💡Okay, so I have this problem about setting admission scores for art students at a high school. There are three different scores they're considering: the professional score (x), the total score of liberal arts (y), and the physical education score (z). I need to represent the conditions given using inequalities and then choose the correct option from A to D.First, let me break down the problem sentence by sentence.1. "The professional score (x) is no less than 95." - "No less than" means that x should be greater than or equal to 95. So, in inequality terms, that would be x ≥ 95.2. "The total score of liberal arts (y) is higher than 380." - "Higher than" means y should be greater than 380. So, y > 380.3. "The physical education score (z) is more than 45." - "More than" also means z should be greater than 45. So, z > 45.Now, I need to match these inequalities with the given options. Let's look at each option one by one.Option A:- x ≥ 95- y ≥ 380- z > 45Hmm, for y, the problem says "higher than 380," which should be y > 380, not y ≥ 380. So, Option A might not be correct because it uses ≥ for y instead of >.Option B:- x ≥ 95- y > 380- z ≥ 45Here, z is given as z ≥ 45, but the problem states "more than 45," which should be z > 45. So, Option B is also incorrect because it uses ≥ for z instead of >.Option C:- x > 95- y > 380- z > 45This option uses x > 95, but the problem says "no less than 95," which includes 95 itself. So, x should be ≥ 95, not just > 95. Therefore, Option C is incorrect because it uses > for x.Option D:- x ≥ 95- y > 380- z > 45This matches all the conditions perfectly:- x is no less than 95: x ≥ 95- y is higher than 380: y > 380- z is more than 45: z > 45So, Option D seems to be the correct one.Wait, just to make sure I didn't make a mistake. Let me double-check each condition:1. Professional score (x): "No less than 95" means 95 or higher. So, x ≥ 95 is correct.2. Liberal arts score (y): "Higher than 380" means strictly greater than 380, so y > 380 is correct.3. Physical education score (z): "More than 45" also means strictly greater than 45, so z > 45 is correct.Yes, that all lines up with Option D. I think I was thorough in checking each condition and making sure the inequalities match the problem statement. It's easy to mix up the inequality signs, especially between ≥ and >, but in this case, the wording clearly indicates which one to use for each score.I don't see any other options that fit all three conditions correctly. Options A and B incorrectly use ≥ for y and z respectively, and Option C incorrectly uses > for x when it should be ≥. So, Option D is the only one that accurately represents all the given conditions with the correct inequalities.Just to think about it another way, if the school is setting an admission line, they might want to include exactly 95 for the professional score because it's "no less than," meaning 95 is acceptable. But for the other scores, since they're "higher than" or "more than," those would exclude the exact numbers (380 and 45) and only accept scores strictly above them. That makes sense in terms of setting a cutoff where you need to surpass certain thresholds.So, yeah, I'm confident now that Option D is the right choice.

💡First, I need to identify the largest and the second largest numbers from the given set: 2.1, 3.2, 32, 12, 3.4, and 2.5.Looking at the numbers, 32 is clearly the largest. Next, I compare the remaining numbers to find the second largest. Among 2.1, 3.2, 12, 3.4, and 2.5, 12 is the next highest.Now, I will multiply the largest number by the second largest number:32 multiplied by 12 equals 384.Therefore, the value of multiplying the first largest number by the second largest number is 384.

