💡Okay, so I have this problem about infinite periodic fractions, which are basically repeating decimals. The fractions are given as a/b and c/d, where a and b are coprime, and c and d are coprime too. The condition is that every finite block of digits in the first fraction after the decimal point appears in the second fraction as well. I need to show that b equals d.Hmm, let's start by recalling what a repeating decimal looks like. A repeating decimal can be written as a fraction where the denominator is a number consisting of all 9s, depending on the period. For example, 0.overline{123} is 123/999. So, if a/b is a repeating decimal, then b must be a factor of some number like 999...9. Similarly for c/d.Given that every finite block in a/b appears in c/d, this suggests that the decimal expansion of a/b is somehow contained within the decimal expansion of c/d. But how does that relate to the denominators b and d?Maybe I should think about the periods of these decimals. The period of a repeating decimal is the smallest number of digits that repeat. For example, 1/7 has a period of 6 because 1/7 = 0.overline{142857}. So, if every block from a/b is in c/d, then the period of a/b must divide the period of c/d or something like that?Wait, but the problem says that every finite block appears, not just the repeating part. So, even the non-repeating part, if there is one, must also be present in the other decimal. But since both are fractions, they can have a non-repeating part followed by a repeating part. For example, 1/6 = 0.1overline{6}.So, maybe I need to consider both the non-repeating part and the repeating part. Let me denote the decimal expansion of a/b as A.overline{B}, where A is the non-repeating part and B is the repeating part. Similarly, c/d would be C.overline{D}.Given that every finite block in A.overline{B} appears in C.overline{D}, this should mean that the structure of A and B is somehow embedded within C and D. But how does that affect the denominators?I remember that the length of the repeating part (the period) is related to the denominator. Specifically, the period is the multiplicative order of 10 modulo the denominator, provided that the denominator is coprime to 10. Since a/b and c/d are in lowest terms, b and d must be coprime to 10 if the decimal is purely repeating. If there's a non-repeating part, then the denominator can have factors of 2 or 5.Wait, so if the decimal has a non-repeating part, that means the denominator has factors of 2 or 5, right? Because 1/2 = 0.5, which terminates, but 1/6 = 0.1overline{6} has a non-repeating part because 6 = 2 * 3.So, perhaps the denominators b and d must have the same factors of 2 and 5? But the problem says that every finite block in a/b appears in c/d. So, if a/b has a non-repeating part, then c/d must also have that non-repeating part or something that can contain it.But the problem is asking to show that b = d. So, maybe the periods have to be the same, and the non-repeating parts also have to be the same, forcing the denominators to be equal.Let me think about the periods first. Suppose the period of a/b is m, and the period of c/d is n. Since every block in a/b appears in c/d, then the period m must divide n? Or maybe n must divide m?Wait, actually, if every block in a/b appears in c/d, then c/d must have a period that is a multiple of m. Because otherwise, the repeating part of c/d might not contain all possible blocks from a/b.But I'm not sure. Maybe it's the other way around. If a/b has a longer period, then c/d must have a period that can accommodate all the blocks from a/b. So, perhaps the period of c/d must be a multiple of the period of a/b.But then, if the periods are related, how does that affect the denominators? The period is related to the denominator's co-factor with 10. So, if the periods are multiples, does that mean the denominators are related?Wait, maybe I should think about the denominators in terms of their prime factors. Since b and d are coprime to a and c respectively, and the decimal expansions are periodic, the denominators must be of the form 2^k * 5^m * q, where q is coprime to 10.The non-repeating part is determined by the maximum of k and m, right? So, if a/b has a non-repeating part of length t, then t is the maximum of the exponents of 2 and 5 