💡Okay, so Vanya has three sets of candies, each containing hard candies, chocolates, and gummy candies. The total number of each type of candy across all three sets is the same. That means the total number of hard candies equals the total number of chocolates, which equals the total number of gummy candies. Let me try to break down the information step by step. First, let's define some variables to make this easier. Let's say:- ( L_1 ) = number of hard candies in the first set- ( S_1 ) = number of chocolates in the first set- ( M_1 ) = number of gummy candies in the first setSimilarly, for the second set:- ( L_2 ) = number of hard candies in the second set- ( S_2 ) = number of chocolates in the second set- ( M_2 ) = number of gummy candies in the second setAnd for the third set:- ( L_3 ) = number of hard candies in the third set- ( S_3 ) = number of chocolates in the third set- ( M_3 ) = number of gummy candies in the third setFrom the problem, we know that the total number of each type of candy is equal. So:[ L_1 + L_2 + L_3 = S_1 + S_2 + S_3 = M_1 + M_2 + M_3 ]Now, let's look at the first set. It says there are equal numbers of chocolates and gummy candies, and 7 more hard candies than chocolates. So:[ S_1 = M_1 ][ L_1 = S_1 + 7 ]Okay, so if I let ( S_1 = x ), then ( M_1 = x ) and ( L_1 = x + 7 ).Moving on to the second set. It says there are equal numbers of hard candies and chocolates, and 15 fewer gummy candies than hard candies. So:[ L_2 = S_2 ][ M_2 = L_2 - 15 ]Again, if I let ( L_2 = y ), then ( S_2 = y ) and ( M_2 = y - 15 ).Now, the third set has no hard candies, so ( L_3 = 0 ). We need to find the total number of candies in the third set, which would be ( S_3 + M_3 ).Since the totals for each type are equal, let's write that out:Total hard candies: ( L_1 + L_2 + L_3 = (x + 7) + y + 0 = x + y + 7 )Total chocolates: ( S_1 + S_2 + S_3 = x + y + S_3 )Total gummy candies: ( M_1 + M_2 + M_3 = x + (y - 15) + M_3 )Since all these totals are equal:1. ( x + y + 7 = x + y + S_3 ) Simplifying this, subtract ( x + y ) from both sides: ( 7 = S_3 ) So, ( S_3 = 7 )2. Now, let's set the total chocolates equal to total gummy candies: ( x + y + 7 = x + y - 15 + M_3 ) Simplify: Subtract ( x + y ) from both sides: ( 7 = -15 + M_3 ) Add 15 to both sides: ( 22 = M_3 ) So, ( M_3 = 22 )Therefore, the third set has ( S_3 = 7 ) chocolates and ( M_3 = 22 ) gummy candies, with ( L_3 = 0 ) hard candies. Adding those together: ( 7 + 22 = 29 )So, the third set has a total of 29 candies.

💡Okay, so I have this problem about a hyperbola and a circle. The hyperbola is given by the equation (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), where (a > 0) and (b > 0). The asymptotes of this hyperbola are tangent to the circle ((x - sqrt{3})^2 + (y - 1)^2 = 1). I need to find the eccentricity of the hyperbola, and the options are A: 2, B: (sqrt{5}), C: (sqrt{3}), and D: (sqrt{2}).First, I remember that the asymptotes of a hyperbola of the form (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) are given by the equations (y = pm frac{b}{a}x). So, the asymptotes are straight lines passing through the origin with slopes (frac{b}{a}) and (-frac{b}{a}).Now, the problem states that these asymptotes are tangent to the given circle. The circle is centered at ((sqrt{3}, 1)) with a radius of 1. For a line to be tangent to a circle, the distance from the center of the circle to the line must be equal to the radius of the circle.So, I need to find the distance from the center ((sqrt{3}, 1)) to each asymptote and set that equal to 1. Since both asymptotes are symmetric with respect to the x-axis, I can just consider one of them, say (y = frac{b}{a}x), and the other one will give the same result.The general formula for the distance from a point ((x_0, y_0)) to the line (Ax + By + C = 0) is (frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}}).Let me rewrite the asymptote equation in the standard form. Starting with (y = frac{b}{a}x), I can subtract (frac{b}{a}x) from both sides to get (-frac{b}{a}x + y = 0). To make it easier, I can multiply both sides by (a) to eliminate the fraction: (-bx + ay = 0). So, the equation becomes (-bx + ay = 0), which can also be written as (bx - ay = 0) if I multiply both sides by -1. It doesn't matter which form I use because the distance formula will account for the absolute value.So, the asymptote equation is (bx - ay = 0). Now, applying the distance formula from the center ((sqrt{3}, 1)) to this line:Distance (d = frac{|b cdot sqrt{3} - a cdot 1|}{sqrt{b^2 + a^2}}).Since the asymptote is tangent to the circle, this distance