💡Okay, so I have this problem where Stacy and Heather are 40 miles apart and walking towards each other. Heather walks at 5 miles per hour, and Stacy walks 1 mile per hour faster, so that's 6 miles per hour. Heather starts 24 minutes after Stacy. I need to find out how far Heather has walked when they meet.First, I should probably figure out how much of a head start Stacy has. Since Heather starts 24 minutes later, I need to convert that into hours because the speeds are in miles per hour. 24 minutes is 24/60 hours, which is 0.4 hours. So, Stacy has a 0.4-hour head start.Now, let's think about the distances each person covers. Let's say Heather walks for 't' hours before they meet. That means Stacy has been walking for 't + 0.4' hours because she started earlier.Heather's distance would be her speed multiplied by time, so 5t miles. Stacy's distance would be her speed multiplied by her time, which is 6(t + 0.4) miles.Since they're walking towards each other, the sum of the distances they cover should equal the initial distance between them, which is 40 miles. So, I can set up the equation:5t + 6(t + 0.4) = 40Let me solve this equation step by step. First, expand the terms:5t + 6t + 2.4 = 40Combine like terms:11t + 2.4 = 40Subtract 2.4 from both sides:11t = 37.6Now, divide both sides by 11:t = 37.6 / 11Calculating that gives:t ≈ 3.418 hoursSo, Heather has been walking for approximately 3.418 hours when they meet. To find out how far she has walked, multiply her time by her speed:Distance = 5 * 3.418 ≈ 17.09 milesTherefore, Heather has walked approximately 17.09 miles from her original starting point when they meet.

💡Okay, so I have this problem here: I need to find the number of lines that pass through the point (1, 2) and have equal intercepts on the coordinate axes. The options are A: 2 lines, B: 1 line, C: 3 lines, D: 4 lines.Hmm, let's start by understanding what it means for a line to have equal intercepts on the coordinate axes. I think that means the x-intercept and the y-intercept are the same. So, if a line crosses the x-axis at (a, 0) and the y-axis at (0, a), then both intercepts are equal to 'a'. Wait, but the intercepts could also be negative, right? So maybe it's the absolute values that are equal? Or do they have to be exactly the same number, including the sign? Hmm, I think they have to be exactly the same, including the sign. So if the x-intercept is (a, 0), then the y-intercept is (0, a). So both are 'a', whether positive or negative.Now, how do I write the equation of such a line? I know that the intercept form of a line is x/a + y/b = 1, where 'a' is the x-intercept and 'b' is the y-intercept. Since the intercepts are equal, a = b. So the equation becomes x/a + y/a = 1, which simplifies to (x + y)/a = 1, or x + y = a.So the equation of the line is x + y = a. Now, this line passes through the point (1, 2). So I can substitute x = 1 and y = 2 into the equation to find 'a'.Let me do that: 1 + 2 = a, so a = 3. Therefore, the equation of the line is x + y = 3. That's one line.Wait, but earlier I thought that the intercepts could be negative. So if a is negative, say a = -k where k is positive, then the equation would be x + y = -k. Let me check if that's possible.If a is negative, say a = -k, then the equation becomes x + y = -k. Does this line pass through (1, 2)? Let's substitute: 1 + 2 = -k, so 3 = -k, which means k = -3. But k was supposed to be positive, so that doesn't make sense. Hmm, maybe I made a mistake here.Alternatively, maybe I should consider the slope-intercept form of the line. The slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept. If the x-intercept is equal to the y-intercept, then the x-intercept is also b. The x-intercept occurs where y = 0, so 0 = mx + b, which gives x = -b/m. So the x-intercept is (-b/m, 0). Since the x-intercept is equal to the y-intercept, which is b, we have -b/m = b.So, -b/m = b. Let's solve for m. Multiply both sides by m: -b = b*m. Then, divide both sides by b (assuming b ≠ 0): -1 = m. So the slope m is -1.Wait, so if m = -1, then the equation is y = -x + b. Now, this line passes through (1, 2). Let's substitute: 2 = -1 + b, so b = 3. Therefore, the equation is y = -x + 3, which is the same as x + y = 3. So that's the same line I found earlier.But earlier I thought maybe a could be negative, but that didn't work out. So does that mean there's only one such line?Wait, but I feel like there might be another line. Maybe if the intercepts are both zero? Because if a line passes through the origin, then both intercepts are zero, which are equal. So let's check if the line passes through (1, 2) and the origin.If the line passes through (0, 0) and (1, 2), then the slope m is (2 - 0)/(1 - 0) = 2. So the equation is y = 2x. Let's check the intercepts: x-intercept is when y = 0, so 0 = 2x, which gives x = 0. Similarly, y-intercept is when x = 0, which is y = 0. So both intercepts are zero, which are equal. So that's another line: y = 2x.Wait, so now I have two lines: y = -x + 3 and y = 2x. So that would be two lines. So the answer should be A: 2 lines.But let me double-check. Is there any other possibility? For example, if the intercepts are both negative, but as I saw earlier, that didn't work because when I tried a negative intercept, it led to a contradiction. So maybe only these two lines.Alternatively, maybe I can think of the general equation of a line with equal intercepts. Let me try that.The general equation is x + y = a, as I had before. But I also considered the case where the line passes through the origin, which is y = 2x. So that's another case.Wait, but in the case of y = 2x, the intercepts are both zero, which are equal. So that's a valid case. So that's another line.So in total, I have two lines: one with positive intercepts (x + y = 3) and one passing through the origin (y = 2x). So that's two lines.Wait, but earlier when I tried to set a = -k, I got a contradiction, so maybe that's not possible. So only these two lines exist.Therefore, the answer should be A: 2 lines.

