🧠:Okay, so I need to find the equation of the tangent line to the curve y = (x⁵ + 1)/(x⁴ + 1) at the point where x₀ = 1. Hmm, let me start by recalling what a tangent line is. The tangent line at a point on a curve is a straight line that just touches the curve at that point and has the same slope as the curve does at that point. So, to find the equation of the tangent line, I need two things: a point on the line and the slope of the line at that point.First, let me find the point on the curve when x₀ = 1. That means I need to plug x = 1 into the equation of the curve to get the corresponding y-coordinate. Let me compute that:y = (1⁵ + 1)/(1⁴ + 1) = (1 + 1)/(1 + 1) = 2/2 = 1. So the point is (1, 1). Got that down.Next, the slope of the tangent line is given by the derivative of the function at x₀ = 1. So I need to compute dy/dx for y = (x⁵ + 1)/(x⁴ + 1) and then evaluate it at x = 1.Since this is a rational function (a quotient of two functions), I should use the quotient rule for derivatives. The quotient rule says that if we have a function f(x)/g(x), its derivative is [f’(x)g(x) - f(x)g’(x)] / [g(x)]². Let me define f(x) = x⁵ + 1 and g(x) = x⁴ + 1. Then f’(x) would be the derivative of x⁵ + 1, which is 5x⁴, and g’(x) is the derivative of x⁴ + 1, which is 4x³.Applying the quotient rule:dy/dx = [5x⁴*(x⁴ + 1) - (x⁵ + 1)*4x³] / (x⁴ + 1)².Now I need to simplify this expression as much as possible before plugging in x = 1. Let's work on the numerator first.First term: 5x⁴*(x⁴ + 1) = 5x⁸ + 5x⁴Second term: (x⁵ + 1)*4x³ = 4x³*(x⁵ + 1) = 4x⁸ + 4x³So the numerator becomes:[5x⁸ + 5x⁴] - [4x⁸ + 4x³] = 5x⁸ + 5x⁴ - 4x⁸ - 4x³ = (5x⁸ - 4x⁸) + 5x⁴ - 4x³ = x⁸ + 5x⁴ - 4x³Therefore, dy/dx = (x⁸ + 5x⁴ - 4x³) / (x⁴ + 1)².Now I need to evaluate this derivative at x = 1. Let's plug in x = 1:Numerator at x=1: (1)^8 + 5*(1)^4 - 4*(1)^3 = 1 + 5*1 - 4*1 = 1 + 5 - 4 = 2.Denominator at x=1: (1^4 + 1)^2 = (1 + 1)^2 = 2^2 = 4.So dy/dx at x=1 is 2/4 = 1/2. Therefore, the slope of the tangent line is 1/2.Now that I have the slope (m = 1/2) and the point (1, 1), I can use the point-slope form of the equation of a line to write the equation of the tangent line. The point-slope form is:y - y₁ = m(x - x₁)Plugging in the values:y - 1 = (1/2)(x - 1)To convert this to slope-intercept form (y = mx + b), I can distribute the 1/2:y - 1 = (1/2)x - 1/2Then add 1 to both sides:y = (1/2)x - 1/2 + 1Simplifying the constants:-1/2 + 1 = 1/2, so:y = (1/2)x + 1/2Let me just double-check my steps to make sure I didn't make any errors. First, calculating the point (1,1) seems straightforward. Then, applying the quotient rule. Let me verify the derivative again:f = x⁵ +1, so f’ = 5x⁴g = x⁴ +1, so g’ = 4x³dy/dx = [5x⁴*(x⁴ +1) - (x⁵ +1)*4x³]/(x⁴ +1)^2Expanding numerator:5x⁴*(x⁴) +5x⁴*1 -4x³*(x⁵) -4x³*1= 5x⁸ +5x⁴ -4x⁸ -4x³Combine like terms:5x⁸ -4x⁸ = x⁸5x⁴ -4x³ remainsSo numerator is x⁸ +5x⁴ -4x³, which at x=1 gives 1 +5 -4 = 2. Denominator is (1 +1)^2 =4. So 2/4=1/2. Correct.Then the equation of the tangent line using point-slope. The steps look correct. Let me check the final equation with x=1 plugged in. y=(1/2)(1)+1/2=1/2 +1/2=1, which matches the point (1,1). That checks out. If I plug in x=0, for example, y=0 +1/2=1/2. Seems reasonable.Another check: maybe compute another point on the tangent line and see if it makes sense. Let me take x=2. The tangent line at x=2 would be y=(1/2)(2)+1/2=1 +1/2=3/2. Now, the original function at x=2 is y=(32 +1)/(16 +1)=33/17≈1.941, which is much higher. But the tangent line at x=1 only approximates the curve near x=1, so at x=2, it's not surprising they differ. So the tangent line equation seems plausible.Is there any other check I can perform? Maybe compute the derivative again or see if there's an alternative way to approach the problem. Alternatively, using the product rule by rewriting the function as (x⁵ +1)*(x⁴ +1)^-1. But the quotient rule was straightforward here.Alternatively, maybe simplify the original function before differentiating, but I don't see an obvious simplification. The numerator and denominator don't factor in a way that cancels out terms. So I think the approach is solid.Thus, the equation of the tangent line is y = (1/2)x + 1/2.