must equal the radius, which is 1. So,(frac{|bsqrt{3} - a|}{sqrt{a^2 + b^2}} = 1).This equation can be written as:(|bsqrt{3} - a| = sqrt{a^2 + b^2}).To eliminate the absolute value, I can square both sides:((bsqrt{3} - a)^2 = (sqrt{a^2 + b^2})^2).Expanding the left side:((bsqrt{3})^2 - 2absqrt{3} + a^2 = a^2 + b^2).Calculating each term:(3b^2 - 2absqrt{3} + a^2 = a^2 + b^2).Now, subtract (a^2 + b^2) from both sides:(3b^2 - 2absqrt{3} + a^2 - a^2 - b^2 = 0).Simplifying:(2b^2 - 2absqrt{3} = 0).Factor out 2b:(2b(b - asqrt{3}) = 0).Since (b > 0), we can divide both sides by 2b:(b - asqrt{3} = 0).So,(b = asqrt{3}).Now, I need to find the eccentricity of the hyperbola. The formula for eccentricity (e) of a hyperbola is (e = frac{c}{a}), where (c) is the distance from the center to each focus, and (c^2 = a^2 + b^2).Given that (b = asqrt{3}), let's compute (c^2):(c^2 = a^2 + b^2 = a^2 + (asqrt{3})^2 = a^2 + 3a^2 = 4a^2).Therefore,(c = sqrt{4a^2} = 2a).So, the eccentricity is:(e = frac{c}{a} = frac{2a}{a} = 2).Therefore, the eccentricity of the hyperbola is 2, which corresponds to option A.Wait, let me double-check my steps to make sure I didn't make a mistake. I started by finding the asymptotes, converted them into standard form, applied the distance formula, set it equal to the radius, squared both sides, expanded, and solved for the relationship between (a) and (b). Then, using that relationship, I found (c) and hence the eccentricity. It seems correct. I don't see any errors in my calculations.Another way to think about it: the asymptotes are tangent to the circle, so they just touch the circle at exactly one point. The distance from the center to the asymptote equals the radius, which is 1. By setting up that equation, I found that (b = asqrt{3}), which leads to (c = 2a), so eccentricity (e = 2). That makes sense because for hyperbolas, eccentricity is always greater than 1, and 2 is a reasonable value given the relationship between (a) and (b).I think my answer is correct.**Final Answer**boxed{A}

💡First, I need to determine the ratio in which Sue's and Jane's trail mixes are combined to achieve a final mixture that is 45% nuts.Let’s denote the amount of Sue's trail mix as ( S ) and the amount of Jane's trail mix as ( J ).From Sue's trail mix:- 30% nuts- 70% dried fruitFrom Jane's trail mix:- 60% nuts- 40% chocolate chipsThe total percentage of nuts in the combined mixture is 45%. I can set up the equation:[0.30S + 0.60J = 0.45(S + J)]Simplifying this equation:[0.30S + 0.60J = 0.45S + 0.45J][0.15J = 0.15S][J = S]This means that Sue and Jane contributed equal amounts of their trail mixes to the combined mixture.Next, I need to calculate the percentage of dried fruit in the combined mixture. Since Jane's trail mix does not contain dried fruit, the dried fruit comes solely from Sue's trail mix.The total amount of dried fruit in the combined mixture is:[0.70S + 0J = 0.70S]Since ( J = S ), the total combined mixture is ( S + J = 2S ).The percentage of dried fruit in the combined mixture is:[frac{0.70S}{2S} = 0.35 = 35%]

💡Okay, so I have this problem about a cyclic sequence of four-digit numbers. It says that the hundreds, tens, and units digits of each term become the thousands, hundreds, and tens digits of the next term, respectively. Also, the tens and units digits of the last term become the thousands and hundreds digits of the first term. An example is given: 2475, 4758, 7582, 5824. Then, T is the sum of all the terms in this sequence. I need to find the largest prime factor that always divides T, and the options are 11, 37, 43, 101, and 111.Alright, let me try to understand this cyclic sequence better. So, each term is a four-digit number, and when we move to the next term, the hundreds, tens, and units digits shift left by one place, becoming the thousands, hundreds, and tens digits of the next term. The thousands digit of the current term must then become the units digit of the next term? Wait, no, because it's a cyclic sequence, so the thousands digit might not directly affect the next term's units digit. Let me think.Wait, the problem says: "the hundreds, tens, and units digits of each term become the thousands, hundreds, and tens digits of the next term respectively." So, if I have a term like ABCD, where A is thousands, B is hundreds, C is tens, D is units, then the next term would be BCD followed by something. But what is the thousands digit of the next term? It's B, the hundreds digit of the current term. Then, the hundreds digit of the next term is C, the tens digit of the current term. The tens digit of the next term is D, the