💡Alright, so I've been given this problem about finding all integers ( n geq 2 ) for which there exists an integer ( m ) and a polynomial ( P(x) ) with integer coefficients satisfying three specific conditions. Let me try to unpack this step by step.First, let's restate the problem in my own words to make sure I understand it. We need to find all integers ( n ) starting from 2 upwards such that there's some integer ( m > 1 ) with ( gcd(m, n) = 1 ), and a polynomial ( P(x) ) with integer coefficients. The conditions are:1. The numbers ( P(0), P^2(0), ldots, P^{m-1}(0) ) are not divisible by ( n ).2. The number ( P^m(0) ) is divisible by ( n ).Here, ( P^k(0) ) means applying the polynomial ( P ) ( k ) times starting from 0. So, ( P^1(0) = P(0) ), ( P^2(0) = P(P(0)) ), and so on.Okay, so I need to find all such ( n ). Let me think about how to approach this.First, let's consider what it means for ( P^m(0) ) to be divisible by ( n ) while none of the previous iterations ( P^k(0) ) for ( k < m ) are divisible by ( n ). This seems similar to the concept of order in group theory, where an element has a certain order if raising it to that power gives the identity, but no smaller power does. Maybe there's an analogy here with polynomials modulo ( n ).Since ( P(x) ) has integer coefficients, evaluating it at integers will give integers. So, ( P(0) ) is just the constant term of the polynomial. Let's denote ( P(0) = a ). Then ( P^2(0) = P(a) ), ( P^3(0) = P(P(a)) ), etc.So, we're looking for a sequence ( a, P(a), P(P(a)), ldots ) such that the first ( m ) terms are not divisible by ( n ), but the ( m )-th term is. Also, ( m ) must be greater than 1 and coprime with ( n ).Hmm, okay. Maybe it's useful to think modulo ( n ). Let's consider the sequence ( P^k(0) ) modulo ( n ). The conditions become:1. ( P(0) notequiv 0 pmod{n} )2. ( P^2(0) notequiv 0 pmod{n} )3. ...4. ( P^{m-1}(0) notequiv 0 pmod{n} )5. ( P^m(0) equiv 0 pmod{n} )So, the sequence ( P^k(0) ) modulo ( n ) has period ( m ), meaning it takes ( m ) steps to reach 0 modulo ( n ). Also, ( m ) must be coprime with ( n ).This seems similar to the concept of the order of an element in a group, where the order is the smallest positive integer ( m ) such that ( g^m = e ). In this case, our "element" is the polynomial ( P ), and the "operation" is composition. The "group" would be the set of integers modulo ( n ) under composition, but I'm not sure if that's a group.Wait, actually, composition of polynomials isn't a group operation because not every polynomial has an inverse under composition. So maybe that's not the right way to think about it.Alternatively, maybe I can think of the sequence ( P^k(0) ) modulo ( n ) as a kind of dynamical system. The key is that it takes exactly ( m ) iterations to reach 0 modulo ( n ), and none before that. So, the period is ( m ).Given that ( m ) must be coprime with ( n ), this might relate to properties of ( n ) in terms of its prime factors.Let me try to consider small values of ( n ) to see if I can spot a pattern.**Case 1: ( n ) is prime**Let's say ( n = p ) is a prime number. Then, we need to find ( m ) and ( P(x) ) such that ( m > 1 ), ( gcd(m, p) = 1 ), and the sequence ( P(0), P^2(0), ldots, P^{m-1}(0) ) are not divisible by ( p ), but ( P^m(0) ) is.Since ( p ) is prime, ( gcd(m, p) = 1 ) implies that ( m ) is not a multiple of ( p ). So, ( m ) can be any integer greater than 1 that is not a multiple of ( p ).Now, can we always find such a polynomial ( P(x) ) for any prime ( p )?Let me try to construct such a polynomial. Let's consider ( P(x) = x + c ) where ( c ) is an integer. Then, ( P(0) = c ), ( P^2(0) = c + c = 2c ), ( P^3(0) = 3c ), and so on. So, ( P^k(0) = kc ).We want ( kc notequiv 0 pmod{p} ) for ( k = 1, 2, ldots, m-1 ), and ( mc equiv 0 pmod{p} ).Since ( p ) is prime, ( mc equiv 0 pmod{p} ) implies that either ( m equiv 0 pmod{p} ) or ( c equiv 0 pmod{p} ). But ( m ) is not a multiple of ( p ) (since ( gcd(m, p) = 1 )), so ( c ) must be a multiple of ( p ). However, if ( c ) is a multiple of ( p ), then ( P(0) = c equiv 0 pmod{p} ), which violates the first condition that ( P(0) ) is not divisible by ( p ).So, using a linear polynomial ( P(x) = x + c ) doesn't seem to work because it forces ( c ) to be a multiple of ( p ), which contradicts the first condition.Maybe I need a different kind of polynomial. Let's try a quadratic polynomial. Suppose ( P(x) = x^2 + c ). Then, ( P(0) = c ), ( P^2(0) = c^2 + c ), ( P^3(0) = (c^2 + c)^2 + c ), and so on.This seems more complicated, but perhaps we can choose ( c ) such that the sequence ( P^k(0) ) modulo ( p ) cycles with period ( m ).Alternatively, maybe a better approach is to consider the multiplicative order of some element modulo ( p ). Since ( p ) is prime, the multiplicative group modulo ( p ) is cyclic of order ( p - 1 ). So, if we can find an element ( a ) in this group with order ( m ), then ( a^m equiv 1 pmod{p} ), but ( a^k notequiv 1 pmod{p} ) for ( k < m ).But in our problem, we need ( P^m(0) equiv 0 pmod{p} ), not 1. So, maybe instead of multiplicative order, we need additive order? But additive order is always 1 in the additive group modulo ( p ), since ( 1 cdot 1 = 1 ), which isn't helpful.Wait, perhaps I'm overcomplicating this. Let's think about the sequence ( P^k(0) ) modulo ( p ). We need it to take exactly ( m ) steps to reach 0. So, starting from some ( a_0 = P(0) ), then ( a_1 = P(a_0) ), ( a_2 = P(a_1) ), etc., until ( a_{m-1} neq 0 pmod{p} ) and ( a_m = 0 pmod{p} ).This is similar to a permutation of the residues modulo ( p ), but since we're mapping integers to integers, it's more of a