💡Okay, so I need to figure out the tens digit of 13 raised to the power of 2017. Hmm, that seems pretty big. I remember that when dealing with large exponents, modular arithmetic can be helpful, especially if we're only interested in the last two digits, which would give us both the units and tens digits. Since the tens digit is the second digit from the right, I think I can find it by calculating 13^2017 modulo 100. That should give me the last two digits, and then I can pick out the tens digit from there.Alright, so first, I need to compute 13^2017 mod 100. I remember that Euler's theorem might be useful here because it deals with exponents in modular arithmetic. Euler's theorem states that if two numbers are coprime, then a^φ(n) ≡ 1 mod n, where φ(n) is Euler's totient function. Since 13 and 100 are coprime (their greatest common divisor is 1), this theorem should apply.Let me calculate φ(100). 100 is 2^2 * 5^2, so φ(100) = 100 * (1 - 1/2) * (1 - 1/5) = 100 * 1/2 * 4/5 = 100 * 2/5 = 40. So φ(100) is 40. That means 13^40 ≡ 1 mod 100.Now, if 13^40 ≡ 1 mod 100, then 13^(40k) ≡ 1^k ≡ 1 mod 100 for any integer k. So, I can express 2017 as a multiple of 40 plus some remainder. Let me divide 2017 by 40 to find the quotient and remainder.2017 divided by 40 is 50 with a remainder of 17 because 40*50 = 2000, and 2017 - 2000 = 17. So, 2017 = 40*50 + 17. Therefore, 13^2017 = 13^(40*50 + 17) = (13^40)^50 * 13^17. Since 13^40 ≡ 1 mod 100, this simplifies to 1^50 * 13^17 ≡ 13^17 mod 100.So, now I just need to compute 13^17 mod 100. Calculating 13^17 directly seems tedious, but maybe I can find a pattern or use exponentiation by squaring to make it easier.Let me try to compute powers of 13 modulo 100 step by step:13^1 ≡ 13 mod 10013^2 = 13*13 = 169 ≡ 69 mod 10013^3 = 13*69. Let's compute that: 13*60=780, 13*9=117, so 780+117=897 ≡ 97 mod 10013^4 = 13*97. 13*90=1170, 13*7=91, so 1170+91=1261 ≡ 61 mod 10013^5 = 13*61. 13*60=780, 13*1=13, so 780+13=793 ≡ 93 mod 10013^6 = 13*93. 13*90=1170, 13*3=39, so 1170+39=1209 ≡ 09 mod 10013^7 = 13*09 = 117 ≡ 17 mod 10013^8 = 13*17. 13*10=130, 13*7=91, so 130+91=221 ≡ 21 mod 10013^9 = 13*21. 13*20=260, 13*1=13, so 260+13=273 ≡ 73 mod 10013^10 = 13*73. Let's compute 13*70=910, 13*3=39, so 910+39=949 ≡ 49 mod 10013^11 = 13*49. 13*40=520, 13*9=117, so 520+117=637 ≡ 37 mod 10013^12 = 13*37. 13*30=390, 13*7=91, so 390+91=481 ≡ 81 mod 10013^13 = 13*81. 13*80=1040, 13*1=13, so 1040+13=1053 ≡ 53 mod 10013^14 = 13*53. 13*50=650, 13*3=39, so 650+39=689 ≡ 89 mod 10013^15 = 13*89. 13*80=1040, 13*9=117, so 1040+117=1157 ≡ 57 mod 10013^16 = 13*57. 13*50=650, 13*7=91, so 650+91=741 ≡ 41 mod 10013^17 = 13*41. 13*40=520, 13*1=13, so 520+13=533 ≡ 33 mod 100Okay, so 13^17 mod 100 is 33. That means the last two digits of 13^2017 are 33. Therefore, the tens digit is 3.Wait, let me double-check my calculations to make sure I didn't make a mistake somewhere. Starting from 13^1 up to 13^17, each step seems correct. I used the previous result multiplied by 13, then took modulo 100 each time. The pattern cycles every 20, but since we only needed up to 17, it's fine. So, yes, 13^17 mod 100 is indeed 33, so the tens digit is 3.I think that's solid. I don't see any errors in my reasoning or calculations. So, the tens digit of 13^2017 is 3.**Final Answer**The tens digit of (13^{2017}) is boxed{3}.