units digit of the current term. So, the units digit of the next term is... Hmm, the problem doesn't specify, but since it's a four-digit number, the units digit must come from somewhere. Maybe it's the thousands digit of the next term? Wait, no, the thousands digit is already B.Wait, maybe the thousands digit of the next term is the hundreds digit of the current term, so the next term is BCD followed by something. But since it's a four-digit number, the units digit must be something else. Maybe the thousands digit of the next term is B, hundreds is C, tens is D, and units is... Maybe it's the thousands digit of the term after that? Hmm, this is confusing.Wait, let's look at the example: 2475, 4758, 7582, 5824. Let's see how this sequence progresses.First term: 2475Second term: 4758Third term: 7582Fourth term: 5824So, starting from 2475, the hundreds, tens, units digits are 4,7,5. The next term is 4758. So, 4 becomes thousands, 7 becomes hundreds, 5 becomes tens, and then 8 is the units digit. So, where does 8 come from? It must be the thousands digit of the next term? Wait, no, because the next term after 4758 is 7582. So, 7 becomes thousands, 5 becomes hundreds, 8 becomes tens, and 2 is units. So, 2 is the thousands digit of the term after 7582, which is 5824.Wait, so 5824 is the fourth term, and then the next term would be 8245, right? Because hundreds, tens, units of 5824 are 8,2,4, so the next term would be 8245. But since it's a cyclic sequence, does it loop back? The problem says "the tens and units digits of the last term become the thousands and hundreds digits of the first term." So, in the example, the last term is 5824. Its tens and units digits are 2 and 4. So, the first term should have thousands and hundreds digits as 2 and 4, which it does: 2475. So, the sequence cycles back.So, in general, for such a cyclic sequence, each term is generated by shifting the hundreds, tens, and units digits to the left, making them the thousands, hundreds, and tens digits of the next term, and the units digit of the next term is the thousands digit of the term after that? Wait, no, because in the example, the units digit of 2475 is 5, which becomes the tens digit of the next term, 4758. Then, the units digit of 4758 is 8, which becomes the tens digit of 7582, and so on. So, actually, each term's units digit becomes the tens digit of the next term, and the thousands digit of the next term is the hundreds digit of the current term.Wait, let me write this down more clearly.Let’s denote each term as a four-digit number: WXYZ.- The next term would be XYZ followed by something. But since it's four digits, it's XYZA, where A is the thousands digit of the next term. But according to the problem, the hundreds, tens, and units digits of each term become the thousands, hundreds, and tens digits of the next term. So, if the current term is WXYZ, then the next term is XYZW? Wait, no, because the thousands digit of the next term is the hundreds digit of the current term, which is X. The hundreds digit of the next term is Y, the tens digit is Z, and the units digit is... Hmm, the problem doesn't specify where the units digit comes from. But in the example, 2475 becomes 4758. So, 2475: thousands=2, hundreds=4, tens=7, units=5. Next term is 4758: thousands=4, hundreds=7, tens=5, units=8. So, the units digit of the next term is 8, which is the thousands digit of the term after that? Wait, no, the term after 4758 is 7582, which has thousands digit 7. So, 8 is not the thousands digit of the next term. Hmm.Wait, maybe the units digit of the next term is the thousands digit of the term after next? That seems complicated. Alternatively, perhaps the units digit cycles back to the thousands digit after four terms. Let me check the example:2475, 4758, 7582, 5824.So, starting from 2475, the next term is 4758, which is shifting 4,7,5 to the left, making them thousands, hundreds, tens, and then 8 is the units digit. Then, 4758 becomes 7582: shifting 7,5,8 to the left, making them thousands, hundreds, tens, and 2 is the units digit. Then, 7582 becomes 5824: shifting 5,8,2 to the left, making them thousands, hundreds, tens, and 4 is the units digit. Then, 5824 becomes 8245: shifting 8,2,4 to the left, making them thousands, hundreds, tens, and 5 is the units digit. But in the example, it stops at 5824, which cycles back to 2475 because the tens and units digits of 5824 are 2 and 4, which become the thousands and hundreds digits of the first term, 2475.So, in this case, the sequence is cyclic with four terms, each shifting the hundreds, tens, units digits to the left, and