in the prime factorization of b.Similarly for c/d. So, if every finite block in a/b appears in c/d, then the non-repeating part of a/b must be contained within the non-repeating part of c/d, or somehow within the repeating part.But wait, the non-repeating part is finite, so if c/d has a non-repeating part, it can only contain a finite number of blocks. But a/b has an infinite number of blocks because it's periodic. So, actually, the non-repeating part of a/b must be contained within the non-repeating part of c/d, and the repeating part must be compatible.Wait, no. Because the non-repeating part is finite, but the repeating part is infinite. So, if every finite block in a/b is in c/d, then the non-repeating part of a/b must be a suffix of the non-repeating part of c/d, or something like that.This is getting a bit tangled. Maybe I should approach it differently.Suppose that a/b and c/d have decimal expansions A.overline{B} and C.overline{D} respectively. The condition is that every finite substring of A.overline{B} appears in C.overline{D}.So, for example, the string A must appear somewhere in C.overline{D}. Similarly, the string AB must appear, and so on.Since A is finite, it must appear in C or in the repeating part D. But if A is longer than C, then part of A must be in C and the rest in D. But since D is repeating, the part of A in D must align with the repeating.Wait, but A is the non-repeating part of a/b. So, if A is longer than C, the non-repeating part of c/d, then part of A would have to be in the repeating part of c/d. But the repeating part is periodic, so that might impose some structure on A.Similarly, the repeating part B of a/b must be such that every finite substring of B is contained in D, the repeating part of c/d. So, perhaps B is a substring of D, or D is a substring of B?But since B is repeating, and D is repeating, if every substring of B is in D, then B must be a substring of D, but also D must be a substring of B because of the symmetry.Wait, but the problem only states that every substring of a/b is in c/d, not the other way around. So, maybe B is a substring of D, but D could be longer.But then, if B is a substring of D, then the period of B divides the period of D. So, the period of a/b divides the period of c/d.But we need to show that the denominators are equal. So, perhaps the periods are equal, and the non-repeating parts are equal as well.Wait, let me think about the denominators. If a/b has a denominator b, and c/d has denominator d, and both are in lowest terms, then b and d must have the same prime factors related to 2 and 5 for the non-repeating parts, and the same co-prime part for the repeating parts.But if the non-repeating parts are different, then the denominators would have different powers of 2 and 5, which would affect the non-repeating part.But since every finite block in a/b is in c/d, the non-repeating part of a/b must be contained within the non-repeating part of c/d or in the repeating part.But the repeating part is periodic, so if the non-repeating part of a/b is longer than that of c/d, then part of it must be in the repeating part, which is periodic. That might force the non-repeating part of a/b to be a suffix of the non-repeating part of c/d.Wait, this is getting complicated. Maybe I should look for a contradiction. Suppose that b ≠ d. Then, since both are denominators of fractions with the given property, perhaps their periods would be different, leading to a contradiction.Alternatively, maybe I can use the fact that if every substring of a/b is in c/d, then the decimal expansion of a/b is a subsequence of the decimal expansion of c/d. So, the structure of a/b is embedded within c/d.But how does that translate to the denominators? Maybe the denominators must have the same prime factors, especially regarding 2 and 5, because those affect the non-repeating part.Wait, if a/b has a non-repeating part, then c/d must have a non-repeating part that can contain all the finite blocks from a/b. But since a/b has an infinite number of blocks, the non-repeating part of c/d must be at least as long as the non-repeating part of a/b.But actually, the non-repeating part is finite, so if a/b has a longer non-repeating part, then c/d must have a non-repeating part that can contain it, but since c/d's