💡Alright, let's tackle this problem step by step. So, we have a 3x3 table where the numbers 1 to 9 are placed, each exactly once. The numbers 1, 2, 3, and 4 are already positioned as shown:| 1 | | 3 ||---|---|---|| | | || 2 | | 4 |We also know that the sum of the numbers adjacent to the cell containing the number 5 is 9. Our goal is to find the sum of the numbers adjacent to the cell containing the number 6.First, let's visualize the table with the given numbers:```1 | | 3--------- | |---------2 | | 4```Now, let's think about where the number 5 could be placed. Since the sum of the numbers adjacent to 5 is 9, we need to consider the possible positions for 5 and see which one satisfies this condition.Let's consider the possible positions for 5:1. **Top row, middle cell**: If 5 is here, its adjacent cells would be 1 (left), 3 (right), and the cell below it. The sum of these adjacent numbers should be 9. So, 1 + 3 + (cell below) = 9. That means the cell below would need to be 5, but 5 is already placed. So this doesn't work.2. **Middle row, middle cell**: If 5 is in the center, its adjacent cells would be 1 (above), 3 (right), 2 (below), and 4 (right). Wait, but 4 is already in the bottom right. So, the sum would be 1 + 3 + 2 + 4 = 10, which is more than 9. So this doesn't work either.3. **Middle row, left cell**: If 5 is here, its adjacent cells would be 1 (above), 2 (below), and the cell to the right. The sum should be 9. So, 1 + 2 + (cell to the right) = 9. That means the cell to the right would need to be 6. Let's see if this works.If 5 is in the middle left, then the cell to its right would be 6. Let's tentatively place 6 there:```1 | | 3---------5 | 6 | ---------2 | | 4```Now, let's check the sum of the numbers adjacent to 5: 1 (above) + 2 (below) + 6 (right) = 9. Perfect, that satisfies the condition.Now, let's try to fill in the remaining numbers. The numbers we've used so far are 1, 2, 3, 4, 5, and 6. The remaining numbers are 7, 8, and 9.Looking at the top row, we have:```1 | | 3```The middle cell in the top row is empty. Let's see what number can go there. The numbers left are 7, 8, and 9. Since 1 and 3 are already in the top row, the middle cell can be any of the remaining numbers. But let's see if we can deduce it based on adjacency.Looking at the middle row:```5 | 6 | ```The right cell in the middle row is empty. Let's see what number can go there. The numbers left are 7, 8, and 9. If we place 7 there, then the sum of the numbers adjacent to 6 would include 5 (left), 7 (right), and the numbers above and below 6.Wait, let's think about the adjacency for 6. The number 6 is in the middle right of the middle row. Its adjacent cells are:- Above: The middle cell of the top row.- Below: The middle cell of the bottom row.- Left: 5.So, the sum of the numbers adjacent to 6 would be 5 (left) + (above) + (below). We need to find what's above and below 6.The above cell is the middle cell of the top row, which is currently empty. The below cell is the middle cell of the bottom row, which is also empty.Let's see if we can place 7 in the middle right of the middle row:```1 | | 3---------5 | 6 | 7---------2 | | 4```Now, let's check the sum of the numbers adjacent to 6: 5 (left) + 7 (right) + (above) + (below). We need to find the numbers above and below 6.The above cell is the middle cell of the top row, which is still empty. The below cell is the middle cell of the bottom row, which is also empty.We have the numbers 8 and 9 left to place. Let's see if we can place them in the remaining cells.Looking at the top row, the middle cell is empty. Let's consider placing 8 there:```1 | 8 | 3---------5 | 6 | 7---------2 | | 4```Now, the bottom middle cell is still empty. The only number left is 9, so we'll place 9 there:```1 | 8 | 3---------5 | 6 | 7---------2 | 9 | 4```Now, let's verify the sum of the numbers adjacent to 6. The adjacent cells are:- Left: 5- Right: 7- Above: 8- Below: 9So, the sum is 5 + 7 + 8 + 9 = 29.Wait, but let's double-check if this placement satisfies all conditions.First, the sum of the numbers adjacent to 5 is 1 + 2 + 6 = 9, which is correct.Second, the numbers 1 to 9 are all used exactly once, which is correct.Third, the sum of the numbers adjacent to 6 is 5 + 7 + 8 + 9 = 29.But let's see if there's another possible placement that might give a different sum.Suppose instead of placing 7 in the middle right of the middle row, we place 8 there:```1 | | 3---------5 | 6 | 8---------2 | | 4```Then, the remaining numbers are 7 and 9.Placing 7 in the top middle:```1 | 7 | 3---------5 | 6 | 8---------2 | | 4```Then, the bottom middle cell would be 9:```1 | 7 | 3---------5 | 6 | 8---------2 | 9 | 4```Now, the sum of the numbers adjacent to 6 would be 5 (left) + 8 (right) + 7 (above) + 9 (below) = 5 + 8 + 7 + 9 = 29.Same result.Alternatively, if we place 9 in the middle right of the middle row:```1 | | 3---------5 | 6 | 9---------2 | | 4```Then, the remaining numbers are 7 and 8.Placing 7 in the top middle:```1 | 7 | 3---------5 | 6 | 9---------2 | | 4```Then, the bottom middle cell would be 8:```1 | 7 | 3---------5 | 6 | 9---------2 | 8 | 4```Now, the sum of the numbers adjacent to 6 would be 5 (left) + 9 (right) + 7 (above) + 8 (below) = 5 + 9 + 7 + 8 = 29.Again, same result.So, regardless of where we place 7, 8, and 9 in the remaining cells, the sum of the numbers adjacent to 6 remains 29.Therefore, the sum of the numbers in the cells adjacent to the cell containing the number 6 is 29.