the units digit of each term is the thousands digit of the term three places ahead? Hmm, maybe not. Alternatively, perhaps each term's units digit is the thousands digit of the next term's next term? This is getting confusing.Wait, maybe it's better to think of the entire sequence as a cycle where each term is generated by shifting the digits left, and the last term connects back to the first term by shifting its last two digits to the front. So, in the example, 5824 has tens and units digits 2 and 4, which become the thousands and hundreds digits of the first term, 2475.So, in general, for such a cyclic sequence, each term is a four-digit number, and the next term is formed by taking the hundreds, tens, and units digits of the current term and appending a new digit to make it four digits. But where does this new digit come from? It seems like it's the thousands digit of the term that comes after the next term. Wait, that might not make sense.Alternatively, perhaps the entire sequence is such that each term is a rotation of the previous term's digits. For example, 2475 rotated left by one digit would be 4752, but in the example, it's 4758. So, that doesn't match. Hmm.Wait, maybe the sequence isn't just a simple rotation. Let me look at the example again:2475, 4758, 7582, 5824.So, 2475 to 4758: 4,7,5 become 4,7,5 in the thousands, hundreds, tens places, and 8 is added as the units digit.4758 to 7582: 7,5,8 become 7,5,8 in thousands, hundreds, tens, and 2 is added as units.7582 to 5824: 5,8,2 become 5,8,2 in thousands, hundreds, tens, and 4 is added as units.5824 to ????: 8,2,4 become 8,2,4 in thousands, hundreds, tens, and what's added as units? It should cycle back to 2475, so the units digit should be 5, making the next term 8245, but in the example, it stops at 5824.Wait, maybe the sequence only has four terms, and the last term connects back to the first term by shifting its last two digits to the front. So, 5824 has tens and units digits 2 and 4, which become the thousands and hundreds digits of the first term, 2475. So, the units digit of the last term is 4, which becomes the hundreds digit of the first term? Wait, no, the first term is 2475, which has thousands digit 2 and hundreds digit 4, which are the tens and units digits of the last term, 5824.So, in general, for a cyclic sequence of four terms, each term is generated by shifting the hundreds, tens, and units digits to the left, making them the thousands, hundreds, and tens digits of the next term, and the units digit of each term is determined by the thousands digit of the term three places ahead? Hmm, maybe not.Alternatively, perhaps each term's units digit is the thousands digit of the term that comes after it. Wait, in the example:2475: units digit 5, next term is 4758, thousands digit 4. Not the same.4758: units digit 8, next term is 7582, thousands digit 7. Not the same.7582: units digit 2, next term is 5824, thousands digit 5. Not the same.5824: units digit 4, next term would be 8245, thousands digit 8. Not the same.So, that doesn't seem to be the case.Wait, maybe the units digit of each term is the thousands digit of the term two places ahead? Let's see:2475: units digit 5, two terms ahead is 7582, thousands digit 7. Not the same.4758: units digit 8, two terms ahead is 5824, thousands digit 5. Not the same.7582: units digit 2, two terms ahead would be 8245, thousands digit 8. Not the same.Hmm, not matching either.Alternatively, maybe the units digit cycles through the thousands digits of the entire sequence. In the example, the thousands digits are 2,4,7,5. The units digits are 5,8,2,4. So, 5,8,2,4 are the units digits, which are the thousands digits shifted by one: 2,4,7,5 shifted left by one would be 4,7,5,2, but that's not matching. Wait, 5,8,2,4 is not a shift of 2,4,7,5.Wait, maybe the units digits are the thousands digits of the previous term? Let's see:2475: units digit 5, previous term would be ???, which is not given.4758: units digit 8, previous term is 2475, thousands digit 2. Not matching.7582: units digit 2, previous term is 4758, thousands digit 4. Not matching.5824: units digit 4, previous term is 7582, thousands digit 7. Not matching.Hmm, this is confusing. Maybe I'm overcomplicating it.Let me try to think differently. The problem says that the hundreds, tens, and units digits of each term become the thousands, hundreds, and tens digits of the next term respectively. So, if I have a term ABCD, the next term is BCDA? Wait, no, because BCDA would be B, C, D, A, but the problem says the hundreds, tens, units become thousands, hundreds, tens. So, the