non-repeating part is finite, it can only contain a finite number of blocks. So, maybe the non-repeating parts have to be the same length?Wait, no. Because even if c/d has a shorter non-repeating part, the non-repeating part of a/b could be split between the non-repeating part of c/d and the repeating part. But the repeating part is periodic, so the part of a/b's non-repeating part that goes into c/d's repeating part must align with the period.This is getting too vague. Maybe I should think about the denominators in terms of their relation to 10. The denominator b determines the period and the non-repeating part.If a/b has a non-repeating part of length k, then b must be divisible by 2^k or 5^k or both. Similarly for c/d.If every finite block in a/b is in c/d, then the non-repeating part of a/b must be compatible with the non-repeating part of c/d. So, maybe the non-repeating parts have to be the same length, forcing the powers of 2 and 5 in b and d to be the same.Similarly, for the repeating parts, since every substring of B is in D, the periods must be related. If B is a substring of D, then the period of B divides the period of D. But since the problem is symmetric (if we swap a/b and c/d, the same condition would hold), then the periods must be equal.Wait, but the problem doesn't state that every block in c/d is in a/b, only that every block in a/b is in c/d. So, maybe the period of a/b divides the period of c/d, but not necessarily the other way around.But then, how do we get b = d?Alternatively, maybe the fact that every block is present forces the denominators to have the same structure, meaning same powers of 2 and 5, and same co-prime part, hence same denominator.Wait, let me think about specific examples. Suppose a/b = 1/3 = 0.overline{3}, and c/d = 1/9 = 0.overline{1}. Does every block in 1/3 appear in 1/9? Well, 3 doesn't appear in 0.overline{1}, so that's not the case. So, in this case, the condition isn't satisfied.Another example: a/b = 1/6 = 0.1overline{6}, c/d = 1/3 = 0.overline{3}. Does every block in 0.1overline{6} appear in 0.overline{3}? The block "1" appears, "16" doesn't appear in 0.overline{3}, so the condition isn't satisfied.Wait, so maybe if a/b has a non-repeating part, c/d must have a non-repeating part that can contain it. So, if a/b has a non-repeating part of length k, then c/d must have a non-repeating part of at least length k.But since c/d's non-repeating part is finite, it can only contain a finite number of blocks. But a/b has an infinite number of blocks because it's periodic. So, actually, the non-repeating part of a/b must be a suffix of the non-repeating part of c/d.Wait, that might make sense. If A is the non-repeating part of a/b, then A must appear somewhere in c/d's expansion. Since c/d's expansion is C followed by repeating D, A must be a suffix of C or a combination of the end of C and the beginning of D.But since D is repeating, if A is longer than C, then part of A would have to be in D, which is periodic. But A is finite, so it can only cover a finite number of periods. Hmm, this is confusing.Maybe I should think about the denominators. If a/b and c/d have the same non-repeating part, then their denominators have the same powers of 2 and 5. And if their repeating parts have periods that are multiples of each other, then their co-prime parts must be the same.Wait, but how do I tie this back to the denominators being equal?Alternatively, maybe I can use the fact that if every substring of a/b is in c/d, then the decimal expansion of a/b is a subsequence of c/d's expansion. Therefore, the structure of a/b is embedded within c/d, which might force their denominators to be related.But I'm not sure how to formalize this.Wait, another approach: consider that the decimal expansions are generated by the fractions, so the denominators determine the possible periods and non-repeating parts. If every substring of a/b is in c/d, then the structure of a/b's decimal is contained within c/d's. Therefore, the denominators must be such that b divides d or vice versa.But since both are in lowest terms, if b divides d, then d must be a multiple of b, but since they are both denominators of fractions with the given property, maybe they have to be equal.Wait, but I'm not sure. Maybe I need to think about the decimal