✅:To determine the equation of the tangent line to the curve ( y = frac{x^5 + 1}{x^4 + 1} ) at ( x_0 = 1 ):1. Find the point on the curve at ( x_0 = 1 ): [ y(1) = frac{1^5 + 1}{1^4 + 1} = frac{2}{2} = 1 quad Rightarrow quad text{Point: } (1, 1) ]2. Compute the derivative ( frac{dy}{dx} ) using the quotient rule: - Let ( f(x) = x^5 + 1 ) and ( g(x) = x^4 + 1 ). - Derivatives: ( f'(x) = 5x^4 ), ( g'(x) = 4x^3 ). - Apply the quotient rule: [ frac{dy}{dx} = frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} = frac{5x^4(x^4 + 1) - (x^5 + 1)(4x^3)}{(x^4 + 1)^2} ] - Simplify the numerator: [ 5x^4(x^4 + 1) - 4x^3(x^5 + 1) = 5x^8 + 5x^4 - 4x^8 - 4x^3 = x^8 + 5x^4 - 4x^3 ] - Thus: [ frac{dy}{dx} = frac{x^8 + 5x^4 - 4x^3}{(x^4 + 1)^2} ]3. Evaluate the derivative at ( x = 1 ): - Numerator at ( x = 1 ): ( 1 + 5 - 4 = 2 ). - Denominator at ( x = 1 ): ( (1 + 1)^2 = 4 ). - Slope ( m ): [ m = frac{2}{4} = frac{1}{2} ]4. Form the tangent line equation using point-slope form: [ y - y_1 = m(x - x_1) quad Rightarrow quad y - 1 = frac{1}{2}(x - 1) ] - Simplify to slope-intercept form: [ y = frac{1}{2}x + frac{1}{2} ]Final Answer: The equation of the tangent line is (boxed{y = frac{1}{2}x + frac{1}{2}}).

🔑:To find the equation of the tangent line to the given curve at the point with abscissa x_0 = 1, we need to follow these steps:1. Find the derivative ( y' ) of the function ( y = frac{x^5 + 1}{x^4 + 1} ) using the quotient rule. The quotient rule states that for a function ( frac{f(x)}{g(x)} ), the derivative is given by: [ left( frac{f(x)}{g(x)} right)' = frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} ]2. Identify ( f(x) ) and ( g(x) ) and their derivatives ( f'(x) ) and ( g'(x) ). Let ( y = frac{f(x)}{g(x)} ) where: [ f(x) = x^5 + 1 quad text{and} quad g(x) = x^4 + 1 ] So, the derivatives are: [ f'(x) = 5x^4 quad text{and} quad g'(x) = 4x^3 ]3. Apply the quotient rule: [ y' = left( frac{x^5 + 1}{x^4 + 1} right)' = frac{(x^5 + 1)' cdot (x^4 + 1) - (x^5 + 1) cdot (x^4 + 1)'}{(x^4 + 1)^2} = frac{5x^4 cdot (x^4 + 1) - (x^5 + 1) cdot 4x^3}{(x^4 + 1)^2} ]4. Simplify the expression: [ y' = frac{5x^8 + 5x^4 - 4x^8 - 4x^3}{(x^4 + 1)^2} = frac{x^8 + 5x^4 - 4x^3}{(x^4 + 1)^2} ]5. Evaluate the derivative at ( x = x_0 = 1 ): [ y' left(1right) = frac{1^8 + 5 cdot 1^4 - 4 cdot 1^3}{(1^4 + 1)^2} = frac{1 + 5 - 4}{(1 + 1)^2} = frac{2}{4} = frac{1}{2} ]6. Determine the value of ( y ) at ( x_0 = 1 ): [ y(1) = frac{1^5 + 1}{1^4 + 1} = frac{1 + 1}{1 + 1} = frac{2}{2} = 1 ]7. Write the equation of the tangent line using the point-slope form: The equation of a line in point-slope form is: [ y - y_0 = m(x - x_0) ] where ( y_0 = 1 ) and ( m = y' left(1right) = frac{1}{2} ). Substituting the values, we get: [ y - 1 = frac{1}{2} (x - 1) ]8. Simplify to get the final equation of the tangent line: [ y - 1 = frac{1}{2}x - frac{1}{2} ] [ y = frac{1}{2}x - frac{1}{2} + 1 ] [ y = frac{1}{2}x + frac{1}{2} ]Hence, the equation of the tangent line to the curve at the point where ( x_0 = 1 ) is:[boxed{y = frac{1}{2}x + frac{1}{2}}]

🧠:Okay, let's try to figure out this problem step by step. So Masha puts either 50 or 100 rubles into her piggy bank every week. Every 4 weeks, she takes the smallest denomination bill from the piggy bank and gives it to her sister. After a year, the total she gave her sister is 1250 rubles. We need to find the minimum amount she could have saved during the year.First, let's understand the timeline. A year has 52 weeks. Since she gives money every 4 weeks, that would happen 52 / 4 = 13 times in a year. So she gives her sister 13 bills, each time the smallest one in the piggy bank. The total from these 13 bills is 1250 rubles. Each bill she gives is either 50 or 100 rubles. Wait, but the problem says she adds a bill of 50 or 100 each week. So each week, she adds one bill. Then every 4 weeks, she removes the smallest bill. So in those 4 weeks, she has added 4 bills (either 50s or 100s). Then she gives away the smallest one. So, over the year, each 4-week period contributes one bill to her sister. Since there are 13 such periods (52 weeks), she gives 13 bills. The total of these 13 bills is 1250. So we need to find how many 50s and 100s she gave to her sister such that their sum is 1250. Let’s denote the number of 50s as x and 100s as y. Then:50x + 100y = 1250And x + y = 13Let me solve this system. From the second equation, x = 13 - y. Substitute into the first equation:50(13 - y) + 