💡Okay, so I need to prove that if a number is divisible by 6, then the sum of the units digit with four times the digits of each of the remaining places of the number is divisible by 6, and the converse is also true. Hmm, that sounds a bit complicated, but let me try to break it down.First, I know that a number is divisible by 6 if and only if it is divisible by both 2 and 3. So, maybe I can use the divisibility rules for 2 and 3 to help with this proof.Let's start by considering a number N. I can represent N in its decimal form as:N = a_k * 10^k + a_{k-1} * 10^{k-1} + ... + a_1 * 10 + a_0where a_0 is the units digit, a_1 is the tens digit, and so on, up to a_k which is the highest place value digit.Now, the problem is asking about the sum of the units digit (a_0) and four times the sum of the remaining digits (a_1 + a_2 + ... + a_k). So, let's denote this sum as S:S = a_0 + 4*(a_1 + a_2 + ... + a_k)I need to show that if N is divisible by 6, then S is also divisible by 6, and vice versa.First, let's assume that N is divisible by 6. That means N ≡ 0 mod 6. Since 6 is 2 * 3, N must be divisible by both 2 and 3.Divisibility by 2: For N to be divisible by 2, the units digit a_0 must be even.Divisibility by 3: For N to be divisible by 3, the sum of its digits (a_0 + a_1 + a_2 + ... + a_k) must be divisible by 3.Now, I need to relate this to the sum S. Let's see:S = a_0 + 4*(a_1 + a_2 + ... + a_k)I can rewrite this as:S = a_0 + 4*(sum of digits excluding a_0)But I also know that the sum of all digits is (a_0 + a_1 + a_2 + ... + a_k). Let's denote this sum as T:T = a_0 + a_1 + a_2 + ... + a_kSo, S = a_0 + 4*(T - a_0) = a_0 + 4T - 4a_0 = 4T - 3a_0Hmm, interesting. So S = 4T - 3a_0Now, since N is divisible by 3, T must be divisible by 3. So, T ≡ 0 mod 3.Also, since N is divisible by 2, a_0 must be even, so a_0 ≡ 0 mod 2.Now, let's see what S is modulo 6.First, since T ≡ 0 mod 3, 4T ≡ 0 mod 3, because 4 is congruent to 1 mod 3, so 4T ≡ T ≡ 0 mod 3.Also, since a_0 is even, 3a_0 is divisible by 6, because 3*even is divisible by 6.So, S = 4T - 3a_0 ≡ 0 - 0 ≡ 0 mod 6.Therefore, S is divisible by 6.Okay, that seems to work for the forward direction. Now, for the converse: if S is divisible by 6, then N is divisible by 6.So, assume S ≡ 0 mod 6.Again, S = 4T - 3a_0 ≡ 0 mod 6.Let's see what this tells us about T and a_0.First, since 4T - 3a_0 ≡ 0 mod 6, we can write:4T ≡ 3a_0 mod 6Now, 4T ≡ 3a_0 mod 6Let's consider this congruence modulo 2 and modulo 3 separately, since 6 is the product of 2 and 3.Modulo 2:4T ≡ 0 mod 2, because 4 is even.3a_0 ≡ a_0 mod 2, since 3 is odd.So, 0 ≡ a_0 mod 2, which means a_0 is even. Therefore, N is divisible by 2.Modulo 3:4T ≡ T mod 3, because 4 ≡ 1 mod 3.3a_0 ≡ 0 mod 3, because 3 is a multiple of 3.So, T ≡ 0 mod 3.Therefore, the sum of the digits T is divisible by 3, which means N is divisible by 3.Since N is divisible by both 2 and 3, it is divisible by 6.So, the converse is also true.Wait, did I cover all the cases? Let me double-check.In the forward direction, I assumed N is divisible by 6, so it's divisible by 2 and 3. Then, I expressed S in terms of T and a_0, and showed that S is divisible by 6.In the converse, I assumed S is divisible by 6, and from that, I deduced that a_0 is even (so N is divisible by 2) and that T is divisible by 3 (so N is divisible by 3). Therefore, N is divisible by 6.Yes, that seems to cover both directions.I think I've got it!