thousands digit of the next term is B, hundreds is C, tens is D, and units is... It doesn't specify, but in the example, it's 2475 to 4758, so units digit becomes 8, which is not related to A=2.Wait, maybe the units digit of the next term is the thousands digit of the term after next? Let's see:In the example, 2475 to 4758: units digit 5 becomes tens digit 5 in 4758, and units digit 8 is new.4758 to 7582: units digit 8 becomes tens digit 8 in 7582, and units digit 2 is new.7582 to 5824: units digit 2 becomes tens digit 2 in 5824, and units digit 4 is new.5824 to ????: units digit 4 becomes tens digit 4 in the next term, which would be 8245, and units digit 5 is new, which is the thousands digit of the first term.So, in this case, the units digit of each term is the thousands digit of the term two places ahead. For example, 2475's units digit is 5, and two terms ahead is 7582, which has thousands digit 7, which is not 5. Wait, no.Wait, 2475's units digit is 5, and the term after next is 7582, which has thousands digit 7. Not matching.Wait, maybe the units digit of each term is the thousands digit of the term three places ahead. 2475's units digit is 5, and three terms ahead would be 5824, which has thousands digit 5. Oh, that matches! Similarly, 4758's units digit is 8, and three terms ahead would be ???, but in the example, it's only four terms, so three terms ahead of 4758 is 5824, which has thousands digit 5, not 8. Hmm, not matching.Wait, maybe it's cyclic, so three terms ahead of 4758 would be 5824, which has thousands digit 5, but 4758's units digit is 8, so that doesn't match.Wait, maybe the units digit of each term is the thousands digit of the term that comes after it. So, 2475's units digit is 5, next term is 4758, thousands digit is 4. Not matching.This is getting too confusing. Maybe I should try to represent the sequence algebraically.Let’s denote the four terms as N1, N2, N3, N4.Each term is a four-digit number, so we can write them as:N1 = 1000a + 100b + 10c + dN2 = 1000b + 100c + 10d + eN3 = 1000c + 100d + 10e + fN4 = 1000d + 100e + 10f + gBut wait, according to the problem, the hundreds, tens, and units digits of each term become the thousands, hundreds, and tens digits of the next term. So, for N1 = abcd, N2 should be bcde, N3 should be cdef, N4 should be defg, and then N5 should be efgh, but since it's cyclic, N5 should be the same as N1. So, N5 = efgh = abcd.Wait, that makes sense. So, if we have four terms, N1, N2, N3, N4, then N5 = N1, which would be efgh = abcd. So, the sequence cycles every four terms.Therefore, the four terms are:N1 = abcdN2 = bcdeN3 = cdefN4 = defgAnd then N5 = efgh = abcd, so e = a, f = b, g = c, h = d.Wait, so that means the digits cycle every four terms. So, the digits are a, b, c, d, a, b, c, d, etc.Wait, let me write this out:N1 = 1000a + 100b + 10c + dN2 = 1000b + 100c + 10d + a (since e = a)N3 = 1000c + 100d + 10a + b (since f = b)N4 = 1000d + 100a + 10b + c (since g = c)Then, N5 = 1000a + 100b + 10c + d = N1, which checks out.So, the four terms are:N1 = 1000a + 100b + 10c + dN2 = 1000b + 100c + 10d + aN3 = 1000c + 100d + 10a + bN4 = 1000d + 100a + 10b + cNow, let's compute the sum T = N1 + N2 + N3 + N4.Let's add them up:N1 = 1000a + 100b + 10c + dN2 = 1000b + 100c + 10d + aN3 = 1000c + 100d + 10a + bN4 = 1000d + 100a + 10b + cAdding them together:T = (1000a + 100b + 10c + d) + (1000b + 100c + 10d + a) + (1000c + 100d + 10a + b) + (1000d + 100a + 10b + c)Let's collect like terms:For a:1000a + a + 10a + 100a = (1000 + 1 + 10 + 100)a = 1111aFor b:100b + 1000b + b + 10b = (100 + 1000 + 1 + 10)b = 1111bFor c:10c + 100c + 1000c + c = (10 + 100 + 1000 + 1)c = 1111cFor d:d + 10d + 100d + 1000d = (1 + 10 + 100 + 1000)d = 1111dSo, T = 1111a + 1111b + 1111c + 1111d = 1111(a + b + c + d)Therefore, T is equal to 1111 multiplied by the sum of the digits a, b, c, d.Now, 1111 is a known number. Let's factorize it:1111 = 101 * 11Because 101 * 11 = 1111.So, T = 11 * 101 * (a + b + c + d)Therefore, T is always divisible by both 11 and 101. Since the question asks for the largest prime factor that always divides T, we need to compare 11 and 101. 101 is larger than 11, so 101 is the largest prime factor that always divides T.Wait, but let me double-check. The sum T is 1111 times the sum of the digits a, b, c, d. So, T is always a multiple of 1111, which factors into 11 and 101. Therefore, regardless of the values of a, b, c, d, T will always be divisible by both 11 and 101. Since 101 is a prime number and larger than 11, it is the largest prime factor that always divides T.So, the answer should be D) 101.