expansions more formally.Let me denote the decimal expansion of a/b as 0.Aoverline{B}, where A is the non-repeating part and B is the repeating part. Similarly, c/d is 0.Coverline{D}.Given that every finite substring of 0.Aoverline{B} appears in 0.Coverline{D}, this includes the entire A and B parts. So, A must appear somewhere in 0.Coverline{D}. Similarly, AB must appear, ABB must appear, and so on.Since A is finite, it must appear in the non-repeating part C or in the repeating part D. But if A is longer than C, then part of A must be in C and the rest in D. But D is repeating, so the part of A in D must align with the period.This suggests that the non-repeating part A is a suffix of C or a combination of the end of C and the beginning of D. But since D is periodic, this would impose that A is a suffix of C concatenated with some number of D's.But A is finite, so this would mean that A is a suffix of C followed by a finite number of D's. However, since D is repeating, the number of D's is arbitrary, which might not necessarily align with A.Wait, maybe I need to consider that the non-repeating part A must be entirely contained within the non-repeating part C, otherwise, the repeating part D would have to contain A, which is finite but could conflict with the periodicity.Alternatively, if A is longer than C, then part of A is in C and the rest is in D. But since D is repeating, the part in D would have to repeat, which might not match A unless A itself is periodic.But A is just a finite string, so it doesn't have to be periodic. Therefore, if A is longer than C, then the part of A in D would have to be a substring of D, which is periodic. So, the part of A in D must be a substring of D, which is repeating.But since D is repeating, any substring of D is just a substring of one period. So, if A is longer than C, then the part of A beyond C must be a substring of D.But since A is finite, this would mean that A can be split into two parts: the part in C and the part in D. The part in D must be a substring of D, which is periodic.But since D is repeating, any substring of D is just a substring of one period. So, the part of A in D must be a substring of one period of D.But A is arbitrary, so this might not necessarily hold unless D is long enough to contain all possible substrings of A beyond C.Wait, but A is fixed once a/b is fixed. So, if A is longer than C, then the part of A beyond C must be a substring of D. But since D is periodic, this substring must repeat every period.Therefore, the part of A beyond C must be a substring of D, which is periodic. So, the part of A beyond C must be a substring of one period of D.But since A is finite, this is possible. So, maybe the non-repeating part of a/b is a combination of the non-repeating part of c/d and a substring of the repeating part of c/d.But how does this affect the denominators?Well, the length of the non-repeating part is determined by the maximum power of 2 and 5 in the denominator. So, if a/b has a non-repeating part of length k, then b must have factors 2^k or 5^k or both.Similarly, c/d has a non-repeating part of length l, determined by the factors of 2 and 5 in d.If the non-repeating part of a/b is longer than that of c/d, then k > l. But since the part of A beyond C must be a substring of D, which is periodic, this might impose that the non-repeating part of a/b can't be too much longer than that of c/d.Wait, but I'm not sure how to connect this to the denominators being equal.Maybe I should think about the denominators in terms of their co-prime parts. The repeating part is determined by the co-prime part of the denominator to 10.So, if a/b has a repeating part B with period m, then m is the multiplicative order of 10 modulo the co-prime part of b. Similarly for c/d.If every substring of B is in D, then the period of B must divide the period of D. Because otherwise, the repeating part of c/d might not contain all substrings of B.But if the period of B divides the period of D, then the co-prime part of b divides the co-prime part of d. Similarly, if we consider the other direction, maybe the co-prime part of d divides the co-prime part of b, leading to them being equal.Wait, but the problem only states that every substring of a/b is in c/d, not the other way around. So, maybe only one