100y = 1250650 - 50y + 100y = 1250650 + 50y = 125050y = 600y = 12So x = 13 - 12 = 1. So she gave her sister 1 bill of 50 and 12 bills of 100. Wait, that seems high. But let's check: 1*50 +12*100 =50 +1200=1250. Yes, correct.But wait, how does this relate to the minimum total she could have accumulated? Because the total she saved is the sum of all the bills she added minus the ones she gave away. Wait, no. Wait, she gives the bills to her sister every 4 weeks, so the total she has at the end of the year is the total she put in minus the 1250 she gave away. But the question is asking for the minimum amount she could have accumulated. So accumulated during this time. Maybe "accumulated" refers to the total she saved, i.e., the amount she put in minus what she gave away. But we need to clarify: is the accumulated amount the total she saved (remaining in the piggy bank) or the total she added over the year? The problem says "the minimum amount of money she could have accumulated during this time". Probably, it's the total she has in the piggy bank at the end of the year, which would be the total she put in minus the 1250 she gave to her sister. But to minimize this amount, she needs to maximize the amount she gave away (but she already gave 1250, so maybe she needs to structure the bills in such a way that she removes as much as possible, but the total removed is fixed at 1250). Wait, but 1250 is fixed. So to minimize the total accumulated, she needs to minimize the total she put in, given that she gave away 1250. Therefore, the problem reduces to finding the minimal total contributions (sum of all weekly bills) such that the sum of the smallest bills taken every 4 weeks is 1250. So the minimal total she put in is the sum of all weekly bills (52 weeks, each week 50 or 100) minus 1250. Wait no, the accumulated amount is the total she saved, which is the sum of all bills she put in minus the sum she gave away. So to minimize the accumulated amount, she needs to minimize the total she put in (since 1250 is fixed). Therefore, we need to find the minimal total she could have put in over the year, given that the sum of the bills she gave to her sister is 1250. Therefore, the problem is to minimize the sum of 52 bills (each 50 or 100) such that the sum of the smallest bills taken every 4 weeks is 1250.So we need to arrange the 52 bills (each week 50 or 100) into 13 groups of 4 weeks each, such that in each group, the smallest bill is either 50 or 100, and the sum of these smallest bills is 1250. We already found that she must have given 1x50 and 12x100. So, in 12 of the 4-week periods, the smallest bill was 100, and in 1 period, the smallest was 50.But how does that affect the total she put in? To minimize the total she put in, we need to minimize the number of 100s, because 100 is more than 50. Wait, no. Wait, if she gives away a 100, that requires that all four bills in that 4-week period are 100s, because the smallest bill is 100. If there's at least one 50 in the 4-week period, the smallest bill would be 50. Therefore, for a 4-week period to give a 100, all four bills must be 100. Otherwise, if there's any 50, the smallest is 50. Therefore, in the 12 periods where she gave 100, each of those periods must have four 100s. And in the 1 period where she gave 50, there must be at least one 50, and the other three can be either 50 or 100. But to minimize the total she put in, she would want to have as many 50s as possible in the periods where the smallest bill is 50. Because 50s are smaller, but since she has to give away the smallest bill, but the rest can be 50s or 100s. Wait, but to minimize the total, she should minimize the total contributions. So in the period where she gives a 50, she needs to have at least one 50, and the other three weeks can be 50s or 100s. But since we want to minimize the total, she should use as many 50s as possible in that period. Because 50 is less than 100. Therefore, in that period with a 50 given, she would have four 50s, right? Because if she has four 50s, that's the minimal total for that period. If she has one 50 and three 100s, the total for that period would be 50 + 3*100 = 350, whereas four 50s would be 200. So definitely, to minimize the total, in the period where she gives 50, she should have all four weeks as 50s. That way, the