💡Okay, so I have this problem where there are four real numbers, x₁, x₂, x₃, and x₄. The conditions given are that the sum of these numbers is zero, and the sum of their seventh powers is also zero. I need to find the value of the expression u, which is defined as x₄ times (x₄ + x₁) times (x₄ + x₂) times (x₄ + x₃). First, I should probably write down what I know. The sum of the numbers is zero, so:x₁ + x₂ + x₃ + x₄ = 0.And the sum of their seventh powers is also zero:x₁⁷ + x₂⁷ + x₃⁷ + x₄⁷ = 0.I need to find u = x₄(x₄ + x₁)(x₄ + x₂)(x₄ + x₃). Hmm, that looks like a product of x₄ with three terms, each of which is x₄ plus one of the other variables. Maybe I can expand this expression to see if it simplifies somehow.Let me try expanding u step by step. First, multiply x₄ with each term:u = x₄(x₄ + x₁)(x₄ + x₂)(x₄ + x₃).If I let’s first compute (x₄ + x₁)(x₄ + x₂)(x₄ + x₃). Maybe expanding this will help. Let's denote A = x₄, and the other terms as B = x₁, C = x₂, D = x₃. So, the expression becomes (A + B)(A + C)(A + D). Expanding this, I can use the formula for expanding three binomials:(A + B)(A + C)(A + D) = A³ + A²(B + C + D) + A(BC + BD + CD) + BCD.So, substituting back, we get:(x₄ + x₁)(x₄ + x₂)(x₄ + x₃) = x₄³ + x₄²(x₁ + x₂ + x₃) + x₄(x₁x₂ + x₁x₃ + x₂x₃) + x₁x₂x₃.Now, since we have the sum x₁ + x₂ + x₃ + x₄ = 0, that means x₁ + x₂ + x₃ = -x₄. So, I can substitute that into the expression:= x₄³ + x₄²(-x₄) + x₄(x₁x₂ + x₁x₃ + x₂x₃) + x₁x₂x₃.Simplify term by term:First term: x₄³.Second term: x₄²(-x₄) = -x₄³.Third term: x₄ times (x₁x₂ + x₁x₃ + x₂x₃).Fourth term: x₁x₂x₃.So, combining the first and second terms: x₄³ - x₄³ = 0.So, the expression simplifies to:x₄(x₁x₂ + x₁x₃ + x₂x₃) + x₁x₂x₃.Therefore, u = x₄ times this expression:u = x₄ [x₄(x₁x₂ + x₁x₃ + x₂x₃) + x₁x₂x₃].Let me write that out:u = x₄²(x₁x₂ + x₁x₃ + x₂x₃) + x₄x₁x₂x₃.Hmm, so u is expressed in terms of x₄ squared times the sum of the products of the other variables, plus x₄ times the product of all three other variables.I wonder if I can relate this to the given conditions. The sum of the variables is zero, and the sum of their seventh powers is zero. Maybe I need to use symmetric polynomials or Newton's identities here.Let me recall that for symmetric polynomials, we have elementary symmetric sums. For four variables, the elementary symmetric sums are:σ₁ = x₁ + x₂ + x₃ + x₄,σ₂ = x₁x₂ + x₁x₃ + x₁x₄ + x₂x₃ + x₂x₄ + x₃x₄,σ₃ = x₁x₂x₃ + x₁x₂x₄ + x₁x₃x₄ + x₂x₃x₄,σ₄ = x₁x₂x₃x₄.Given that σ₁ = 0, since the sum of the variables is zero.Now, I need to