💡Okay, so I have this series to compute:[ 5(1+5(1+5(1+5(1+5(1+5(1+5(1+5(1+5(1+5(1+5(1+5)))))))))) ]Wow, that's a lot of nested parentheses. It looks like a repeated pattern where each term is 5 times something that includes another 5 times something else, and so on. I need to figure out a way to simplify this expression or find a pattern that can help me compute the sum without having to expand all those parentheses manually.First, let me try to understand the structure of this expression. It starts with 5 multiplied by (1 + 5 times something). Inside that something, it's again 1 + 5 times something else, and this continues all the way down. So, it's like a chain of 5s and 1s. Maybe I can represent this recursively or find a formula for it.Let me denote the entire expression as S. So,[ S = 5(1 + 5(1 + 5(1 + cdots ))) ]But wait, how many times does this pattern repeat? Looking at the original expression, I can count the number of 5s. Let's see: starting from the outermost 5, each subsequent term adds another 5. Counting them, I think there are 12 fives in total. So, the expression is 12 layers deep.Hmm, so if there are 12 layers, maybe I can express this as a geometric series. A geometric series is a series where each term is a constant multiple of the previous term. In this case, each term seems to be multiplied by 5. Let me see if I can break down the expression into a sum of powers of 5.Starting from the innermost part, the expression is 1 + 5. Then, moving one layer out, it's 1 + 5*(1 + 5). Then, another layer out, it's 1 + 5*(1 + 5*(1 + 5)), and so on. So, each time, we're adding another term that's a multiple of 5.If I expand this step by step, maybe I can see a pattern.Let's start from the innermost part:1. Innermost: 1 + 5 = 62. Next layer: 1 + 5*6 = 1 + 30 = 313. Next layer: 1 + 5*31 = 1 + 155 = 1564. Next layer: 1 + 5*156 = 1 + 780 = 7815. Next layer: 1 + 5*781 = 1 + 3905 = 39066. Next layer: 1 + 5*3906 = 1 + 19530 = 195317. Next layer: 1 + 5*19531 = 1 + 97655 = 976568. Next layer: 1 + 5*97656 = 1 + 488280 = 4882819. Next layer: 1 + 5*488281 = 1 + 2441405 = 244140610. Next layer: 1 + 5*2441406 = 1 + 12207030 = 1220703111. Next layer: 1 + 5*12207031 = 1 + 61035155 = 6103515612. Finally, multiply by 5: 5*61035156 = 305175780Wait, so after expanding each layer step by step, I end up with 305,175,780. That seems like a lot, but maybe there's a formulaic way to get there without expanding all 12 layers.Thinking about it, each layer adds another power of 5. So, the first term is 5^1, the next is 5^2, and so on, up to 5^12. But actually, when I expanded it, I saw that each step was 1 + 5 times the previous result. That might mean that the entire expression can be represented as a geometric series.Let me recall the formula for the sum of a geometric series:[ S = a frac{r^n - 1}{r - 1} ]Where:- ( a ) is the first term,- ( r ) is the common ratio,- ( n ) is the number of terms.In this case, the first term ( a ) is 5, the common ratio ( r ) is also 5, and the number of terms ( n ) is 12 because there are 12 layers of multiplication by 5.Plugging these values into the formula:[ S = 5 frac{5^{12} - 1}{5 - 1} ]Simplifying the denominator:[ S = 5 frac{5^{12} - 1}{4} ]Now, I need to compute ( 5^{12} ). Let's calculate that step by step:- ( 5^1 = 5 )- ( 5^2 = 25 )- ( 5^3 = 125 )- ( 5^4 = 625 )- ( 5^5 = 3125 )- ( 5^6 = 15625 )- ( 5^7 = 78125 )- ( 5^8 = 390625 )- ( 5^9 = 1953125 )- ( 5^{10} = 9765625 )- ( 5^{11} = 48828125 )- ( 5^{12} = 244140625 )So, ( 5^{12} = 244,140,625 ).Now, subtract 1:[ 244,140,625 - 1 = 244,140,624 ]Multiply by 5:[ 5 times 244,140,624 = 1,220,703,120 ]Then, divide by 4:[ frac{1,220,703,120}{4} = 305,175,780 ]So, the sum of the series is 305,175,780.Wait, that matches the result I got when I expanded each layer step by step. That's reassuring. So, both methods give me the same answer, which is good.I think I've got it. The series is a geometric series with 12 terms, each term being 5 raised to the power of its position, starting from 1. Using the geometric series formula, I can compute the sum without having to expand all the parentheses manually, which would be time-consuming and error-prone.Just to double-check, let me verify the calculations:- ( 5^{12} = 244,140,625 ) (correct)- ( 244,140,625 - 1 = 244,140,624 ) (correct)- ( 5 times 244,140,624 = 1,220,703,120 ) (correct)- ( 1,220,703,120 / 4 = 305,175,780 ) (correct)Everything checks out. So, I'm confident that the sum of the series is indeed 305,175,780.

💡First, I need to determine the total number of parrots in the pet store. Since there are 8 cages and each cage contains 2 parrots, I can calculate the total number of parrots by multiplying the number of cages by the number of parrots per cage.Next, I'll subtract the total number of parrots from the overall total number of birds to find out how many parakeets there are in total.Finally, to find out how many parakeets are in each cage, I'll divide the total number of parakeets by the number of cages.

💡Okay, so I've got this equation to solve: (x - 2)/(x - 3) = 2/(x - 3). Hmm, it's a fractional equation, and I need to find the value of x that makes this true. Let me think about how to approach this.First, I remember that when you have fractions equal to each other, you can cross-multiply to get rid of the denominators. But wait, in this case, both sides have the same denominator, which is (x - 3). That might make things easier. Maybe I can just set the numerators equal to each other?So, if (x - 2)/(x - 3) equals 2/(x - 3), then the numerators should be equal as long as the denominators are not zero. That means x - 3 cannot be zero, so x cannot be 3. I should keep that in mind because if I get x = 3 as a solution, it would actually be invalid.Alright, so setting the numerators equal: x - 2 = 2. Solving for x, I add 2 to both sides: x = 4. Okay, that seems straightforward. But wait, I need to make sure that x = 4 doesn't make the denominator zero. Plugging it back in: (4 - 3) is 1, which is not zero. So, x = 4 is a valid solution.Let me double-check my work to make sure I didn't make any mistakes. Starting with the original equation: (x - 2)/(x - 3) = 2/(x - 3). If I plug x = 4 into both sides, the left side becomes (4 - 2)/(4 - 3) = 2/1 = 2. The right side is 2/(4 - 3) = 2/1 = 2. Both sides are equal, so x = 4 works.I also need to confirm that there are no other solutions. Since the denominators are the same and we set the numerators equal, there shouldn't be any other solutions unless there's something I'm missing. Maybe I should consider if there's a case where the denominators are zero, but we already established that x can't be 3 because that would make the denominator zero, which is undefined.So, after checking everything, I'm confident that x = 4 is the correct solution. It doesn't make any part of the equation undefined, and it satisfies the original equation when plugged back in. I think that's it!