total for that period is 4*50=200, and she gives away one 50. Then, in the other 12 periods where she gives 100, each of those periods must have all four weeks as 100s (since if there was a 50, the smallest would be 50, which contradicts giving 100). So those 12 periods each contribute 4*100=400. Therefore, the total contributions would be 12*400 + 1*200 = 4800 + 200 = 5000. Then, the total she gave away is 1250, so the accumulated amount is 5000 - 1250 = 3750. But wait, is that correct? Let me check. If she has 12 periods with four 100s, each of those periods she gives away one 100. So 12*100=1200. Then one period with four 50s, giving away one 50. Total given: 1200 +50=1250. Total contributions: 12*4*100 +1*4*50=4800 +200=5000. So accumulated amount is 5000 -1250=3750. But wait, maybe there's a way to have fewer 100s. Because maybe in the periods where she gives 100, she doesn't need to have all four weeks as 100s. Wait, no. Because if in a 4-week period there's at least one 50, then the smallest bill is 50, so she would have to give 50. Therefore, to give 100, all four weeks must be 100. Therefore, in those 12 periods where she gives 100, they must be four 100s each. And in the one period where she gives 50, she must have at least one 50, but to minimize the total, she should have all four 50s. But what if in that one period, instead of four 50s, she had three 50s and one 100? Then the total for that period would be 3*50 +1*100=250, which is more than 200, so that's worse. Similarly, any mix would result in a higher total. Therefore, the minimal total for that period is 200. So the total contributions are indeed 5000. Therefore, the accumulated amount is 5000 -1250=3750. But wait, the question says "the minimum amount of money she could have accumulated during this time". So is the accumulated amount the total she saved (i.e., remaining in the piggy bank), which would be total put in minus total given away. So yes, 5000 -1250=3750. But is this the minimal possible? Let me see if there's a way to have a lower total. Wait, maybe there's a different arrangement where in some periods, even though she gives 50, she doesn't have to have all four 50s. But then the total for those periods would be higher, which would make the total contributions higher, leading to a higher accumulated amount. Wait, but we need to minimize the accumulated amount, which is total put in minus 1250. So to minimize accumulated amount, we need to minimize the total put in. So yes, making the total put in as small as possible. Therefore, making the periods where she gives 50 as low as possible. So having four 50s in that period. But let's check if there's another way. Suppose that in the period where she gives 50, she has one 50 and three 100s. Then the total for that period is 50 +3*100=350. She gives away 50, so total given remains 1250. But total contributions become 12*400 +350=4800 +350=5150. Then accumulated amount is 5150 -1250=3900, which is higher than 3750. So worse. Therefore, no, that's not better. Alternatively, if she has two 50s and two 100s in that period. Total for the period: 2*50 +2*100=300. She gives away 50. Total contributions: 12*400 +300=4800 +300=5100. Accumulated amount:5100 -1250=3850. Still worse. So indeed, the minimal total put in is 5000, leading to 3750 accumulated. Wait, but let's think again. Suppose that in the period where she gives 50, she has one 50 and the rest 100s. But then, when she takes the smallest bill (50), the remaining three bills in that period are 100s. Do those 100s stay in the piggy bank? Wait, no. Wait, does she remove only the smallest bill each time, or all instances of the smallest bill? The problem says "she takes the smallest denomination bill from the piggy bank and gives it to her sister." So it's singular: "the smallest denomination bill", so just one bill. Wait, but in the piggy bank at the end of 4 weeks, she has four bills (each week she adds one). So when she gives the smallest one, she removes one bill (the smallest) and the other three remain. So each time she gives one bill (either 50 or 100). So if in a period she has, say, one 50 and three 100s, she gives the 50, and the three 100s stay in the piggy bank. Similarly, if she has four 50s, she gives one 50, and the remaining three 50s stay. Therefore, the total