function on the residues.I recall that in finite fields, functions can have various properties, including being permutations. Maybe we can use a permutation polynomial that maps a certain number of steps before hitting 0.Alternatively, perhaps we can construct ( P(x) ) such that it acts as a shift in some way. For example, if we can define ( P(x) ) such that ( P(x) = x + 1 ) when ( x neq p - 1 ), and ( P(p - 1) = 0 ). But this would require ( P(x) ) to have different definitions depending on ( x ), which isn't a polynomial.Wait, polynomials are fixed functions; they can't have conditional definitions. So, I need a polynomial that, when iterated, will eventually reach 0 after ( m ) steps, without hitting 0 earlier.Maybe I can use a polynomial that increments by 1 each time, but wraps around modulo ( p ). But as I saw earlier, linear polynomials don't work because they force ( c ) to be a multiple of ( p ), which conflicts with the first condition.Alternatively, maybe a higher-degree polynomial can achieve this. For example, consider ( P(x) = x + 1 ) modulo ( p ). Then, ( P^k(0) = k ). So, ( P^m(0) = m ). To have ( m equiv 0 pmod{p} ), we need ( m ) to be a multiple of ( p ). But ( gcd(m, p) = 1 ), so ( m ) can't be a multiple of ( p ). Therefore, this approach doesn't work.Wait, but if ( m ) is coprime with ( p ), then ( m ) can't be a multiple of ( p ), so ( P^m(0) = m ) can't be 0 modulo ( p ). Therefore, using ( P(x) = x + 1 ) doesn't satisfy the third condition.Hmm, maybe I need a different kind of polynomial. Let's think about polynomials that can have cycles. For example, consider ( P(x) = x + c ) modulo ( p ). The sequence ( P^k(0) ) will be ( kc ) modulo ( p ). For this to reach 0 after ( m ) steps, we need ( mc equiv 0 pmod{p} ). Since ( m ) is coprime with ( p ), ( c ) must be 0 modulo ( p ), which again causes ( P(0) = c equiv 0 pmod{p} ), violating the first condition.So, linear polynomials seem problematic because they force ( c ) to be 0 modulo ( p ), which conflicts with the first condition.What about quadratic polynomials? Let's try ( P(x) = x^2 + c ). Then, ( P(0) = c ), ( P^2(0) = c^2 + c ), ( P^3(0) = (c^2 + c)^2 + c ), and so on. This seems more flexible, but it's also more complicated.Suppose we choose ( c = 1 ). Then, ( P(0) = 1 ), ( P^2(0) = 1 + 1 = 2 ), ( P^3(0) = 4 + 1 = 5 ), etc. This sequence is just the Fibonacci sequence modulo ( p ). Depending on ( p ), this sequence might eventually reach 0.But I'm not sure if we can control the period ( m ) in this way. It might be difficult to ensure that ( P^m(0) equiv 0 pmod{p} ) for a specific ( m ) coprime with ( p ).Maybe a better approach is to consider the multiplicative inverse. If we can find a polynomial ( P(x) ) such that ( P(x) equiv x + 1 pmod{p} ), but adjusted so that after ( m ) steps, it wraps around to 0.Wait, but as I saw earlier, linear polynomials don't work because they force ( c ) to be 0 modulo ( p ). So, maybe I need a polynomial that's not linear.Alternatively, perhaps I can use a polynomial that's a permutation polynomial modulo ( p ), meaning it permutes the residues modulo ( p ). If such a polynomial has a cycle of length ( m ), then iterating it ( m ) times would bring us back to the starting point, but we need it to reach 0 after ( m ) steps.Wait, that's a bit different. If the polynomial permutes the residues, then starting from some ( a ), iterating it ( m ) times would bring us back to ( a ), not necessarily to 0. So, unless 0 is part of the cycle, which it isn't because we start at ( P(0) neq 0 ).Hmm, maybe I'm overcomplicating this. Let's think about the problem differently. Since ( P^m(0) equiv 0 pmod{n} ), and ( P^k(0) notequiv 0 pmod{n} ) for ( k < m ), this suggests that the sequence ( P^k(0) ) modulo ( n ) has a period of ( m ). So, the order of ( P ) modulo ( n ) is ( m ).But what does it mean for a polynomial to have an order modulo ( n )? I'm not sure if this is a standard concept, but perhaps we can define it similarly to the order of an element in a group.In any case, the key seems to be that ( m ) must divide the order of some structure related to ( n ). Since ( m ) is coprime with ( n ), this might impose certain restrictions on ( n ).Wait, actually, ( m ) being coprime with ( n ) is a crucial condition. This means that ( m ) shares no common factors with ( n ), which might relate to the structure of the multiplicative group modulo ( n ).But I'm not sure how to connect this directly. Maybe I should consider the prime factorization of ( n ). If ( n ) is a prime power, say ( p^e ), then perhaps the problem reduces to finding such ( m ) and ( P(x) ) modulo ( p^e ).Let me try to think about ( n ) being a prime power first. Suppose ( n = p^e ) for some prime ( p ) and integer ( e geq 1 ). Then, we need to find ( m ) and ( P(x) ) such that:1. ( m > 1 ) and ( gcd(m, p^e) = 1 ), which implies ( m ) is not divisible by ( p ).2. ( P(0), P^2(0), ldots, P^{m-1}(0) notequiv 0 pmod{p^e} ).3. ( P^m(0) equiv 0 pmod{p^e} ).Maybe I can construct such a polynomial for prime powers and then use the Chinese Remainder Theorem to extend it to composite ( n ).Let's try to construct ( P(x) ) for ( n = p ). Suppose ( p ) is an odd prime. Let's consider the polynomial ( P(x) = x + 1 ). Then, ( P^k(0) = k ). So, ( P^m(0) = m ). To have ( m equiv 0 pmod{p} ), we need ( m ) to be a multiple of ( p ). But ( gcd(m, p) = 1 ), so ( m ) can't be a multiple of ( p ). Therefore, this doesn't work.Wait, but if ( P(x) = x + c ), then ( P^k(0) = kc ). To have ( kc equiv 0 pmod{p} ) for ( k = m ), we need ( c equiv 0 pmod{p} ), which again causes ( P(0) = c equiv 0 pmod{p} ), violating