

💡First, substitute ( a = -2 ) into each expression in the set:- ( -3a = -3(-2) = 6 )- ( 4a = 4(-2) = -8 )- ( frac{24}{a} = frac{24}{-2} = -12 )- ( a^2 = (-2)^2 = 4 )- The last element in the set is ( 1 ).Next, list the evaluated set:[{ -3a, 4a, frac{24}{a}, a^2, 1 } = { 6, -8, -12, 4, 1 }]Now, identify the largest number in the set:- Comparing the numbers ( 6, -8, -12, 4, 1 ), we see that ( 6 ) is the largest.Finally, match the largest number to its corresponding expression:- The number ( 6 ) corresponds to the expression ( -3a ).Thus, the largest number in the set when ( a = -2 ) is ( -3a ).

💡First, I need to determine the total number of 5-letter words that can be formed using the letters A, B, C, D, E, and F. Since there are 6 letters and each position in the word can be any of these letters, the total number of possible words is (6^5).Next, I calculate the number of 5-letter words that consist only of consonants. The consonants available are B, C, D, and F, which totals 4 letters. Therefore, the number of all-consonant words is (4^5).To find the number of 5-letter words that contain at least one vowel, I subtract the number of all-consonant words from the total number of possible words. This gives me the final count of words with at least one vowel.

💡Alright, so I have this problem where there's a sequence of functions defined recursively. It starts with f₁(x) = √(x² + 48), and then each subsequent function is defined as f_{n+1}(x) = √(x² + 6f_n(x)). The question is asking me to find all real solutions to the equation f_n(x) = 2x for each positive integer n.Hmm, okay. Let me try to break this down. First, I need to understand what these functions look like and how they behave. Maybe if I compute the first few functions, I can spot a pattern or figure out a general approach.Starting with n=1: f₁(x) = √(x² + 48). So, setting this equal to 2x gives me √(x² + 48) = 2x. To solve for x, I can square both sides to eliminate the square root. That would give me x² + 48 = (2x)², which simplifies to x² + 48 = 4x². Subtracting x² from both sides, I get 48 = 3x², so x² = 16, which means x = ±4. But wait, since we're dealing with square roots, which are defined to be non-negative, does that mean x has to be positive? Let me think. If x is negative, then 2x would be negative, but the square root is always non-negative. So, 2x must also be non-negative, meaning x must be non-negative. Therefore, the only solution is x = 4.Okay, so for n=1, x=4 is the solution. Let me check that: f₁(4) = √(16 + 48) = √64 = 8, and 2x = 8. Yep, that works.Now, moving on to n=2: f₂(x) = √(x² + 6f₁(x)). Plugging in f₁(x) gives me √(x² + 6√(x² + 48)). Setting this equal to 2x, I have √(x² + 6√(x² + 48)) = 2x. Again, I'll square both sides to get rid of the outer square root: x² + 6√(x² + 48) = 4x². Subtracting x² from both sides, I get 6√(x² + 48) = 3x². Dividing both sides by 3, I have 2√(x² + 48) = x². Now, I can square both sides again: 4(x² + 48) = x⁴. Expanding that, I get 4x² + 192 = x⁴. Rearranging terms, x⁴ - 4x² - 192 = 0.Hmm, this looks like a quadratic in terms of x². Let me set y = x², so the equation becomes y² - 4y - 192 = 0. Using the quadratic formula, y = [4 ± √(16 + 768)] / 2 = [4 ± √784]/2 = [4 ± 28]/2. So, y = (4 + 28)/2 = 32/2 = 16, or y = (4 - 28)/2 = -24/2 = -12. Since y = x² can't be negative, we discard y = -12. So, y = 16, which means x² = 16, so x = ±4. Again, since 2x must be non-negative, x must be positive, so x = 4.Alright, so for n=2, x=4 is also the solution. Let me verify: f₂(4) = √(16 + 6f₁(4)) = √(16 + 6*8) = √(16 + 48) = √64 = 8, and 2x = 8. Yep, that works too.Hmm, interesting. So, for both n=1 and n=2, x=4 is the solution. Maybe this pattern continues? Let's test n=3 to see.For n=3: f₃(x) = √(x² + 6f₂(x)). Plugging in f₂(x) gives me √(x² + 6√(x² + 6√(x² + 48))). Setting this equal to 2x, I have √(x² + 6√(x² + 6√(x² + 48))) = 2x. Squaring both sides: x² + 6√(x² + 6√(x² + 48)) = 4x². Subtracting x²: 6√(x² + 6√(x² + 48)) = 3x². Dividing by 3: 2√(x² + 6√(x² + 48)) = x². Squaring again: 4(x² + 6√(x² + 48)) = x⁴. Expanding: 4x² + 24√(x² + 48) = x⁴. Rearranging: x⁴ - 4x² - 24√(x² + 48) = 0.This seems more complicated. Maybe I can substitute x=4 and see if it works: x=4, so x²=16. Plugging into the equation: 256 - 64 - 24√(16 + 48) = 256 - 64 - 24√64 = 192 - 24*8 = 192 - 192 = 0. Yep, it works. So x=4 is a solution. But are there any other solutions?Let me see. Suppose x ≠ 4. Is there a possibility of another solution? Maybe I can assume x > 4 or x < 4 and see what happens.If x > 4, then 2x > 8, and f₁(x) = √(x² + 48) > √(16 + 48) = 8. Similarly, f₂(x) would be larger than f₁(x), so f₂(x) > 8, and so on. So, as n increases, f_n(x) increases for x > 4. But 2x is also increasing. Hmm, not sure if that helps.If x < 4, then 2x < 8, and f₁(x) = √(x² + 48) > √(0 + 48) = √48 ≈ 6.928, which is greater than 2x if x < 4. Similarly, f₂(x) would be larger than f₁(x), so f₂(x) > 6.928, which is still greater than 2x for x < 4. So, it seems like for x < 4, f_n(x) > 2x, and for x > 4, f_n(x) increases faster than 2x, but we already saw that x=4 is the only solution where they meet.Wait, but in the case of n=3, when I plugged in x=4, it worked, but when I tried to solve the equation, it got complicated. Maybe I can use induction to show that x=4 is the only solution for all n.Let me try to set up an induction argument. For the base case, n=1, we've already shown that x=4 is the only solution. Now, assume that for some k ≥ 1, x=4 is the only solution to f_k(x) = 2x. Then, for n=k+1, f_{k+1}(x) = √(x² + 6f_k(x)). Setting this equal to 2x gives √(x² + 6f_k(x)) = 2x. Squaring both sides: x² + 6f_k(x) = 4x². So, 6f_k(x) = 3x², which simplifies to 2f_k(x) = x². But from the inductive hypothesis, f_k(x) = 2x only when x=4. So, plugging x=4 into 2f_k(x) = x², we get 2*8 = 16, which is true. Therefore, x=4 is the only solution for n=k+1 as well.Thus, by induction, x=4 is the only real solution for all positive integers n.Wait, but in the case of n=3, when I tried to solve the equation, I ended up with a complicated expression. Does this induction argument hold even though solving for higher n becomes more complex? I think it does because the inductive step relies on the relationship between f_{k+1}(x) and f_k(x), and since we're assuming f_k(x) = 2x only at x=4, it carries over to f_{k+1}(x).So, putting it all together, it seems that no matter what n is, the only real solution to f_n(x) = 2x is x=4.