direction holds, meaning that the co-prime part of b divides the co-prime part of d.But then, how do we get equality?Wait, maybe if we consider that the decimal expansion of a/b is a subsequence of c/d's expansion, then the structure of a/b is embedded within c/d. Therefore, the denominators must be related in such a way that b divides d or vice versa.But since both fractions are in lowest terms, if b divides d, then d must be a multiple of b. But since the decimal expansion of a/b is contained within c/d, maybe d must be a multiple of b, but also, considering the other direction, b must be a multiple of d, leading to b = d.Wait, that might make sense. If b divides d and d divides b, then b = d.But how do I formalize this?Maybe I can use the fact that if every substring of a/b is in c/d, then the decimal expansion of a/b is a subsequence of c/d's expansion. Therefore, the denominators must be related such that b divides d.But since both are in lowest terms, if b divides d, then d = b * k for some integer k. But then, the decimal expansion of c/d would have a period that is a multiple of the period of a/b.But since every substring of a/b is in c/d, the periods must be compatible. If the period of a/b is m, then the period of c/d must be a multiple of m. But since c/d is in lowest terms, the period is determined by the multiplicative order of 10 modulo the co-prime part of d.Similarly, the period of a/b is determined by the multiplicative order of 10 modulo the co-prime part of b.If the co-prime part of d is a multiple of the co-prime part of b, then the period of c/d would be a multiple of the period of a/b.But if we also consider that the non-repeating parts must align, then the powers of 2 and 5 in b and d must be the same, otherwise, the non-repeating parts would differ.Wait, if the non-repeating parts are different, then the finite blocks in a/b that include the non-repeating part would not be present in c/d unless the non-repeating parts are the same.Therefore, the non-repeating parts must be the same, meaning that the powers of 2 and 5 in b and d must be the same. So, b and d have the same factors of 2 and 5.Similarly, the co-prime parts must be the same because the periods must be compatible, and since every substring is present, the co-prime parts must be equal.Therefore, combining both, b and d must be equal.Wait, that seems plausible. So, the non-repeating parts force the powers of 2 and 5 to be the same, and the repeating parts force the co-prime parts to be the same, hence b = d.I think that's the way to go. So, to summarize:1. The non-repeating part of a/b must be contained within the non-repeating part of c/d or in the repeating part. But since the non-repeating part is finite, and the repeating part is periodic, the non-repeating parts must be the same length, forcing the powers of 2 and 5 in b and d to be equal.2. The repeating parts must have periods that are multiples of each other, but since every substring is present, the periods must be equal, forcing the co-prime parts of b and d to be equal.Therefore, b and d must be equal.Yeah, that makes sense. I think I've got it.

💡First, I need to determine the number of chickens based on the given ratio. The ratio of ducks to parrots to chickens is 2:3:5, which sums up to 10 parts. Since Peter has 60 birds in total, each part represents 6 birds. Therefore, the number of chickens is 5 parts multiplied by 6, which equals 30 chickens.Next, I calculate the cost to feed all the chickens. Each chicken requires 2 worth of feed. Multiplying the number of chickens (30) by the cost per chicken (2) gives the total feeding cost.

💡First, I need to calculate the total travel time between each pair of stations. I'll convert all travel times from hours to minutes for consistency.Next, I'll add up all the travel times to get the total travel duration.Then, I'll account for the break times at the specified stations. I'll convert these break times into minutes as well.Finally, I'll sum the total travel time and the total break time to find the overall time Kira takes to travel from the first to the seventh station.