the first condition.So, linear polynomials don't seem to work. Maybe a higher-degree polynomial can help. Let's try ( P(x) = x^2 + c ). Then, ( P(0) = c ), ( P^2(0) = c^2 + c ), ( P^3(0) = (c^2 + c)^2 + c ), and so on.Suppose we choose ( c = 1 ). Then, ( P(0) = 1 ), ( P^2(0) = 2 ), ( P^3(0) = 5 ), ( P^4(0) = 26 ), etc. This sequence is similar to the Fibonacci sequence, and it might eventually reach 0 modulo ( p ) for some ( m ). But I'm not sure if we can control ( m ) to be coprime with ( p ).Alternatively, maybe we can choose ( c ) such that ( P(x) ) has a specific behavior modulo ( p ). For example, if ( P(x) equiv x + 1 pmod{p} ), but with higher terms that vanish modulo ( p ). But I'm not sure how to ensure that ( P^m(0) equiv 0 pmod{p} ) without causing ( P(0) equiv 0 pmod{p} ).Wait, perhaps I need to use a polynomial that's not linear but still has a controlled behavior. Maybe something like ( P(x) = x + x^p ). But modulo ( p ), ( x^p equiv x pmod{p} ) by Fermat's little theorem, so ( P(x) equiv 2x pmod{p} ). Then, ( P^k(0) equiv 2^k x pmod{p} ). So, ( P^m(0) equiv 2^m x pmod{p} ). To have ( 2^m x equiv 0 pmod{p} ), we need ( x equiv 0 pmod{p} ), which again causes ( P(0) equiv 0 pmod{p} ), violating the first condition.Hmm, this is tricky. Maybe I need to think differently. Instead of trying to construct ( P(x) ) directly, perhaps I can use properties of the multiplicative group modulo ( p ).Since ( gcd(m, p) = 1 ), ( m ) is invertible modulo ( p ). So, maybe I can find an element ( a ) in the multiplicative group modulo ( p ) such that ( a^m equiv 1 pmod{p} ), but ( a^k notequiv 1 pmod{p} ) for ( k < m ). This would mean that ( a ) has order ( m ) in the multiplicative group.But in our problem, we need ( P^m(0) equiv 0 pmod{p} ), not 1. So, maybe instead of working in the multiplicative group, I need to work in the additive group.In the additive group modulo ( p ), every element has order ( p ), so that doesn't seem helpful.Wait, perhaps I can combine both additive and multiplicative structures. For example, consider a polynomial that acts as a multiplicative shift and an additive shift. But I'm not sure how to formalize this.Alternatively, maybe I can use a polynomial that's a permutation polynomial with a specific cycle structure. For example, if ( P(x) ) permutes the residues modulo ( p ) and has a cycle of length ( m ), then iterating it ( m ) times would bring us back to the starting point. But we need it to reach 0 after ( m ) steps, not return to the starting point.Hmm, perhaps I'm overcomplicating this. Let me try to think about specific examples.Let's take ( p = 5 ) and see if I can find ( m ) and ( P(x) ) such that ( m > 1 ), ( gcd(m, 5) = 1 ), and ( P^m(0) equiv 0 pmod{5} ) while ( P^k(0) notequiv 0 pmod{5} ) for ( k < m ).Possible values for ( m ) are 2, 3, 4, 6, etc., since ( gcd(m, 5) = 1 ).Let's try ( m = 2 ). We need ( P(0) notequiv 0 pmod{5} ) and ( P(P(0)) equiv 0 pmod{5} ).Let me choose ( P(x) = x + 1 ). Then, ( P(0) = 1 ), ( P^2(0) = 2 ). Neither is 0 modulo 5, so this doesn't work.Wait, but we need ( P^2(0) equiv 0 pmod{5} ). So, ( P(P(0)) equiv 0 pmod{5} ). Let's set ( P(0) = a ), then ( P(a) equiv 0 pmod{5} ).So, we need ( P(a) equiv 0 pmod{5} ), but ( a notequiv 0 pmod{5} ).Let me choose ( a = 1 ). Then, ( P(1) equiv 0 pmod{5} ). So, ( P(x) equiv 0 pmod{5} ) when ( x = 1 ). Let's define ( P(x) = (x - 1) cdot Q(x) ), where ( Q(x) ) is some polynomial with integer coefficients.But then, ( P(0) = (-1) cdot Q(0) ). We need ( P(0) notequiv 0 pmod{5} ), so ( Q(0) notequiv 0 pmod{5} ).Let me choose ( Q(x) = 1 ). Then, ( P(x) = x - 1 ). Then, ( P(0) = -1 equiv 4 pmod{5} ), which is not 0. ( P^2(0) = P(4) = 4 - 1 = 3 ), which is not 0. ( P^3(0) = P(3) = 3 - 1 = 2 ), still not 0. ( P^4(0) = 1 ), and ( P^5(0) = 0 ). So, ( m = 5 ), but ( gcd(5, 5) = 5 neq 1 ), which violates the condition that ( gcd(m, n) = 1 ).So, this doesn't work. Maybe I need a different ( Q(x) ).Let me try ( Q(x) = 2 ). Then, ( P(x) = 2(x - 1) ). Then, ( P(0) = -2 equiv 3 pmod{5} ), ( P^2(0) = P(3) = 2(3 - 1) = 4 ), ( P^3(0) = P(4) = 2(4 - 1) = 6 equiv 1 pmod{5} ), ( P^4(0) = P(1) = 2(1 - 1) = 0 pmod{5} ). So, ( m = 4 ), and ( gcd(4, 5) = 1 ). This works!So, for ( n = 5 ), ( m = 4 ), and ( P(x) = 2(x - 1) ) satisfies all the conditions.Great, so for ( n = 5 ), such a polynomial exists. Let me check another prime, say ( p = 7 ).We need ( m > 1 ) with ( gcd(m, 7) = 1 ). Possible ( m ) values are 2, 3, 4, 5, 6, 8, etc.Let me try ( m = 3 ). I need ( P(0) notequiv 0 pmod{7} ), ( P^2(0) notequiv 0 pmod{7} ), and ( P^3(0) equiv 0 pmod{7} ).Let me define ( P(x) = (x - a) cdot Q(x) ), where ( a ) is such that ( P(a) equiv 0 pmod{7} ), but ( P(0) notequiv 0 pmod{7} ).Let me choose ( a = 1 ). Then, ( P(x) = (x - 1) cdot Q(x) ). Let me choose ( Q(x) = 2 ). Then, ( P(x) = 2(x - 1) ).Compute the sequence:- ( P(0) = 2(-1) = -2 equiv 5 pmod{7} )- ( P^2(0) = P(5) = 2(5 - 1) = 8 equiv 1 pmod{7} )- ( P^3(0) = P(1) = 2(1 - 1) = 0 pmod{7} )So, ( m = 3 ), and ( gcd(3, 7) = 1 ). This works!So, for ( n = 7 ), such a polynomial exists.Wait, so it seems that for any prime ( p ), we can choose ( m = p - 1 ) and construct a polynomial ( P(x) = (x - 1) cdot Q(x) ) with ( Q(0) ) not divisible by ( p ). Then, ( P^{p-1}(0) equiv 0 pmod{p} ), and none of the previous iterations are 0 modulo ( p ).But wait, in the case of ( p = 5 ), ( m = 4 ), which is ( p - 1 ). For ( p = 7 ), ( m = 6 ), but I chose ( m = 3 ) instead. So, maybe ( m ) can be any divisor of ( p - 1 ) as long as it's coprime with ( p ).Wait, ( p - 1 ) is the order of the multiplicative group modulo ( p ). So, if ( m ) divides ( p - 1 ), then there exists an element of order ( m ) in the multiplicative group. But in our case, we're dealing with additive sequences, not multiplicative.Hmm, perhaps I'm conflating additive and multiplicative orders here. Let me think again.In the examples above, for ( p = 5 ), ( m = 4 ) worked, which is ( p - 1 ). For ( p = 7 ), ( m = 3 ) worked, which is a divisor of ( p - 1 = 6 ). So, maybe ( m ) can be any divisor of ( p - 1 ) that is coprime with ( p ).But ( m ) must be coprime with ( p ), which it is since ( m ) divides ( p - 1 ) and ( p ) is prime.So, perhaps for any prime ( p ), there exists such an ( m ) and ( P(x) ). Therefore, all prime numbers ( p ) satisfy the condition.Now, what about composite numbers? Let's consider ( n = 6 ). We need to find ( m > 1 ) with ( gcd(m, 6) = 1 ), so ( m ) must be coprime with 6, i.e., ( m ) can be 5, 7, 11, etc.But let's try ( m = 5 ). We need a polynomial ( P(x) ) such that:1. ( P(0), P^2(0), P^3(0), P^4(0) ) are not divisible by 6.2. ( P^5(0) ) is divisible by 6.Since 6 is composite, we can use the Chinese Remainder Theorem. If we can find such a polynomial modulo 2 and modulo 3, then we can combine them to get a polynomial modulo 6.So, let's first consider modulo 2. We need ( P^5(0) equiv 0 pmod{2} ), and ( P^k(0) notequiv 0 pmod{2} ) for ( k < 5 ).But modulo 2, the possible residues are 0 and 1. If ( P(0) equiv 1 pmod{2} ), then ( P^k(0) equiv 1 pmod{2} ) for all ( k ), which never reaches 0. If ( P(0) equiv 0 pmod{2} ), then ( P^1(0) equiv 0 pmod{2} ), which violates the condition that ( P^k(0) notequiv 0 pmod{2} ) for ( k < 5 ).Wait, this seems impossible. Because modulo 2, the only possibilities are 0 and 1. If ( P(0) equiv 1 pmod{2} ), then all iterations stay at 1. If ( P(0) equiv 0 pmod{2} ), then it's already 0. So, there's no way to have ( P^5(0) equiv 0 pmod{2} ) without having ( P^k(0) equiv 0 pmod{2} ) for some ( k < 5 ).Therefore, it's impossible to satisfy the conditions for ( n = 6 ).Wait, but maybe I'm missing something. Perhaps the polynomial can behave differently modulo 2 and modulo 3, and when combined, it works modulo 6.Let me try to construct such a polynomial. Suppose ( P(x) equiv x + 1 pmod{2} ) and ( P(x) equiv x + 1 pmod{3} ). Then, by the Chinese Remainder Theorem, ( P(x) equiv x + 1 pmod{6} ).Then, ( P^k(0) = k pmod{6} ). So, ( P^5(0) = 5 pmod{6} ), which is not 0. So, this doesn't work.Alternatively, maybe choose different behaviors modulo 2 and 3. For example, modulo 2, have ( P(x) equiv x + 1 pmod{2} ), and modulo 3, have ( P(x) equiv x + 2 pmod{3} ).Then, ( P(0) equiv 1 pmod{2} ) and ( 2 pmod{3} ). So, ( P(0) equiv 5 pmod{6} ).( P^2(0) = P(5) equiv P(1) pmod{2} ) and ( P(2) pmod{3} ).Modulo 2: ( P(1) = 1 + 1 = 2 equiv 0 pmod{2} ).Modulo 3: ( P(2) = 2 + 2 = 4 equiv 1 pmod{3} ).So, ( P^2(0) equiv 0 pmod{2} ) and ( 1 pmod{3} ), which is 4 modulo 6.But ( P^2(0) equiv 4 pmod{6} ), which is not 0 modulo 6, but it is 0 modulo 2. However, the condition requires that ( P^k(0) notequiv 0 pmod{6} ) for ( k < 5 ). But ( P^2(0) equiv 4 pmod{6} ), which is not 0 modulo 6, so it's okay.Continue:( P^3(0) = P(4) ).Modulo 2: ( P(4) = 4 + 1 = 5 equiv 1 pmod{2} ).Modulo 3: ( P(4) = 4 + 2 = 6 equiv 0 pmod{3} ).So, ( P^3(0) equiv 1 pmod{2} ) and ( 0 pmod{3} ), which is 3 modulo 6.Again, not 0 modulo 6.( P^4(0) = P(3) ).Modulo 2: ( P(3) = 3 + 1 = 4 equiv 0 pmod{2} ).Modulo 3: ( P(3) = 3 + 2 = 5 equiv 2 pmod{3} ).So, ( P^4(0) equiv 0 pmod{2} ) and ( 2 pmod{3} ), which is 2 modulo 6.Still not 0 modulo 6.( P^5(0) = P(2) ).Modulo 2: ( P(2) = 2 + 1 = 3 equiv 1 pmod{2} ).Modulo 3: ( P(2) = 2 + 2 = 4 equiv 1 pmod{3} ).So, ( P^5(0) equiv 1 pmod{2} ) and ( 1 pmod{3} ), which is 1 modulo 6. Not 0.Hmm, this approach isn't working. Maybe I need a different polynomial.Alternatively, perhaps I can use a polynomial that's 0 modulo 2 after 5 steps and 0 modulo 3 after 5 steps, but not before. But I'm not sure how to construct such a polynomial.Wait, maybe I can use the fact that 5 is coprime with both 2 and 3. So, perhaps I can construct a polynomial that has order 5 modulo 2 and modulo 3.But modulo 2, the multiplicative group has order 1, so any element has order 1. Similarly, modulo 3, the multiplicative group has order 2. So, it's impossible to have an element of order 5 in either group.Therefore, it's impossible to have such a polynomial for ( n = 6 ).This suggests that composite numbers might not satisfy the condition, but I need to verify this.Wait, let's try ( n = 4 ). Maybe it's possible.We need ( m > 1 ) with ( gcd(m, 4) = 1 ). So, ( m ) can be 3, 5, 7, etc.Let me try ( m = 3 ). We need ( P(0), P^2(0) notequiv 0 pmod{4} ), and ( P^3(0) equiv 0 pmod{4} ).Let me choose ( P(x) = x + 1 ). Then, ( P(0) = 1 ), ( P^2(0) = 2 ), ( P^3(0) = 3 ). None of these are 0 modulo 4. So, this doesn't work.Alternatively, let me choose ( P(x) = x + 2 ). Then, ( P(0) = 2 ), ( P^2(0) = 4 equiv 0 pmod{4} ). But ( m = 2 ), which is not coprime with 4. So, this violates the condition.Wait, but ( m = 3 ). Let me try a different polynomial. Suppose ( P(x) = x^2 + 1 ). Then, ( P(0) = 1 ), ( P^2(0) = 1 + 1 = 2 ), ( P^3(0) = 4 + 1 = 5 equiv 1 pmod{4} ). This cycles without reaching 0.Alternatively, ( P(x) = x^2 + 2 ). Then, ( P(0) = 2 ), ( P^2(0) = 4 + 2 = 6 equiv 2 pmod{4} ), ( P^3(0) = 4 + 2 = 6 equiv 2 pmod{4} ). This also doesn't reach 0.Hmm, maybe it's impossible for ( n = 4 ) as well.Wait, let me try ( P(x) = x + 3 ). Then, ( P(0) = 3 ), ( P^2(0) = 6 equiv 2 pmod{4} ), ( P^3(0) = 5 equiv 1 pmod{4} ), ( P^4(0) = 4 equiv 0 pmod{4} ). So, ( m = 4 ), but ( gcd(4, 4) = 4 neq 1 ). So, this doesn't satisfy the condition.Alternatively, ( P(x) = 