💡Alright, so I've got this problem here about a set S, which consists of all odd integers greater than 1. That means S includes numbers like 3, 5, 7, 9, and so on. The problem defines a function δ(x) for each x in S, which is the unique integer such that 2^δ(x) < x < 2^(δ(x)+1). So, δ(x) is basically the exponent of the highest power of 2 that's less than x. For example, if x is 5, then δ(x) would be 2 because 2^2 = 4 is less than 5, and 2^3 = 8 is greater than 5.Next, there's this operation defined for a and b in S: a ∗ b = 2^(δ(a)−1) * (b − 3) + a. The problem has two parts: first, to prove that a ∗ b is also in S, meaning it's an odd integer greater than 1. Second, to prove that this operation is associative, meaning (a ∗ b) ∗ c = a ∗ (b ∗ c) for any a, b, c in S.Starting with part (a): proving that a ∗ b is in S. So, I need to show two things: that a ∗ b is odd and that it's greater than 1.First, let's look at the expression for a ∗ b: 2^(δ(a)−1) * (b − 3) + a. Since a and b are both in S, they are odd numbers greater than 1. Let's consider the term (b − 3). Since b is odd, subtracting 3 (which is also odd) gives us an even number. So, (b − 3) is even, which means it can be written as 2k for some integer k.Now, 2^(δ(a)−1) is clearly a power of 2, so multiplying it by (b − 3) gives us 2^(δ(a)−1) * 2k = 2^(δ(a)) * k, which is still even because it's a multiple of 2. Then, we add a, which is odd. So, even + odd = odd. Therefore, a ∗ b is odd.Next, we need to check that a ∗ b is greater than 1. Since a is in S, a > 1. The term 2^(δ(a)−1) * (b − 3) is non-negative because both 2^(δ(a)−1) and (b − 3) are non-negative (since b > 3, because b is in S and greater than 1, but actually b could be 3, so let's check that). If b = 3, then (b − 3) = 0, so a ∗ b = a, which is still greater than 1. If b > 3, then (b − 3) is positive, so 2^(δ(a)−1) * (b − 3) is positive, and adding a (which is greater than 1) keeps the result greater than 1. So, a ∗ b is indeed greater than 1.Therefore, part (a) is proven: a ∗ b is in S.Moving on to part (b): proving associativity, i.e., (a ∗ b) ∗ c = a ∗ (b ∗ c). This seems a bit more involved. Let's break it down step by step.First, let's compute (a ∗ b) ∗ c. From part (a), we know that a ∗ b is in S, so we can apply the ∗ operation again. Let's denote d = a ∗ b. Then, (a ∗ b) ∗ c = d ∗ c = 2^(δ(d)−1) * (c − 3) + d.Similarly, let's compute a ∗ (b ∗ c). Let's denote e = b ∗ c. Then, a ∗ (b ∗ c) = a ∗ e = 2^(δ(a)−1) * (e − 3) + a.To show that these two expressions are equal, we need to express both in terms of a, b, and c, and then show they simplify to the same thing.First, let's find expressions for d and e.We have d = a ∗ b = 2^(δ(a)−1) * (b − 3) + a.Similarly, e = b ∗ c = 2^(δ(b)−1) * (c − 3) + b.Now, let's substitute these into the expressions for (a ∗ b) ∗ c and a ∗ (b ∗ c).Starting with (a ∗ b) ∗ c = d ∗ c:d ∗ c = 2^(δ(d)−1) * (c − 3) + d.But d = 2^(δ(a)−1) * (b − 3) + a, so δ(d) is the exponent such that 2^(δ(d)) < d < 2^(δ(d)+1). We need to find δ(d). Since d = 2^(δ(a)−1) * (b − 3) + a, and a is in S, which is between 2^(δ(a)) and 2^(δ(a)+1). Similarly, (b − 3) is even, so 2^(δ(a)−1) * (b − 3) is a multiple of 2^(δ(a)−1). Adding a, which is between 2^(δ(a)) and 2^(δ(a)+1), the sum d should be between 2^(δ(a)) and 2^(δ(a)+2). Therefore, δ(d) is likely