💡Okay, so I have this problem where I need to show that the length of the angle bisector of angle C in triangle ABC is (2ab cos(C/2))/(a + b). Hmm, let me try to visualize this. Triangle ABC with sides BC = a and CA = b. So, angle C is at vertex C, and we're supposed to find the length of the bisector of angle C.First, I remember that an angle bisector in a triangle divides the angle into two equal parts. So, the bisector of angle C will split it into two angles of measure C/2 each. I also recall that there's a formula for the length of an angle bisector, but I'm not exactly sure what it is. Maybe I can derive it?Let me denote the angle bisector as CD, where D is the point where the bisector meets side AB. So, CD is the angle bisector, and we need to find its length. I think I can use the Angle Bisector Theorem here, which states that the ratio of the lengths of the two segments created by the bisector on the opposite side is equal to the ratio of the other two sides of the triangle.So, according to the Angle Bisector Theorem, AD/DB = AC/BC = b/a. That means AD = (b/(a + b)) * AB and DB = (a/(a + b)) * AB. But wait, I don't know the length of AB yet. Maybe I can express AB in terms of a, b, and angle C using the Law of Cosines.Law of Cosines says that AB² = AC² + BC² - 2*AC*BC*cos(C). So, AB² = b² + a² - 2ab cos(C). Therefore, AB = sqrt(a² + b² - 2ab cos(C)). Hmm, that seems a bit complicated, but maybe it will simplify later.Alternatively, maybe I can use the formula for the length of the angle bisector directly. I think the formula is something like CD = (2ab cos(C/2))/(a + b). Wait, that's exactly what I'm supposed to prove! So, maybe I can derive this formula using some trigonometric identities or area considerations.Let me think about the areas of the triangles involved. The angle bisector CD divides triangle ABC into two smaller triangles, ADC and BDC. Both of these triangles share the same height from point C to side AB. The areas of these triangles can be expressed in terms of the lengths AD, DB, and the angles at C.The area of triangle ADC can be written as (1/2)*AC*CD*sin(C/2), and similarly, the area of triangle BDC is (1/2)*BC*CD*sin(C/2). Since CD is the angle bisector, both these areas should be proportional to the lengths AC and BC, which are b and a respectively.So, the ratio of the areas of ADC to BDC should be equal to the ratio of AC to BC, which is b/a. Let me write that down:Area of ADC / Area of BDC = b/a.Substituting the expressions for the areas:[(1/2)*b*CD*sin(C/2)] / [(1/2)*a*CD*sin(C/2)] = b/a.Simplifying, the (1/2) and CD and sin(C/2) terms cancel out, leaving b/a = b/a. Hmm, that doesn't really help me find CD. Maybe I need a different approach.What if I consider the entire area of triangle ABC? The area can be expressed in two ways: one using the angle C and sides a and b, and another using the angle bisector CD and the segments AD and DB.Using the standard formula, the area of ABC is (1/2)*a*b*sin(C). On the other hand, the area can also be expressed as the sum of the areas of ADC and BDC, which are (1/2)*b*CD*sin(C/2) and (1/2)*a*CD*sin(C/2) respectively.So, adding these two areas together:Area of ABC = (1/2)*b*CD*sin(C/2) + (1/2)*a*CD*sin(C/2) = (1/2)*CD*sin(C/2)*(a + b).Now, setting this equal to the standard area formula:(1/2)*CD*sin(C/2)*(a + b) = (1/2)*a*b*sin(C).Simplifying, we can cancel out the (1/2) terms:CD*sin(C/2)*(a + b) = a*b*sin(C).Now, I can solve for CD:CD = (a*b*sin(C)) / [(a + b)*sin(C/2)].Hmm, that's close, but not exactly the formula I'm supposed to prove. I have sin(C) in the numerator, but I need cos(C/2). Maybe I can use a trigonometric identity to express sin(C) in terms of sin(C/2) and cos(C/2).I remember that sin(C) = 2*sin(C/2)*cos(C/2). Let me substitute that into the equation:CD = (a*b*2*sin(C/2)*cos(C/2)) / [(a + b)*sin(C/2)].Now, the sin(C/2) terms cancel out:CD = (2*a*b*cos(C/2)) / (a + b).Yes! That's the formula I was supposed to prove. So, the length of the angle bisector CD is indeed (2ab cos(C/2))/(a + b).I think that makes sense. I used the area approach, considering the areas of the two smaller triangles created by the angle bisector and equating it to the area of the whole triangle. Then, by using the double-angle identity for sine, I was able to simplify the expression to get the desired formula. It feels a bit involved, but I think each step logically follows from the previous one. I don't see any mistakes in my reasoning, so I'm confident this is correct.

