💡Okay, so I need to find the remainder when 7 raised to the 27th power is divided by 1,000. Hmm, that sounds a bit tricky at first glance, but maybe I can break it down step by step.First, I know that calculating 7^27 directly would result in a huge number, and dividing that by 1,000 to find the remainder isn't practical. There must be a smarter way to do this without dealing with massive numbers.I remember something about modular arithmetic, which deals with remainders. Maybe I can use that here. The idea is to find patterns or cycles in the powers of 7 modulo 1,000. If I can find a repeating cycle, I can use that to simplify the calculation.Let me start by calculating the first few powers of 7 modulo 1,000 to see if there's a pattern.7^1 = 7 → 7 mod 1,000 = 77^2 = 49 → 49 mod 1,000 = 497^3 = 343 → 343 mod 1,000 = 3437^4 = 343 * 7 = 2,401 → 2,401 mod 1,000 = 4017^5 = 401 * 7 = 2,807 → 2,807 mod 1,000 = 8077^6 = 807 * 7 = 5,649 → 5,649 mod 1,000 = 6497^7 = 649 * 7 = 4,543 → 4,543 mod 1,000 = 5437^8 = 543 * 7 = 3,801 → 3,801 mod 1,000 = 8017^9 = 801 * 7 = 5,607 → 5,607 mod 1,000 = 6077^10 = 607 * 7 = 4,249 → 4,249 mod 1,000 = 2497^11 = 249 * 7 = 1,743 → 1,743 mod 1,000 = 7437^12 = 743 * 7 = 5,201 → 5,201 mod 1,000 = 2017^13 = 201 * 7 = 1,407 → 1,407 mod 1,000 = 4077^14 = 407 * 7 = 2,849 → 2,849 mod 1,000 = 8497^15 = 849 * 7 = 5,943 → 5,943 mod 1,000 = 9437^16 = 943 * 7 = 6,601 → 6,601 mod 1,000 = 6017^17 = 601 * 7 = 4,207 → 4,207 mod 1,000 = 2077^18 = 207 * 7 = 1,449 → 1,449 mod 1,000 = 4497^19 = 449 * 7 = 3,143 → 3,143 mod 1,000 = 1437^20 = 143 * 7 = 1,001 → 1,001 mod 1,000 = 1Oh, interesting! At the 20th power, we get back to 1 modulo 1,000. That means the cycle repeats every 20 powers. So, 7^20 ≡ 1 mod 1,000.Now, since 27 is the exponent we're dealing with, I can express 27 as 20 + 7. That is, 7^27 = 7^(20+7) = 7^20 * 7^7.From what I found earlier, 7^20 ≡ 1 mod 1,000, so 7^27 ≡ 1 * 7^7 mod 1,000.I already calculated 7^7 earlier, which was 543 mod 1,000.Therefore, 7^27 mod 1,000 = 543.Wait, let me double-check that. I don't want to make a mistake here.So, 7^20 is 1 mod 1,000, which is correct because 7^20 = (7^10)^2, and 7^10 was 249 mod 1,000, and 249^2 is 62,001, which is 1 mod 1,000. So that's correct.Then, 7^27 = 7^(20+7) = 7^20 * 7^7 ≡ 1 * 543 ≡ 543 mod 1,000.Yes, that seems right. So the remainder when 7^27 is divided by 1,000 is 543.