2x + 1 ). Then, ( P(0) = 1 ), ( P^2(0) = 2*1 + 1 = 3 ), ( P^3(0) = 2*3 + 1 = 7 equiv 3 pmod{4} ), ( P^4(0) = 2*3 + 1 = 7 equiv 3 pmod{4} ). This cycles without reaching 0.Hmm, maybe it's impossible for ( n = 4 ) as well.This suggests that composite numbers might not satisfy the condition, while prime numbers do. But I need to verify this more carefully.Wait, let's consider ( n = 9 ), which is ( 3^2 ). Let me see if I can find ( m ) and ( P(x) ) such that ( m > 1 ), ( gcd(m, 9) = 1 ), and the sequence ( P^k(0) ) modulo 9 reaches 0 at ( k = m ) and not before.Possible ( m ) values are 2, 4, 5, 7, 8, etc.Let me try ( m = 2 ). We need ( P(0) notequiv 0 pmod{9} ), ( P^2(0) equiv 0 pmod{9} ).Let me choose ( P(x) = x + 3 ). Then, ( P(0) = 3 ), ( P^2(0) = 6 ), which is not 0 modulo 9. So, this doesn't work.Alternatively, ( P(x) = x + 6 ). Then, ( P(0) = 6 ), ( P^2(0) = 12 equiv 3 pmod{9} ), ( P^3(0) = 9 equiv 0 pmod{9} ). So, ( m = 3 ), but ( gcd(3, 9) = 3 neq 1 ). So, this violates the condition.Wait, maybe I need a different polynomial. Let me try ( P(x) = x^2 + 3 ). Then, ( P(0) = 3 ), ( P^2(0) = 9 + 3 = 12 equiv 3 pmod{9} ), ( P^3(0) = 9 + 3 = 12 equiv 3 pmod{9} ). This cycles without reaching 0.Alternatively, ( P(x) = x^2 + 6 ). Then, ( P(0) = 6 ), ( P^2(0) = 36 + 6 = 42 equiv 6 pmod{9} ), ( P^3(0) = 36 + 6 = 42 equiv 6 pmod{9} ). Again, cycles without reaching 0.Hmm, maybe it's impossible for ( n = 9 ) as well.Wait, perhaps I need to use a different approach. Let me think about the structure of ( n ). If ( n ) is a prime power, say ( p^e ), then perhaps the multiplicative group modulo ( p^e ) has order ( (p-1)p^{e-1} ). So, if ( m ) divides ( (p-1)p^{e-1} ) and is coprime with ( p ), then ( m ) must divide ( p-1 ).Therefore, for ( n = p^e ), ( m ) must be a divisor of ( p-1 ). So, if ( p-1 ) has divisors coprime with ( p ), which it does since ( p ) is prime, then such ( m ) exists.But in the case of ( n = 4 ), which is ( 2^2 ), the multiplicative group modulo 4 has order 2, which is ( 2 ). So, ( m ) must divide 2 and be coprime with 4, which means ( m = 1 ) or ( m = 2 ). But ( m > 1 ), so ( m = 2 ). However, ( gcd(2, 4) = 2 neq 1 ), so it's impossible.Similarly, for ( n = 9 ), the multiplicative group modulo 9 has order 6. So, ( m ) must divide 6 and be coprime with 9, which means ( m ) can be 1, 2, 4, 5, etc. But ( m > 1 ), so possible ( m = 2, 4, 5 ). However, in our earlier attempts, we couldn't find such a polynomial for ( m = 2 ) or ( m = 4 ).Wait, maybe I need to think about higher-degree polynomials. Let me try ( P(x) = x + 3 ) modulo 9. Then, ( P(0) = 3 ), ( P^2(0) = 6 ), ( P^3(0) = 9 equiv 0 pmod{9} ). So, ( m = 3 ), but ( gcd(3, 9) = 3 neq 1 ). So, this doesn't work.Alternatively, ( P(x) = x + 6 ). Then, ( P(0) = 6 ), ( P^2(0) = 12 equiv 3 pmod{9} ), ( P^3(0) = 9 equiv 0 pmod{9} ). Again, ( m = 3 ), which is not coprime with 9.Hmm, maybe it's impossible for ( n = 9 ) as well.This suggests that for prime powers ( p^e ), it's possible to find such ( m ) and ( P(x) ) only if ( p ) is odd. Because for ( p = 2 ), ( n = 4 ) didn't work, and for ( p = 3 ), ( n = 9 ) didn't work either.Wait, but earlier for ( p = 5 ) and ( p = 7 ), it worked. So, maybe it's possible for odd primes and their powers, but not for ( p = 2 ).Let me check ( n = 8 ). Let's see if it's possible.We need ( m > 1 ) with ( gcd(m, 8) = 1 ), so ( m ) can be 3, 5, 7, etc.Let me try ( m = 3 ). We need ( P(0), P^2(0) notequiv 0 pmod{8} ), and ( P^3(0) equiv 0 pmod{8} ).Let me choose ( P(x) = x + 2 ). Then, ( P(0) = 2 ), ( P^2(0) = 4 ), ( P^3(0) = 6 ), which is not 0 modulo 8.Alternatively, ( P(x) = x + 4 ). Then, ( P(0) = 4 ), ( P^2(0) = 8 equiv 0 pmod{8} ). But ( m = 2 ), which is not coprime with 8.Wait, maybe a different polynomial. Let me try ( P(x) = x^2 + 2 ). Then, ( P(0) = 2 ), ( P^2(0) = 4 + 2 = 6 ), ( P^3(0) = 36 + 2 = 38 equiv 6 pmod{8} ). This cycles without reaching 0.Alternatively, ( P(x) = x^2 + 4 ). Then, ( P(0) = 4 ), ( P^2(0) = 16 + 4 = 20 equiv 4 pmod{8} ), ( P^3(0) = 16 + 4 = 20 equiv 4 pmod{8} ). Cycles without reaching 0.Hmm, maybe it's impossible for ( n = 8 ) as well.This suggests that for ( p = 2 ), even powers like ( 4, 8 ) don't work, but for odd primes, it works.Wait, but earlier for ( p = 3 ), ( n = 9 ) didn't work. So, maybe it's only for primes, not their powers.Wait, let me try ( n = 25 ), which is ( 5^2 ). Let me see if I can find ( m ) and ( P(x) ) such that ( m > 1 ), ( gcd(m, 25) = 1 ), and ( P^m(0) equiv 0 pmod{25} ) while ( P^k(0) notequiv 0 pmod{25} ) for ( k < m ).Let me choose ( m = 4 ), which is coprime with 25. Let me try to construct ( P(x) ) such that ( P^4(0) equiv 0 pmod{25} ) and none of the previous iterations are 0 modulo 25.Let me define ( P(x) = 2(x - 1) ) as before. Then, modulo 5, we saw that ( P^4(0) equiv 0 pmod{5} ). Let's see what happens modulo 25.Compute the sequence:- ( P(0) = 2(-1) = -2 equiv 23 pmod{25} )- ( P^2(0) = P(23) = 2(23 - 1) = 2*22 = 44 equiv 19 pmod{25} )- ( P^3(0) = P(19) = 2(19 - 1) = 2*18 = 36 equiv 11 pmod{25} )- ( P^4(0) = P(11) = 2(11 - 1) = 2*10 = 20 pmod{25} )- ( P^5(0) = P(20) = 2(20 - 1) = 2*19 = 38 equiv 13 pmod{25} )- ( P^6(0) = P(13) = 2(13 - 1) = 2*12 = 24 pmod{25} )- ( P^7(0) = P(24) = 2(24 - 1) = 2*23 = 46 equiv 21 pmod{25} )- ( P^8(0) = P(21) = 2(21 - 1) = 2*20 = 40 equiv 15 pmod{25} )- ( P^9(0) = P(15) = 2(15 - 1) = 2*14 = 28 equiv 3 pmod{25} )- ( P^{10}(0) = P(3) = 2(3 - 1) = 4 pmod{25} )- ( P^{11}(0) = P(4) = 2(4 - 1) = 6 pmod{25} )- ( P^{12}(0) = P(6) = 2(6 - 1) = 10 pmod{25} )- ( P^{13}(0) = P(10) = 2(10 - 1) = 18 pmod{25} )- ( P^{14}(0) = P(18) = 2(18 - 1) = 34 equiv 9 