contributions are 52 bills (each week she puts one in). At each 4-week period, she removes one, so over 13 periods, she removes 13 bills. Therefore, the total remaining in the piggy bank is 52 -13=39 bills. But the accumulated amount is the total of these 39 bills. Wait, but the problem says "the minimum amount of money she could have accumulated during this time". So it's the total money she has in the piggy bank at the end of the year, which is the sum of all 52 bills she put in minus the sum of the 13 bills she gave away (1250). So to minimize this accumulated amount, we need to minimize the total she put in, because 1250 is fixed. Therefore, the problem reduces to finding the minimal total of 52 bills (each 50 or 100) such that the sum of the 13 smallest bills (each taken every 4 weeks) is 1250. But in this case, the previous reasoning may not hold, because the way she removes the bills affects which bills are left. For example, if she removes a 50 from a group, the remaining bills in that group stay in the piggy bank. Therefore, those remaining bills could affect subsequent removals. Wait, but the problem says "at the end of every 4 weeks, she takes the smallest denomination bill from the piggy bank and gives it to her sister". So each time, after 4 weeks, she looks at all the bills currently in the piggy bank and takes the smallest one? Or only the bills from the last 4 weeks? This is crucial. If she takes the smallest bill from the entire piggy bank at each 4-week interval, then previous bills would affect the choice. However, if she takes the smallest bill only from the last 4 weeks' bills, then it's per 4-week block. The problem statement is a bit ambiguous. Let's re-read:"Masha has a piggy bank where she puts a bill of 50 or 100 rubles every week. At the end of every 4 weeks, she takes the smallest denomination bill from the piggy bank and gives it to her sister."The phrase "from the piggy bank" suggests that it's from the entire collection of bills in the piggy bank at that time. So, for example, after the first 4 weeks, she adds 4 bills, then takes the smallest one from those 4 and gives it away. Then, in the next 4 weeks, she adds 4 more bills, making a total of 4 +4=8, then takes the smallest one from all 8? Or only from the new 4? This is critical. If it's from the entire piggy bank each time, then the problem becomes more complex because earlier bills remain and can be taken later. However, if it's only from the bills added in the last 4 weeks, then each 4-week group is independent. The problem statement says "at the end of every 4 weeks, she takes the smallest denomination bill from the piggy bank and gives it to her sister." The natural reading is that at each 4-week interval, she takes the smallest bill currently in the piggy bank (which includes all bills saved up to that point). However, that might not be the case. Let's think of an example. Suppose she adds 4 bills in the first 4 weeks: say three 100s and one 50. Then she gives the 50 to her sister. Then in the next 4 weeks, she adds four more bills. Now the piggy bank has the remaining three 100s from the first 4 weeks plus the new four bills. Then she takes the smallest from all of them. If the new four bills are all 100s, she still has three 100s from before and four new 100s, so she takes one 100. But if in the new four weeks she adds a 50, then the smallest bill would be 50. Alternatively, if she only considers the bills added in the last 4 weeks each time, then it's separate. The problem statement isn't entirely clear. However, the key is whether the bills accumulate over time or each 4-week block is handled separately. Given that it's a piggy bank where she continues to add bills every week, and at each 4-week mark she removes the smallest bill from the piggy bank, it's likely that the piggy bank contains all the bills she has added so far, minus the ones she has already given away. Therefore, each time she removes the smallest bill from the entire collection. But this complicates the problem significantly because the removal of a bill affects the available bills for future removals. For example, if she has a 50 in the piggy bank, and in subsequent weeks she adds 100s, the presence of that 50 means she might have to remove it in a later 4-week period. However, if we consider that