find expressions for σ₂, σ₃, and σ₄. Maybe using Newton's identities, which relate power sums to elementary symmetric sums.Newton's identities say that for each k, the power sum S_k = x₁ᵏ + x₂ᵏ + x₃ᵏ + x₄ᵏ can be expressed in terms of the elementary symmetric sums.Given that S₁ = σ₁ = 0,S₂ = σ₁S₁ - 2σ₂ = -2σ₂,S₃ = σ₁S₂ - σ₂S₁ + 3σ₃ = 3σ₃,S₄ = σ₁S₃ - σ₂S₂ + σ₃S₁ - 4σ₄ = -σ₂S₂ - 4σ₄,But since σ₁ = 0, some terms drop out.Similarly, for higher powers, but we have S₇ = 0.This might get complicated, but maybe I can find relations between σ₂, σ₃, and σ₄.Wait, let's see:Given that S₁ = 0,S₂ = -2σ₂,S₃ = 3σ₃,S₄ = -σ₂S₂ - 4σ₄,But S₂ = -2σ₂, so S₄ = -σ₂(-2σ₂) - 4σ₄ = 2σ₂² - 4σ₄.Similarly, S₅ can be expressed as:S₅ = σ₁S₄ - σ₂S₃ + σ₃S₂ - σ₄S₁,But σ₁ = 0, so S₅ = -σ₂S₃ + σ₃S₂.Substituting S₂ = -2σ₂ and S₃ = 3σ₃,S₅ = -σ₂(3σ₃) + σ₃(-2σ₂) = -3σ₂σ₃ - 2σ₂σ₃ = -5σ₂σ₃.Similarly, S₆ = σ₁S₅ - σ₂S₄ + σ₃S₃ - σ₄S₂,Again, σ₁ = 0, so S₆ = -σ₂S₄ + σ₃S₃ - σ₄S₂.Substituting S₄ = 2σ₂² - 4σ₄, S₃ = 3σ₃, S₂ = -2σ₂,S₆ = -σ₂(2σ₂² - 4σ₄) + σ₃(3σ₃) - σ₄(-2σ₂)= -2σ₂³ + 4σ₂σ₄ + 3σ₃² + 2σ₂σ₄= -2σ₂³ + 6σ₂σ₄ + 3σ₃².Similarly, S₇ can be expressed as:S₇ = σ₁S₆ - σ₂S₅ + σ₃S₄ - σ₄S₃,Again, σ₁ = 0, so S₇ = -σ₂S₅ + σ₃S₄ - σ₄S₃.We know S₇ = 0, so:0 = -σ₂S₅ + σ₃S₄ - σ₄S₃.Substituting S₅ = -5σ₂σ₃, S₄ = 2σ₂² - 4σ₄, S₃ = 3σ₃,We get:0 = -σ₂(-5σ₂σ₃) + σ₃(2σ₂² - 4σ₄) - σ₄(3σ₃)= 5σ₂²σ₃ + 2σ₂²σ₃ - 4σ₃σ₄ - 3σ₃σ₄= (5σ₂²σ₃ + 2σ₂²σ₃) + (-4σ₃σ₄ - 3σ₃σ₄)= 7σ₂²σ₃ - 7σ₃σ₄.Factor out 7σ₃:0 = 7σ₃(σ₂² - σ₄).So, either σ₃ = 0 or σ₂² = σ₄.So, we have two cases:Case 1: σ₃ = 0.Case 2: σ₂² = σ₄.Let me analyze each case.Case 1: σ₃ = 0.If σ₃ = 0, then from S₃ = 3σ₃, we have S₃ = 0.Also, from S₅ = -5σ₂σ₃, since σ₃ = 0, S₅ = 0.Similarly, from S₇ = 0, which is given.So, in this case, σ₃ = 0.Now, let's think about the polynomial whose roots are x₁, x₂, x₃, x₄. Since σ₁ = 0, the polynomial is:t⁴ + σ₂t² + σ₄.Because the coefficients of t³ and t are zero (since σ₁ = 0 and σ₃ = 0).So, f(t) = t⁴ + σ₂t² + σ₄.Given that σ₂² = σ₄ in Case 2, but in Case 1, σ₃ = 0, so σ₄ is independent.Wait, in Case 1, σ₃ = 0, but σ₂ and σ₄ can be anything, as long as the polynomial is t⁴ + σ₂t² + σ₄.But we also have that the sum of the seventh powers is zero.Wait, but in Case 1, σ₃ = 0, so S₃ = 0, S₅ = 0, S₇ = 0.But how does that affect the roots?Alternatively, maybe the roots come in pairs of inverses or something.Wait, since the polynomial is t⁴ + σ₂t² + σ₄, which is a biquadratic, so roots come in pairs ±a, ±b.So, suppose the roots are a, -a, b, -b.Then, x₁ = a, x₂ = -a, x₃ = b, x₄ = -b.Then, σ₁ = a - a + b - b = 0, which is consistent.