💡Alright, so I have this problem here: "How many non-similar triangles have angles whose degree measures are distinct positive integers in an arithmetic progression where the common difference is an even number?" Hmm, okay, let's break this down step by step.First, I know that in any triangle, the sum of the interior angles is always 180 degrees. That's a fundamental rule. Also, non-similar triangles mean that their angles aren't the same, so each set of angles we find must be unique in terms of their measures.The problem mentions that the angles are distinct positive integers in an arithmetic progression. An arithmetic progression is a sequence where each term after the first is obtained by adding a constant difference. In this case, the common difference is an even number. So, if I denote the three angles as ( a - d ), ( a ), and ( a + d ), where ( d ) is the common difference, which is even.Since the sum of the angles in a triangle is 180 degrees, I can write the equation:[ (a - d) + a + (a + d) = 180 ]Simplifying this, the ( -d ) and ( +d ) cancel out, so I get:[ 3a = 180 ]Dividing both sides by 3:[ a = 60 ]Okay, so the middle angle is 60 degrees. That makes sense because the angles are in an arithmetic progression, so they are equally spaced around 60 degrees.Now, the three angles are ( 60 - d ), ( 60 ), and ( 60 + d ). Since all angles must be positive, each of these must be greater than 0. So, ( 60 - d > 0 ), which implies that ( d < 60 ). Also, since the angles must be distinct, ( d ) cannot be zero. So, ( d ) must be a positive even integer less than 60.Wait, the problem says the common difference is an even number. So, ( d ) must be even. That means ( d ) can be 2, 4, 6, ..., up to 58. Let me list these out to see how many there are.Starting from 2 and going up by 2 each time until 58. So, the sequence is 2, 4, 6, ..., 58. To find how many terms are in this sequence, I can use the formula for the nth term of an arithmetic sequence:[ a_n = a_1 + (n - 1)d ]Here, ( a_n = 58 ), ( a_1 = 2 ), and the common difference ( d = 2 ). Plugging these into the formula:[ 58 = 2 + (n - 1) times 2 ]Subtract 2 from both sides:[ 56 = (n - 1) times 2 ]Divide both sides by 2:[ 28 = n - 1 ]Add 1 to both sides:[ n = 29 ]So, there are 29 possible values for ( d ). Each value of ( d ) gives a unique set of angles, meaning each corresponds to a non-similar triangle. Therefore, there are 29 such triangles.But wait, let me double-check to make sure I didn't miss anything. The angles must be positive, so ( 60 - d > 0 ) which gives ( d < 60 ). Since ( d ) is even, the maximum ( d ) can be is 58, which I included. The minimum ( d ) is 2, which also makes sense because ( d ) can't be 0 or 1 (since it's even and positive). So, yes, 29 seems correct.I think that's all. I don't see any other constraints or possibilities I missed. So, the number of non-similar triangles is 29.

💡Okay, so I have this problem about an arithmetic sequence. Let me try to understand it step by step. The sequence is denoted as {a_n}, and it's an arithmetic sequence. That means each term increases by a constant difference, right? So, if I denote the first term as a_1 and the common difference as d, then the nth term can be written as a_n = a_1 + (n-1)d.Now, the problem gives me some conditions:1. a_1 ≥ 12. a_2 ≤ 53. a_5 ≥ 8I need to find the sum of the first 15 terms, S_15, and determine its maximum value M and minimum value m. Then, I have to find M + m.First, let me recall the formula for the sum of the first n terms of an arithmetic sequence. It's S_n = n/2 * [2a_1 + (n-1)d]. So, for S_15, it would