she gives away the smallest bill at each 4-week interval, but the piggy bank is cumulative, then the problem becomes finding a sequence of 52 bills (each 50 or 100) such that at each 4-week interval, the smallest bill present at that time is removed, and the sum of these 13 removed bills is 1250. We need to arrange the bills such that the total sum of all 52 is minimized. This is more complex. Let's consider this interpretation. At each 4-week mark (weeks 4, 8, 12, ..., 52), she removes the smallest bill present in the piggy bank at that time. The total of these 13 removed bills is 1250. We need to find the minimal total sum of all 52 bills. To minimize the total sum, we need to maximize the number of 50s removed (since 50 is less than 100), but the total removed is fixed at 1250. Wait, but earlier we found that she must have removed 1x50 and 12x100. But if the piggy bank is cumulative, maybe we can have more 50s removed, but the problem states that the total given is 1250. Wait, but 1250 must be the sum of the 13 removed bills. If she could remove more 50s, the sum would be less than 1250, but she can't. Therefore, she must have exactly 1x50 and 12x100 removed. Therefore, even in this cumulative case, the same applies. However, arranging the sequence such that only one 50 is removed and twelve 100s are removed, while minimizing the total contributions. But if the piggy bank is cumulative, then removing a 50 early on would eliminate the possibility of having to remove it later. Conversely, if she keeps 50s in the piggy bank, they could be removed in later periods, forcing her to give more 50s, which would lower the total given, but we need the total given to be 1250. Wait, this seems contradictory. If she gives more 50s, the total given would be less than 1250. Therefore, to reach 1250, she must give mostly 100s. Therefore, she must remove 12 100s and 1 50. But how to arrange the sequence so that only one 50 is removed and the rest are 100s. But if the piggy bank is cumulative, to avoid removing more 50s, she must ensure that after the first removal, there are no 50s left in the piggy bank. Because if there's a 50 present at any subsequent removal, she would have to remove it, which would increase the number of 50s given, hence lowering the total given, which we can't have. Therefore, the only way to have only one 50 removed is to introduce a 50 at some point, remove it, and then ensure no other 50s are present in the piggy bank at any subsequent removal. But how? Because she adds a bill every week. If she adds a 50 in the first 4 weeks, removes it at week 4, and then from week 5 onwards only adds 100s, then at each subsequent removal (weeks 8, 12, ..., 52), the piggy bank contains only 100s, so she removes 100 each time. That would give 1x50 +12x100=1250. But let's check the timeline. Weeks 1-4: Add four bills. To have a 50 removed at week 4, there must be at least one 50 in those four. To minimize the total, she should add one 50 and three 100s. Then the total for weeks 1-4 is 50 + 3*100=350. She removes the 50, so remaining in the piggy bank are three 100s. Weeks 5-8: She adds four bills. To ensure that at week 8, the smallest bill is 100, she must add only 100s. Because if she adds any 50s, then at week 8, the piggy bank has three 100s from weeks 1-4 plus four new bills. If any of those new bills is 50, the smallest bill is 50, which she would have to remove, contradicting the need to remove only one 50. Therefore, weeks 5-8 must be all 100s. Total for weeks 5-8: 4*100=400. At week 8, the piggy bank has three 100s (remaining from weeks 1-4) + four 100s (weeks 5-8) = seven 100s. She removes one 100. Remaining: six 100s. Similarly, weeks 9-12: add four 100s. Total: 400. At week 12, piggy bank has six 100s + four 100s= ten 100s. Remove one 100. Remaining: nine 100s. This pattern continues until week 52. Each 4-week block after the first must be all 100s to avoid introducing any 50s. Therefore, the total contributions would be: Weeks 1-4: 1*50 +3*100=350Weeks 5-52: 48 weeks left, which is 12 blocks of 4 weeks. Each block is 4*100=400. So 12*400=4800Total contributions:350 +4800=5150Total given to sister:1*50 +12*100=1250Accumulated amount:5150 -1250=3900But wait, earlier when we considered non-cumulative piggy bank (each 4-week block independent), the