💡Okay, so I've got this chess tournament problem with four students: Andrey, Vanya, Dima, and Sasha. Each of them plays two games against each other. The scoring is straightforward: 1 point for a win, 0.5 points for a draw, and 0 for a loss. First, I need to figure out how many points each student scored based on the given conditions. Let me list out what I know:1. Each student scored a different number of points.2. The ranking is Andrey first, Dima second, Vanya third, and Sasha fourth.3. Andrey won the same number of games as Sasha.Alright, so let's break this down step by step.**Total Number of Games and Points:**There are four participants, and each plays two games against each of the other three. So, each student plays 6 games in total. Since each game awards a total of 1 point (either 1 for a win and 0 for a loss, or 0.5 each for a draw), the total number of points distributed in the tournament is 12 (because there are 12 games in total: 4 students * 3 opponents * 2 games each / 2 to avoid double-counting). So, the sum of all points should be 12.**Unique Scores:**Each student has a different score, so we need four distinct point totals. Andrey is first, Dima second, Vanya third, and Sasha fourth. So, their points should be in descending order without any overlaps.**Andrey and Sasha's Wins:**Andrey and Sasha won the same number of games. Since each game is worth 1 point for a win, the number of wins is equal to the number of points from wins. So, if Andrey has 'w' wins, Sasha also has 'w' wins.**Possible Points for Sasha:**Let me think about Sasha's possible points. Since Sasha is last, their score should be the lowest. The minimum points Sasha could have is 0, but since they played 6 games, it's more likely they have some points. Also, considering that Andrey and Sasha have the same number of wins, and Andrey is first, Sasha's points can't be too low because Andrey's points would be significantly higher.Let me consider the maximum number of points Sasha could have. If Sasha had, say, 3 points, that would mean they won 3 games. But since Andrey also has 3 wins, Andrey would have at least 3 points, but Andrey is first, so Andrey must have more points than Dima, who is second. This might not add up because if Sasha has 3 points, Andrey also has 3 points, but they need to have different total points. So, Sasha's points must be such that when added to Andrey's, they still allow for distinct scores for Dima and Vanya.Let me try to narrow it down. If Sasha has 2 points, that means they won 2 games and lost the other 4. Then Andrey also has 2 wins. But Andrey could have more points from draws. Similarly, Sasha could have some draws as well.Wait, but if Sasha has 2 points, that could be from 2 wins and 0 draws, or 1 win and 2 draws, or 0 wins and 4 draws. But since Sasha is last, it's more likely they have fewer points, so maybe 2 points from 2 wins and 4 losses.But let's see. If Sasha has 2 points, Andrey also has 2 wins, but Andrey could have more points from draws. Let's assume Andrey has 2 wins and some draws. Let's say Andrey has 2 wins and 4 draws. But wait, that would be 6 games, but 2 wins and 4 draws would give Andrey 2 + (4 * 0.5) = 2 + 2 = 4 points.If Andrey has 4 points, then Sasha has 2 points. Now, the remaining points are 12 - 4 - 2 = 6 points for Dima and Vanya.Since Dima is second and Vanya is third, Dima must have more points than Vanya. Let's say Dima has 3.5 points and Vanya has 2.5 points. That adds up to 6 points, which fits.So, the points would be:- Andrey: 4 points- Dima: 3.5 points- Vanya: 2.5 points- Sasha: 2 pointsLet me check if this makes sense. Each has a different score, Andrey is first, Dima second, Vanya third, and Sasha fourth. Andrey and Sasha both have 2 wins (since 2 wins give 2 points, but Andrey also has draws to make it 4 points). Sasha has 2 points from 2 wins and 4 losses, or maybe 1 win and 2 draws, but since Andrey has 2 wins, Sasha must also have 2 wins. So, Sasha has 2 wins and 4 losses, totaling 2 points. Andrey has 2 wins and 4 draws, totaling 4 points.Dima has 3.5 points, which could be from 3 wins and 1 draw and 2 losses, but wait, 3 wins would be 3 points, plus 1 draw (0.5) would make 3.5. But Dima only played 6 games, so 3 wins, 1 draw, and 2 losses make sense.Vanya has 2.5 points, which could be from 2 draws and 2 wins? Wait, no, 2.5 points would be from 5 draws, but that's not possible because Vanya played 6 games. Alternatively, 2 wins and 1 draw and 3 losses: 2 + 0.5 = 2.5 points.Wait, but if Vanya has 2 wins, that's 2 points, and 1 draw, that's 0.5, totaling 2.5. But then, how many wins does Dima have? Dima has 3 wins, which is 3 points, plus 1 draw, totaling 3.5.Andrey has 2 wins and 4 draws, which is 4 points. Sasha has 2 wins and 4 losses, which is 2 points.Let me check the total points:Andrey: 4Dima: 3.5Vanya: 2.5Sasha: 2Total: 4 + 3.5 + 2.5 + 2 = 12 points, which matches the total points possible.Also, each has a different score, and the order is correct. Andrey and Sasha both have 2 wins, which satisfies the condition that they won the same number of games.I think this makes sense. So, the points are:Andrey: 4 pointsDima: 3.5 pointsVanya: 2.5 pointsSasha: 2 points