pmod{25} )- ( P^{15}(0) = P(9) = 2(9 - 1) = 16 pmod{25} )- ( P^{16}(0) = P(16) = 2(16 - 1) = 30 equiv 5 pmod{25} )- ( P^{17}(0) = P(5) = 2(5 - 1) = 8 pmod{25} )- ( P^{18}(0) = P(8) = 2(8 - 1) = 14 pmod{25} )- ( P^{19}(0) = P(14) = 2(14 - 1) = 26 equiv 1 pmod{25} )- ( P^{20}(0) = P(1) = 2(1 - 1) = 0 pmod{25} )So, ( m = 20 ), but ( gcd(20, 25) = 5 neq 1 ). So, this doesn't satisfy the condition.Wait, but I was trying ( m = 4 ). It seems that ( P^4(0) equiv 20 pmod{25} ), which is not 0. So, maybe ( m = 20 ) is the order, but it's not coprime with 25.Hmm, maybe I need a different polynomial. Let me try ( P(x) = 2(x - 1) ) again, but see if I can find a smaller ( m ).Wait, in the sequence above, ( P^{20}(0) equiv 0 pmod{25} ), and none of the previous terms were 0 modulo 25. So, ( m = 20 ), but ( gcd(20, 25) = 5 neq 1 ). Therefore, this doesn't satisfy the condition.Alternatively, maybe there's a polynomial with a smaller ( m ) that's coprime with 25. Let me try ( m = 4 ).Let me define ( P(x) = x + 5 ). Then, ( P(0) = 5 ), ( P^2(0) = 10 ), ( P^3(0) = 15 ), ( P^4(0) = 20 ), ( P^5(0) = 25 equiv 0 pmod{25} ). So, ( m = 5 ), but ( gcd(5, 25) = 5 neq 1 ). So, this doesn't work.Alternatively, ( P(x) = x + 10 ). Then, ( P(0) = 10 ), ( P^2(0) = 20 ), ( P^3(0) = 30 equiv 5 pmod{25} ), ( P^4(0) = 15 ), ( P^5(0) = 25 equiv 0 pmod{25} ). Again, ( m = 5 ), which is not coprime with 25.Hmm, maybe it's impossible for ( n = 25 ) as well.This suggests that for prime powers ( p^e ) where ( p ) is an odd prime, it's possible to find such ( m ) and ( P(x) ) only when ( e = 1 ), i.e., when ( n ) is prime. For higher powers, it seems impossible because the required ( m ) would share a common factor with ( n ).Therefore, putting it all together, it seems that the integers ( n geq 2 ) for which such ( m ) and ( P(x) ) exist are exactly the prime numbers.**Final Answer**The integers ( n ) are all prime numbers. Thus, the solution is boxed{n text{ is prime}}.

💡Alright, so I've got this problem here about a function f(x). Let me try to parse it step by step. First, it says that f(x) is an even function. Okay, so that means f(-x) = f(x) for all x. Got that. Even functions are symmetric about the y-axis, right? So whatever the graph looks like on the positive side, it mirrors on the negative side.Next, it gives me another condition: f(1 + x) = f(3 - x). Hmm, that seems like some kind of symmetry or periodicity condition. Maybe I can manipulate this to find a period or something about the function.Then, it tells me that when x is between -2 and 0, f(x) = 3^x. So for negative values from -2 to 0, the function is just an exponential function with base 3. That's straightforward.The question is asking for a_n = f(n) where n is a positive integer, specifically n = 2011. So I need to find f(2011). Since 2011 is a large number, I probably need to find some periodicity or pattern in the function to simplify this.Alright, let's start by understanding the condition f(1 + x) = f(3 - x). Maybe I can substitute a variable to make this clearer. Let me set y = 1 + x. Then, x = y - 1. Substituting into the equation, we get f(y) = f(3 - (y - 1)) = f(4 - y). So, f(y) = f(4 - y). That means that for any y, the function value at y is the same as at 4 - y. So, f(y) = f(4 - y). That looks like a reflection around the line y = 2. So, the function is symmetric about the line x = 2. Interesting. So, it's symmetric about both x = 0 (because it's even) and x = 2. Wait, if a function is symmetric about two different vertical lines, does that imply it's periodic? I think so. The distance between x = 0 and x = 2 is 2 units. So, if it's symmetric about both, then the period should be twice the distance between these lines, so 4 units. Let me verify that.If f(y) = f(4 - y), and since f is even, f(4 - y) = f(y - 4). So, f(y) = f(y - 4). That means f is periodic with period 4. So, f(y + 4) = f(y) for all y. So, the function repeats every 4 units. That's good to know.So, knowing that f is periodic with period 4, I can find f(2011) by finding the remainder when 2011 is divided by 4. Let me compute that.2011 divided by 4: 4*502 = 2008, so 2011 - 2008 = 3. So, 2011 = 4*502 + 3. Therefore, f(2011) = f(3). So, I need to find f(3).But I don't have an expression for f(x) when x is positive. However, since f is even, f(3) = f(-3). So, f(3) = f(-3). Now, I need to find f(-3). But the function is defined for x between -2 and 0 as 3^x. So, -3 is outside that interval. Hmm, but maybe I can use the periodicity again to bring it into the defined interval.Since the period is 4, f(-3) = f(-3 + 4) = f(1). So, f(-3) = f(1). Therefore, f(3) = f(-3) = f(1). So, f(3) = f(1). Now, I need to find f(1).But again, f(1) is not directly given. However, maybe I can use the condition f(1 + x) = f(3 - x). Let me set x = 0 here. Then, f(1 + 0) = f(3 - 0), so f(1) = f(3). Wait, that's just restating what I already know.Alternatively, maybe I can find another value. Let's see. Since f is even, f(1) = f(-1). And f(-1) is within the interval [-2, 0], so f(-1) = 3^{-1} = 1/3. Therefore, f(1) = f(-1) = 1/3. So, f(3) = f(1) = 1/3.Therefore, f(2011) = f(3) = 1/3. So, the answer should be 1/3, which is option D.Wait, let me double-check my steps to make sure I didn't make a mistake.1. f is even: f(-x) = f(x).2. f(1 + x) = f(3 - x) implies f(y) = f(4 - y), so symmetric about x = 2.3. Combining with evenness, f(y) = f(4 - y) = f(y - 4), so period 4.4. 2011 mod 4 is 3, so f(2011) = f(3).5. f(3) = f(-3) because f is even.6. f(-3) = f(1) because f is periodic with period 4.7. f(1) = f(-1) because f is even.8. f(-1) = 3^{-1} = 1/3.Yes, that seems correct. So, the answer is 1/3.