accumulated amount was 3750. But that required that in each 4-week block where a 100 is given, all four bills are 100s. However, in the cumulative case, the accumulated amount is 3900. Which is higher. But the problem statement says "the minimum amount of money she could have accumulated during this time". Depending on the interpretation of whether the piggy bank is cumulative or each 4-week block is independent, the answer differs. The key is to clarify whether the "smallest denomination bill" is taken from the entire piggy bank or just the last 4 weeks. The problem statement says "from the piggy bank", which suggests the entire piggy bank. However, in typical piggy bank scenarios, once you put money in, it's all mixed together. However, the problem might be interpreted as each 4-week block being separate, especially since the total given is 1250, which with 13 removals, matches the 1x50 +12x100. But let's check both interpretations. First interpretation: cumulative piggy bank. Total contributions 5150, accumulated 3900.Second interpretation: non-cumulative, each 4-week block is separate. Then total contributions 5000, accumulated 3750. Which interpretation is correct? The problem statement says "at the end of every 4 weeks, she takes the smallest denomination bill from the piggy bank and gives it to her sister." If the piggy bank is cumulative, then after each 4 weeks, the piggy bank contains all previous bills not yet removed. But if it's non-cumulative, then each 4-week block is separate, and after giving the bill, the rest are kept. But the problem doesn't specify that she empties the piggy bank each time. Alternatively, perhaps each time she adds 4 bills, then removes the smallest one, and the other three stay in the piggy bank. Wait, but that would mean the piggy bank accumulates 3 bills each 4 weeks. Over 13 periods, that would be 3*13=39 bills. But she adds 4 each time, removes 1, so net 3. That matches. Then the total accumulated would be 39 bills. But the problem says "the minimum amount of money she could have accumulated during this time". The accumulated amount is the total in the piggy bank at the end, which is the sum of all bills added minus the ones given away. But in this case, whether the piggy bank is cumulative or not depends on whether the bills from previous blocks remain. If each 4-week block's remaining bills are kept, then it's cumulative. If each 4-week block is independent (i.e., she clears the piggy bank each time), then it's not. The problem statement doesn't mention clearing the piggy bank, so the default assumption is that it's cumulative. Therefore, the first interpretation where she accumulates bills over time is correct. But this leads to a higher accumulated amount (3900) compared to the non-cumulative case (3750). However, we need to check whether the non-cumulative case is possible. If each 4-week block is independent, meaning that after each 4 weeks, she adds four bills, removes the smallest one, and the other three are discarded or not part of the piggy bank anymore. But the problem says she "puts a bill of 50 or 100 rubles every week" into the piggy bank. There's no mention of removing bills except when giving to her sister. Therefore, the piggy bank retains all bills except the ones given to the sister. Therefore, it's cumulative. Therefore, the correct interpretation is cumulative. Hence, the earlier calculation of 5150 total contributions leading to 3900 accumulated is correct. But wait, but there's a mistake here. Let's re-examine. If in weeks 1-4, she adds one 50 and three 100s (total 350), gives away the 50, leaving three 100s. Then in weeks 5-8, she adds four 100s (total 400), so piggy bank now has 3 +4=7 bills, all 100s. She gives away one 100. Then in weeks 9-12, adds four 100s, total in piggy bank: 6 +4=10 100s. Gives away one 100. This continues. Total contributions: weeks 1-4:350, weeks 5-52:48 weeks, each week 100. So 48*100=4800. Total contributions:350 +4800=5150. Total given away:1250. Accumulated:5150 -1250=3900. But is there a way to have fewer total contributions? Suppose instead that in weeks 1-4, she adds four 50s. Then she gives away one 50, leaving three 50s. Then in weeks 5-8, she must add four 100s (to avoid having 50s in the piggy bank). But then at week 8, the piggy bank has three 50s and