be S_15 = 15/2 * [2a_1 + 14d].But before I can find S_15, I need to figure out the possible values of a_1 and d based on the given conditions.Let me write down the conditions in terms of a_1 and d.1. a_1 ≥ 12. a_2 = a_1 + d ≤ 53. a_5 = a_1 + 4d ≥ 8So, from condition 2: a_1 + d ≤ 5From condition 3: a_1 + 4d ≥ 8I can write these inequalities as:1. a_1 ≥ 12. a_1 + d ≤ 53. a_1 + 4d ≥ 8I need to find the range of possible values for a_1 and d that satisfy all three inequalities.Let me try to express these inequalities in terms of d.From condition 2: a_1 ≤ 5 - dFrom condition 3: a_1 ≥ 8 - 4dSo combining these with condition 1: 1 ≤ a_1 ≤ 5 - d and a_1 ≥ 8 - 4d.So, for a_1 to satisfy both 1 ≤ a_1 ≤ 5 - d and a_1 ≥ 8 - 4d, the lower bound of a_1 is the maximum of 1 and 8 - 4d, and the upper bound is 5 - d.Therefore, we have:max(1, 8 - 4d) ≤ a_1 ≤ 5 - dBut since a_1 must satisfy both inequalities, the lower bound must be less than or equal to the upper bound. So:max(1, 8 - 4d) ≤ 5 - dThis gives two cases:Case 1: 1 ≤ 5 - d, which is always true because d is a real number, but let's see.Case 2: 8 - 4d ≤ 5 - dLet me solve 8 - 4d ≤ 5 - d:8 - 4d ≤ 5 - dSubtract 5 from both sides:3 - 4d ≤ -dAdd 4d to both sides:3 ≤ 3dDivide both sides by 3:1 ≤ dSo, d must be at least 1.So, combining this with the previous inequalities, d ≥ 1.Now, let's see the possible range for d.From condition 2: a_1 + d ≤ 5, and since a_1 ≥ 1, then:1 + d ≤ 5So, d ≤ 4.So, d is between 1 and 4, inclusive.So, d ∈ [1, 4]Now, I need to find the maximum and minimum values of S_15.S_15 = 15/2 * [2a_1 + 14d] = 15/2 * [2a_1 + 14d] = 15*(a_1 + 7d)So, S_15 = 15*(a_1 + 7d)Therefore, to maximize S_15, I need to maximize a_1 + 7d, and to minimize S_15, I need to minimize a_1 + 7d.So, let's find the maximum and minimum of a_1 + 7d given the constraints on a_1 and d.We have:1. a_1 ≥ 12. a_1 ≤ 5 - d3. a_1 ≥ 8 - 4d4. d ∈ [1, 4]So, a_1 is bounded below by max(1, 8 - 4d) and above by 5 - d.Therefore, to maximize a_1 + 7d, we need to maximize a_1 and d as much as possible.Similarly, to minimize a_1 + 7d, we need to minimize a_1 and d as much as possible.Let me first find the maximum of a_1 + 7d.To maximize a_1 + 7d, we need to take a_1 as large as possible and d as large as possible.From condition 2, a_1 ≤ 5 - d. So, to maximize a_1, set a_1 = 5 - d.So, a_1 + 7d = (5 - d) + 7d = 5 + 6d.Now, since d is in [1,4], to maximize 5 + 6d, set d as large as possible, which is d = 4.So, a_1 + 7d = 5 + 6*4 = 5 + 24 = 29.Therefore, the maximum S_15 is 15*29 = 435.Wait, but let me check if a_1 = 5 - d when d =4 is allowed.a_1 = 5 -4 =1, which is ≥1, so it's okay.Also, a_5 = a_1 +4d =1 +16=17, which is ≥8, so that's okay.So, M = 435.Now, for the minimum of a_1 +7d.To minimize a_1 +7d, we need to take a_1 as small as possible and d as small as possible.From condition 3, a_1 ≥8 -4d.So, to minimize a_1 +7d, set a_1 =8 -4d.So, a_1 +7d = (8 -4d) +7d =8 +3d.Now, we need to find the minimum of 8 +3d, where d ∈ [1,4].Since 3d is increasing in d, the minimum occurs at d=1.So, a_1 +7d =8 +3*1=11.Therefore, the minimum S_15 is 15*11=165.Wait, let me verify.If d=1, then a_1=8 -4*1=4.Check the conditions:a_1=4 ≥1, okay.a_2 =a_1 +d=4 +1=5 ≤5, okay.a_5=a_1 +4d=4 +4=8 ≥8, okay.So, that's valid.Therefore, m=165.Thus, M=435 and m=165, so M + m=435 +165=600.So, the answer is 600, which is option B.Wait, let me double-check my calculations.For M:a_1=1, d=4.S_15=15/2*(2*1 +14*4)=15/2*(2 +56)=15/2*58=15*29=435.For m:a_1=4, d=1.S_15=15/2*(2*4 +14*1)=15/2*(8 +14)=15/2*22=15*11=165.Yes, that's correct.So, M + m=435 +165=600.Therefore, the answer is B:600.