💡Okay, so I have this logarithmic inequality to solve: log base a of the square root of 2 is less than 1. Hmm, logarithms can be tricky, especially when the base is a variable. Let me think about how to approach this.First, I remember that the logarithm function log_a(x) has different behaviors depending on whether the base a is greater than 1 or between 0 and 1. If a is greater than 1, the function is increasing, which means that as x increases, log_a(x) also increases. On the other hand, if a is between 0 and 1, the function is decreasing, so as x increases, log_a(x) decreases.Given that, I need to consider two separate cases for the base a: one where a is greater than 1 and another where a is between 0 and 1.Let me write down the inequality again:log_a(sqrt(2)) < 1I can rewrite the right side of the inequality as log_a(a) because log_a(a) is always equal to 1, regardless of the value of a (as long as a is positive and not equal to 1, which it isn't in this case because we're considering a as a real number greater than 0 and not equal to 1).So now, the inequality becomes:log_a(sqrt(2)) < log_a(a)Now, let's analyze each case.**Case 1: a > 1**Since a is greater than 1, the logarithm function is increasing. That means if log_a(sqrt(2)) < log_a(a), then the argument of the logarithm on the left must be less than the argument on the right. So, sqrt(2) < a.Therefore, in this case, a must be greater than sqrt(2).**Case 2: 0 < a < 1**Here, the logarithm function is decreasing. So, if log_a(sqrt(2)) < log_a(a), then the argument on the left must be greater than the argument on the right. That is, sqrt(2) > a.But wait, in this case, a is already less than 1, and sqrt(2) is approximately 1.414, which is greater than 1. So, if a is less than 1, then sqrt(2) is definitely greater than a. Therefore, the inequality holds true for all a in this interval.So, putting both cases together, the solution set for a is all real numbers a such that 0 < a < 1 or a > sqrt(2).Let me double-check my reasoning.In Case 1, when a > 1, the function is increasing, so the inequality log_a(sqrt(2)) < log_a(a) simplifies to sqrt(2) < a, which makes sense because as the base increases, the logarithm of a fixed number decreases.In Case 2, when 0 < a < 1, the function is decreasing, so the inequality log_a(sqrt(2)) < log_a(a) implies that sqrt(2) > a. Since sqrt(2) is greater than 1 and a is less than 1, this inequality is automatically true for all a in this interval.Therefore, the range of a is 0 < a < 1 or a > sqrt(2).I think that covers all possibilities. I don't see any mistakes in my reasoning. So, the final answer should be the union of these two intervals.

💡First, I need to define the variables for the width and length of the garden. Let’s denote the width as ( w ) feet. According to the problem, the length exceeds three times the width by 30 feet, so the length can be expressed as ( 3w + 30 ) feet.Next, I'll use the formula for the perimeter of a rectangle, which is ( 2 times (text{length} + text{width}) ). Given that the perimeter is 800 feet, I can set up the equation:[2 times (3w + 30 + w) = 800]Simplifying inside the parentheses:[2 times (4w + 30) = 800]Distributing the 2:[8w + 60 = 800]Subtracting 60 from both sides:[8w = 740]Dividing both sides by 8:[w = 92.5 text{ feet}]Now that I have the width, I can find the length:[text{Length} = 3w + 30 = 3 times 92.5 + 30 = 277.5 + 30 = 307.5 text{ feet}]Finally, to find the area of the garden, I'll multiply the width by the length:[text{Area} = 92.5 times 307.5 = 28,443.75 text{ square feet}]

💡First, I need to subtract the corresponding components of vector (overrightarrow{b}) from vector (overrightarrow{a}).For the x-component:(2 - 1 = 1)For the y-component:(3 - 1 = 2)So, the result of (overrightarrow{a} - overrightarrow{b}) is ((1, 2)), which corresponds to option A.

💡To find the coefficient of the term containing (x^{3}) in the expansion of ((x - 4)^{5}), I will use the binomial theorem. The binomial theorem states that ((a + b)^{n} = sum_{r=0}^{n} C_{n}^{r} a^{n - r} b^{r}), where (C_{n}^{r}) is the binomial coefficient.In this case, (a = x), (b = -4), and (n = 5). The general term in the expansion is (T_{r+1} = C_{5}^{r} x^{5 - r} (-4)^{r}).I need to find the term where the power of (x) is 3. This means I need to solve for (r) in the equation (5 - r = 3), which gives (r = 2).Substituting (r = 2) into the general term formula, I get:[T_{3} = C_{5}^{2} (-4)^{2} x^{3}]Calculating the binomial coefficient (C_{5}^{2}) gives 10, and ((-4)^{2}) equals 16. Multiplying these together with (x^{3}) results in:[T_{3} = 10 times 16 times x^{3} = 160x^{3}]Therefore, the coefficient of the term containing (x^{3}) is 160.