σ₂ = (a)(-a) + (a)(b) + (a)(-b) + (-a)(b) + (-a)(-b) + (b)(-b).Wait, let's compute σ₂:σ₂ = x₁x₂ + x₁x₃ + x₁x₄ + x₂x₃ + x₂x₄ + x₃x₄.Substituting:= (a)(-a) + (a)(b) + (a)(-b) + (-a)(b) + (-a)(-b) + (b)(-b)= -a² + ab - ab - ab + ab - b²Simplify:- a² - b².So, σ₂ = - (a² + b²).Similarly, σ₃ = x₁x₂x₃ + x₁x₂x₄ + x₁x₃x₄ + x₂x₃x₄.Substituting:= (a)(-a)(b) + (a)(-a)(-b) + (a)(b)(-b) + (-a)(b)(-b)= -a²b + a²b - ab² + ab²= 0.So, σ₃ = 0, which is consistent with Case 1.σ₄ = x₁x₂x₃x₄ = (a)(-a)(b)(-b) = a²b².So, σ₄ = (ab)².Now, in this case, the roots are ±a, ±b.So, let's compute u = x₄(x₄ + x₁)(x₄ + x₂)(x₄ + x₃).Assuming x₄ is one of the roots, say x₄ = -b.Then, x₁ = a, x₂ = -a, x₃ = b.So, u = (-b)(-b + a)(-b - a)(-b + b).Wait, let's compute each term:x₄ + x₁ = -b + a,x₄ + x₂ = -b - a,x₄ + x₃ = -b + b = 0.So, u = (-b)(-b + a)(-b - a)(0) = 0.Similarly, if x₄ is any of the roots, say x₄ = a, then:u = a(a + x₁)(a + x₂)(a + x₃).But x₁, x₂, x₃ would be -a, b, -b.So, u = a(a - a)(a + b)(a - b) = a(0)(a + b)(a - b) = 0.Similarly, if x₄ is b or -b, we get zero.Therefore, in Case 1, u = 0.Case 2: σ₂² = σ₄.So, σ₄ = σ₂².From the polynomial f(t) = t⁴ + σ₂t² + σ₄, substituting σ₄ = σ₂², we get:f(t) = t⁴ + σ₂t² + σ₂².This can be written as (t² + σ₂/2)² + (σ₂² - σ₂²/4) = (t² + σ₂/2)² + (3σ₂²/4).Wait, actually, let me factor it differently.Wait, f(t) = t⁴ + σ₂t² + σ₂².Let me set y = t², then f(t) becomes y² + σ₂ y + σ₂².This quadratic in y is y² + σ₂ y + σ₂².The discriminant is σ₂² - 4σ₂² = -3σ₂², which is negative unless σ₂ = 0.So, if σ₂ ≠ 0, the roots are complex, but the problem states that x_i are real numbers. Therefore, the only possibility is that σ₂ = 0.If σ₂ = 0, then f(t) = t⁴ + 0 + 0 = t⁴.So, all roots are zero: x₁ = x₂ = x₃ = x₄ = 0.Therefore, u = x₄(x₄ + x₁)(x₄ + x₂)(x₄ + x₃) = 0*(0 + 0)*(0 + 0)*(0 + 0) = 0.So, in Case 2, u = 0.Therefore, in both cases, u = 0.So, the value of u is 0.**Final Answer**boxed{0}

💡First, I need to determine the total amount of meal available. The meal can cater to either 70 adults or 90 children, which means the total meal is equivalent to 70 adult meals or 90 child meals.Next, I'll calculate how much meal is consumed when 42 adults have their meal. Since 1 adult meal is equivalent to 90/70 child meals, 42 adults consume 42 * (90/70) = 54 child meals.Now, I'll find out how much meal is left after 42 adults have eaten. The total meal is 90 child meals, so the remaining meal is 90 - 54 = 36 child meals.Therefore, the remaining meal can cater to 36 children.