💡Okay, so I have this problem here: The arithmetic mean of an even number of consecutive even integers is y. I need to find the sum of the smallest and largest of these integers in terms of y. Hmm, let me try to figure this out step by step.First, let's break down what the problem is saying. We have an even number of consecutive even integers. So, for example, if we have 4 integers, they might be 2, 4, 6, 8. Or if we have 6 integers, they might be 10, 12, 14, 16, 18, 20. The key points here are that the number of integers is even, and each integer is even and consecutive, meaning each is 2 more than the previous one.The arithmetic mean of these integers is given as y. The arithmetic mean is just the average, so if I have n integers, their mean is the sum of the integers divided by n. So, if I can express the sum of these integers in terms of y and n, maybe I can find a relationship that helps me find the sum of the smallest and largest integers.Let me denote the number of integers as n, which is even. Let's say the first integer is a. Since they are consecutive even integers, the sequence would be a, a+2, a+4, ..., up to a + 2(n-1). So, the last term is a + 2(n-1).Now, the sum of these integers can be calculated using the formula for the sum of an arithmetic series. The formula is: sum = (number of terms)/2 * (first term + last term). So, in this case, the sum S would be:S = n/2 * (a + [a + 2(n-1)]) = n/2 * (2a + 2(n-1)) = n/2 * 2(a + n - 1) = n(a + n - 1)So, the sum S is n(a + n - 1). Given that the arithmetic mean is y, we have:y = S / n = [n(a + n - 1)] / n = a + n - 1So, from this, we can express a in terms of y and n:a = y - n + 1Now, the problem asks for the sum of the smallest and largest integers. The smallest integer is a, and the largest integer is a + 2(n-1). So, their sum is:a + [a + 2(n-1)] = 2a + 2(n - 1)Substituting the expression for a from above:2(y - n + 1) + 2(n - 1) = 2y - 2n + 2 + 2n - 2 = 2yWait, that's interesting. The terms involving n cancel out, leaving us with 2y. So, the sum of the smallest and largest integers is 2y.Let me verify this with an example to make sure I didn't make a mistake. Suppose we have 4 consecutive even integers. Let's choose 2, 4, 6, 8. The mean is (2 + 4 + 6 + 8)/4 = 20/4 = 5. So, y = 5. The smallest integer is 2, the largest is 8, and their sum is 10, which is 2*5 = 10. That checks out.Another example: 6 consecutive even integers starting at 10: 10, 12, 14, 16, 18, 20. The mean is (10 + 12 + 14 + 16 + 18 + 20)/6 = 90/6 = 15. So, y = 15. The smallest is 10, the largest is 20, their sum is 30, which is 2*15 = 30. That also works.Wait, let me try with a different starting point. Suppose we have 2 consecutive even integers: 4 and 6. The mean is (4 + 6)/2 = 5. So, y = 5. The sum of the smallest and largest is 4 + 6 = 10, which is 2*5 = 10. Perfect.What if n is larger? Let's say n=8: 12, 14, 16, 18, 20, 22, 24, 26. The mean is (12 + 14 + 16 + 18 + 20 + 22 + 24 + 26)/8. Let's compute that: 12+14=26, 16+18=34, 20+22=42, 24+26=50. So, 26+34=60, 42+50=92. Total sum is 60+92=152. Mean is 152/8=19. So, y=19. The smallest is 12, largest is 26, sum is 38, which is 2*19=38. Yep, that works too.So, it seems that regardless of the number of terms (as long as it's even), the sum of the smallest and largest integers is always twice the mean. That makes sense because in a symmetric distribution, the mean is the midpoint, so the average of the smallest and largest would be the mean. Therefore, their sum would be twice the mean.Wait, let me think about that symmetry. If the integers are evenly spaced, which they are since they are consecutive even integers, then the mean is the average of the first and last terms. So, (first + last)/2 = mean. Therefore, first + last = 2*mean. That's a more straightforward way to see it.So, in general, for any arithmetic sequence, the mean is equal to the average of the first and last terms. Therefore, the sum of the first and last terms is twice the mean. Since in this case, the mean is y, the sum is 2y.That's a much simpler way to think about it. I guess I overcomplicated it initially by trying to express a in terms of y and n, but recognizing the property of arithmetic sequences makes it much clearer.So, to summarize, because the integers form an arithmetic sequence with a common difference of 2, the mean is the average of the first and last terms. Therefore, the sum of the first and last terms is twice the mean, which is 2y.I think that solidifies the conclusion. It's always good to check with examples and also to see if there's a more straightforward reasoning path once you recognize the underlying principles, like the properties of arithmetic sequences.**Final Answer**The sum of the smallest and largest integers is boxed{2y}.