δ(a) + δ(b) - 1, but I need to verify this.Wait, let's think about the magnitude. Since a is between 2^(δ(a)) and 2^(δ(a)+1), and (b − 3) is between 0 and 2^(δ(b)) (since b < 2^(δ(b)+1), so b − 3 < 2^(δ(b)+1) − 3, which is less than 2^(δ(b)+1)). Therefore, 2^(δ(a)−1) * (b − 3) is less than 2^(δ(a)−1) * 2^(δ(b)+1) = 2^(δ(a) + δ(b)). Adding a, which is less than 2^(δ(a)+1), the total d is less than 2^(δ(a)+1) + 2^(δ(a) + δ(b)). Depending on the relationship between δ(a) and δ(b), this could be dominated by the larger term.But perhaps a better approach is to consider that δ(d) = δ(a) + δ(b) - 1. Let's assume that for now and see if it holds.So, assuming δ(d) = δ(a) + δ(b) - 1, then:d ∗ c = 2^(δ(d)−1) * (c − 3) + d = 2^(δ(a) + δ(b) - 2) * (c − 3) + d.Substituting d:= 2^(δ(a) + δ(b) - 2) * (c − 3) + 2^(δ(a)−1) * (b − 3) + a.Now, let's compute a ∗ (b ∗ c) = a ∗ e:e = b ∗ c = 2^(δ(b)−1) * (c − 3) + b.So, a ∗ e = 2^(δ(a)−1) * (e − 3) + a.Substituting e:= 2^(δ(a)−1) * (2^(δ(b)−1) * (c − 3) + b - 3) + a.= 2^(δ(a)−1) * 2^(δ(b)−1) * (c − 3) + 2^(δ(a)−1) * (b - 3) + a.= 2^(δ(a) + δ(b) - 2) * (c − 3) + 2^(δ(a)−1) * (b - 3) + a.Comparing this with the expression for (a ∗ b) ∗ c:= 2^(δ(a) + δ(b) - 2) * (c − 3) + 2^(δ(a)−1) * (b − 3) + a.They are exactly the same! Therefore, (a ∗ b) ∗ c = a ∗ (b ∗ c).Wait, but I assumed δ(d) = δ(a) + δ(b) - 1. Is this always true? Let's verify.Given d = a ∗ b = 2^(δ(a)−1) * (b − 3) + a.We know that a is between 2^(δ(a)) and 2^(δ(a)+1).Similarly, (b − 3) is between 0 and 2^(δ(b)+1) - 3, which is less than 2^(δ(b)+1).Therefore, 2^(δ(a)−1) * (b − 3) is less than 2^(δ(a)−1) * 2^(δ(b)+1) = 2^(δ(a) + δ(b)).Adding a, which is less than 2^(δ(a)+1), the total d is less than 2^(δ(a)+1) + 2^(δ(a) + δ(b)).Now, depending on whether δ(a) + δ(b) is greater than δ(a)+1, which it is if δ(b) >1, which it is since b >1 and δ(b) ≥1.So, 2^(δ(a)+ δ(b)) dominates 2^(δ(a)+1), so d is less than 2^(δ(a)+ δ(b)+1).But we need to find δ(d) such that 2^(δ(d)) < d < 2^(δ(d)+1).Given that d is approximately 2^(δ(a)+ δ(b)), so δ(d) should be δ(a) + δ(b) -1.Yes, because 2^(δ(a)+ δ(b) -1) < d < 2^(δ(a)+ δ(b)).Therefore, δ(d) = δ(a) + δ(b) -1.Thus, the assumption holds, and the associativity follows.Therefore, part (b) is proven: (a ∗ b) ∗ c = a ∗ (b ∗ c).So, both parts (a) and (b) are proven.

💡First, calculate the total amount Francie saved from her initial allowance of 5 per week for 8 weeks:[5 , text{dollars/week} times 8 , text{weeks} = 40 , text{dollars}]Let ( x ) be the number of weeks she received her increased allowance of 6 per week. The total amount saved from the increased allowance is:[6 , text{dollars/week} times x , text{weeks} = 6x , text{dollars}]The total savings before spending is:[40 , text{dollars} + 6x , text{dollars}]Francie spends half of her total savings on new clothes, so the remaining half is used to buy the video game and the 3 left over. Let ( S ) be the total savings:[frac{S}{2} = 35 , text{dollars} + 3 , text{dollars}]Solving for ( S ):[S = (35 + 3) times 2 = 38 times 2 = 76 , text{dollars}]The total savings from both allowances is 76. Subtracting the initial allowance savings:[76 , text{dollars} - 40 , text{dollars} = 36 , text{dollars}]Determine the number of weeks she received the increased allowance:[frac{36 , text{dollars}}{6 , text{dollars/week}} = 6 , text{weeks}]