💡Alright, so I've got this problem about two shooters, A and B, who have taken five tests each. The scores are given, and there are two parts to this problem. First, I need to fill in a table with some statistics for both A and B, and then explain why the coach chose A over B. The second part is about how the variance changes when B takes an additional shot.Okay, let's start with part (1). The table has four columns: Mean, Mode, Median, and Variance. For A, the Mean is already given as 8, the Median is 8, and the Variance is 0.4. I need to find the Mode for A. For B, the Mode is given as 9, the Variance is 3.2, but I need to find the Mean and the Median.First, let's recall what each of these terms means:- **Mean**: This is the average of all the scores. You add up all the scores and divide by the number of scores.- **Mode**: This is the score that appears most frequently.- **Median**: This is the middle score when all the scores are arranged in order.- **Variance**: This measures how spread out the scores are from the mean.Starting with A's scores: 8, 8, 7, 8, 9.For the Mode, since 8 appears three times, which is more than any other number, the Mode for A is 8.For B's scores: 5, 9, 7, 10, 9.First, let's find the Mean for B. Adding up all the scores: 5 + 9 + 7 + 10 + 9 = 40. Then, divide by the number of scores, which is 5. So, 40 divided by 5 is 8. Therefore, the Mean for B is 8.Next, the Median for B. To find the Median, we need to arrange the scores in order. So, let's sort B's scores: 5, 7, 9, 9, 10. The middle number here is 9. So, the Median for B is 9.Now, looking back at the table, for A, the Mean is 8, Mode is 8, Median is 8, and Variance is 0.4. For B, the Mean is 8, Mode is 9, Median is 9, and Variance is 3.2.The coach chose A to participate in the shooting competition. Why? Well, both A and B have the same Mean score of 8, so their average performance is the same. However, looking at the Variance, A has a much lower Variance of 0.4 compared to B's 3.2. Variance measures how consistent the scores are. A lower Variance means that A's scores are closer to the Mean, indicating more consistent performance. On the other hand, B has a higher Variance, meaning B's scores are more spread out, which could indicate less reliability or more inconsistency in performance. Therefore, the coach probably chose A because A is more consistent, which is important in a competition where reliability is key.Moving on to part (2). If B shoots one more time and hits 8 points, how does this affect the Variance of B's shooting scores? Will it increase, decrease, or remain the same?First, let's recall that Variance is calculated as the average of the squared differences from the Mean. So, if we add another score, we need to see how it affects the overall spread of the scores.Originally, B's scores are: 5, 9, 7, 10, 9. The Mean is 8, and the Variance is 3.2.If B scores an additional 8, the new set of scores becomes: 5, 9, 7, 10, 9, 8. Now, let's calculate the new Mean. Adding up all the scores: 5 + 9 + 7 + 10 + 9 + 8 = 48. Divided by 6, the Mean is still 8.Now, let's calculate the new Variance. The formula for Variance is:[text{Variance} = frac{1}{N} sum_{i=1}^{N} (x_i - mu)^2]Where ( N ) is the number of scores, ( x_i ) are the individual scores, and ( mu ) is the Mean.Calculating each term:- ( (5 - 8)^2 = (-3)^2 = 9 )- ( (9 - 8)^2 = 1^2 = 1 )- ( (7 - 8)^2 = (-1)^2 = 1 )- ( (10 - 8)^2 = 2^2 = 4 )- ( (9 - 8)^2 = 1^2 = 1 )- ( (8 - 8)^2 = 0^2 = 0 )Adding these up: 9 + 1 + 1 + 4 + 1 + 0 = 16Now, divide by the number of scores, which is 6:[text{Variance} = frac{16}{6} approx 2.6667]Originally, the Variance was 3.2, and now it's approximately 2.6667. So, the Variance has decreased.Therefore, adding another score of 8 to B's results decreases the Variance because the new score is exactly the Mean, which doesn't add any new spread to the data. In fact, it brings the overall spread down because it's closer to the existing Mean.So, to summarize:1. For A, the Mode is 8. For B, the Mean is 8 and the Median is 9. The coach chose A because A has a lower Variance, indicating more consistent performance.2. Adding another score of 8 to B's results decreases the Variance.**Final Answer**(1) The completed table is:- A: Mode = boxed{8}- B: Mean = boxed{8}, Median = boxed{9}The coach chose A because A has a lower variance, indicating more consistent performance.(2) The variance of B's shooting score will boxed{text{decrease}}.

💡First, I'll assume the total weight of the original solution is 100 units to simplify the calculations.The original solution is 10% sugar by weight, which means it contains 10 units of sugar.One fourth of this solution is replaced, so one fourth of the sugar is removed. One fourth of 10 units is 2.5 units. After removing one fourth of the solution, we're left with 7.5 units of sugar.Next, we add the second solution, which is 38% sugar by weight. Since we removed one fourth of the original solution, we add one fourth of the total weight back in, which is 25 units.The amount of sugar in the second solution is 38% of 25 units, which is 0.38 * 25 = 9.5 units.Now, we add the sugar from the second solution to the remaining sugar from the first solution:7.5 units (from the first solution) + 9.5 units (from the second solution) = 17 units of sugar in the final solution.The final solution has a total weight of 100 units. Therefore, the percentage of sugar in the final solution is:(17 units of sugar / 100 units of solution) * 100% = 17%So, the sugar percentage of the final solution is 17%.