💡Okay, so I have this problem where someone bought a plot of land that's shaped like an equilateral triangle, bounded by three straight roads. They want to build a house in such a way that there are three straight driveways leading to it from each of those roads. The goal is to figure out where to build the house so that the total cost of laying the driveways is minimized.First, I need to visualize the scenario. An equilateral triangle has all sides equal and all angles equal to 60 degrees. The three roads form the sides of this triangle. The house needs to be somewhere inside this triangle, and from the house, there should be three straight paths (driveways) leading to each of the three roads.So, the problem is essentially asking for the point inside the equilateral triangle where the sum of the distances from this point to each of the three sides is minimized. This is because the cost of laying the driveways would be proportional to their lengths, assuming the cost per unit length is the same for all driveways.I recall that in an equilateral triangle, there's a special point called the centroid, which is the intersection of the medians. The centroid is also the center of mass of the triangle. I wonder if this point has any significance in minimizing distances to the sides.Wait, there's also something called the incenter, which is the point where the angle bisectors meet, and it's equidistant from all sides of the triangle. In an equilateral triangle, the centroid, incenter, circumcenter, and orthocenter all coincide at the same point. So, maybe this common point is the one that minimizes the sum of distances to the sides.But I'm not entirely sure. Maybe I should think about this more carefully. If I consider any point inside the triangle, the sum of the perpendicular distances from that point to the three sides is constant and equal to the height of the triangle. Is that true?Let me recall. For an equilateral triangle with side length 'a', the height 'h' can be calculated using the Pythagorean theorem. If we split the triangle down the middle, we get a right triangle with base 'a/2', height 'h', and hypotenuse 'a'. So, h = sqrt(a^2 - (a/2)^2) = sqrt(3a^2/4) = (a*sqrt(3))/2.Now, if I take any point inside the triangle, and drop perpendiculars to each of the three sides, the sum of these three distances should be equal to the height 'h'. That seems to be a property of equilateral triangles. So, no matter where I place the house inside the triangle, the total length of the driveways will always be equal to the height of the triangle.Wait, does that mean that the total cost is the same regardless of where I build the house? That doesn't sound right. Maybe I'm misunderstanding the problem.Perhaps the cost isn't just the sum of the distances, but something else. Maybe the cost depends on the individual lengths of the driveways, and we need to minimize the maximum length of any single driveway. Or maybe it's about minimizing the total length, but considering some weights or costs associated with each road.But the problem statement says "to minimize the cost of laying the driveways," and it doesn't specify any particular weighting or constraints. So, I think it's safe to assume that the cost is directly proportional to the total length of the driveways.Given that, and the property that the sum of the perpendicular distances from any interior point to the sides is constant, it seems that the total cost would be the same no matter where the house is built inside the triangle.But that contradicts my initial thought that the centroid might be the optimal point. Maybe I need to revisit this.Alternatively, perhaps the problem is not about the sum of the distances, but about something else, like the product of the distances, or the maximum distance. Let me think.If it's about the product of the distances, then the centroid might indeed be the point that minimizes or maximizes this product. But I'm not sure. If it's about the maximum distance, then the centroid might be the point that minimizes the maximum distance to the sides.But the problem specifically mentions "three straight driveways lead to it from each of those roads." So, it's about having three driveways, each leading to a different road. So, the lengths of these driveways would be the perpendicular distances from the house to each of the three roads.Given that, and the fact that the sum of these distances is constant, it seems that the total cost would be the same regardless of where the house is built. Therefore, the position of the house doesn't affect the total cost.But that seems counterintuitive. Usually, in optimization problems, there is a specific point that minimizes or maximizes a certain quantity. Maybe I'm missing something here.Wait, perhaps the problem is not about the sum of the distances, but about something else. Maybe the cost is proportional to the sum of the squares of the distances, or some other function.Alternatively, maybe the problem is about minimizing the maximum distance from the house to any of the roads, which would be a different optimization problem.But the problem statement doesn't specify, so I have to make an assumption. Given that it's about laying driveways, and driveways are typically built as straight paths from the house to the road, the cost would likely be proportional to the length of each driveway. Therefore, the total cost would be the sum of the lengths of the three driveways.Given that, and the property that the sum of the perpendicular distances from any interior point to the sides of an equilateral triangle is constant, it follows that the total cost is the same regardless of where the house is built.But that seems to suggest that the position of the house doesn't matter, which might be the case. However, I want to double-check this.Let me consider a simple case. Suppose the house is built at one of the vertices of the triangle. Then, two of the driveways would have zero length (since the house is already on two roads), and the third driveway would have a length equal to the height of the triangle. So, the total cost would be equal to the height.On the other hand, if the house is built at the centroid, then each driveway would have a length equal to one-third of the height. So, the total cost would be three times one-third of the height, which is again equal to the height.Wait, so in both cases, the total cost is the same. That seems to confirm the earlier property that the sum of the distances is constant.Therefore, regardless of where the house is built inside the triangle, the total length of the driveways will always be equal to the height of the triangle. Hence, the cost is the same everywhere.But then, why does the problem ask where to build the house to minimize the cost? If the cost is the same everywhere, then it doesn't matter where you build it.Maybe I'm misunderstanding the problem. Perhaps the cost isn't just the sum of the lengths, but something else, like the sum of the squares of the lengths, or the maximum length.If it's the sum of the squares, then the centroid would minimize this sum. If it's the maximum length, then the centroid would minimize the maximum distance.But since the problem doesn't specify, I have to go with the most straightforward interpretation, which is that the cost is proportional to the total length of the driveways.Given that, and the fact that the sum of the distances is constant, the position of the house doesn't affect the total cost. Therefore, the house can be built anywhere inside the triangle, and the cost will be the same.However, I'm still a bit confused because usually, optimization problems have a unique solution. Maybe the problem is more about finding a specific point, like the centroid, even though the total cost is the same everywhere.Alternatively, perhaps the problem is considering the cost as the sum of the squares of the distances, in which case the centroid would indeed be the point that minimizes this sum.Let me think about that. If we consider the sum of the squares of the distances from a point to the three sides, then the centroid would minimize this sum. This is because the centroid is the point that minimizes the sum of squared distances to the vertices, but I'm not sure about the sides.Wait, actually, in general, the centroid minimizes the sum of squared distances to the vertices, but for the sides, it might be different.Alternatively, maybe the problem is about minimizing the maximum distance from the house to any of the roads. In that case, the centroid would be the point that minimizes the maximum distance, as it's equidistant to all sides.But again, the problem doesn't specify, so I have to make an assumption.Given that, and considering that the sum of the distances is constant, I think the most straightforward answer is that the house can be built anywhere inside the triangle, as the total cost will be the same.But to be thorough, let me consider the sum of the squares of the distances. Suppose we have an equilateral triangle with side length 'a', and we place the house at a point (x, y) inside the triangle. The distances to the three sides can be expressed in terms of x and y, and then we can compute the sum of the squares.However, this might be more complicated than necessary, and since the problem doesn't specify, I think it's safe to stick with the initial interpretation.Therefore, the conclusion is that the house can be built anywhere inside the equilateral triangle, as the total cost of laying the driveways will be the same regardless of the position.