four 100s. The smallest bill is 50, so she has to give away a 50. But we need to give only one 50 in total. Therefore, this approach would result in giving two 50s, which would make the total given 2*50 +11*100=50 +1100=1150, which is less than 1250. Not allowed. Therefore, to have only one 50 given away, she must introduce a single 50 in the first 4 weeks, remove it, and then never have any 50s in the piggy bank afterwards. Hence, the total contributions are 5150, accumulated 3900. But wait, let's think differently. Suppose she adds a 50 not in the first 4 weeks, but later. For example, in weeks 1-4: four 100s. Give away 100. Then weeks 5-8: adds three 100s and one 50. At week 8, piggy bank has three 100s from weeks 1-4 plus four new bills: three 100s and one 50. So total bills: seven, including one 50. She must give away the 50. Then total given is 100 +50=150. Then weeks 9-52 need to give 11*100=1100. Total given:150 +1100=1250. But the problem is that the piggy bank after week 8 still has three 100s from weeks 1-4 and three 100s from weeks 5-8 (since one 50 was removed). So total six 100s. Then weeks 9-12: add four 100s. Now ten 100s. Remove one 100. And so on. Total contributions: weeks 1-4:4*100=400weeks 5-8:3*100 +50=350weeks 9-52:44 weeks, 11*4=44 weeks. All 100s. 44*100=4400Total contributions:400 +350 +4400=5150Same as before. Total given:100 +50 +11*100=1250. Accumulated:5150 -1250=3900. So no difference when introducing the 50 later. Alternatively, introducing the 50 even later. But each time you introduce a 50, you have to give it away, which would add to the total given, but we need exactly one 50 given. Therefore, the 50 must be introduced in a single 4-week block, and then no other 50s can be introduced. Therefore, regardless of when the 50 is introduced, the total contributions would be the same:5150. Therefore, the minimal accumulated amount is 3900. But wait, this contradicts the previous non-cumulative approach where accumulated was 3750. So which is correct? The key is the interpretation of the piggy bank. If it's cumulative, answer is 3900. If each 4-week block is independent (i.e., bills are removed after each 4 weeks except the one given), then answer is 3750. But the problem statement doesn't specify, which is ambiguous. However, typically, a piggy bank accumulates money over time unless stated otherwise. Therefore, the cumulative interpretation is more natural. However, let's check the problem statement again: "At the end of every 4 weeks, she takes the smallest denomination bill from the piggy bank and gives it to her sister." The phrase "from the piggy bank" suggests taking from the entire collection, not just the recent additions. Therefore, cumulative. Hence, the answer should be 3900. But let's verify with another angle. Suppose that the piggy bank is cumulative. To give 12*100 +1*50=1250, she must have removed 12 hundreds and one 50. To remove a 100, there must be no 50s in the piggy bank at that time. Therefore, after the one time she removes a 50, all subsequent additions must be 100s to avoid any 50s. Therefore, the minimal total contributions are achieved by adding one 50 in a single 4-week period, three 100s in that same period, then adding 100s in all other weeks. The total contributions: First 4 weeks:1*50 +3*100=350Next 48 weeks:48*100=4800Total:350+4800=5150Total given:1250Accumulated:5150 -1250=3900Yes, this seems consistent. However, there's another possibility. What if she adds more 50s but arranges the removals such that they are not taken. For example, suppose she adds two 50s in different 4-week periods but manages to remove 100s each time. But this is impossible because if there's a 50 in the piggy bank at the time of removal, the smallest bill is 50, so she must remove it. Therefore, any 50 present in the piggy bank at the time of removal would force her to remove it. Hence, to have only one 50 removed, she can only have 50s in one 4-week period, and must remove that 50, after which no 50s are added. Therefore, the minimal total is indeed 5150, leading to accumulated 3900. But wait, let's consider the alternative where she adds multiple 50s but in such a way that they are all removed in one go. For example, suppose in the first 4 weeks she adds four 50s. Then she removes one 50, leaving three 50s. Then in weeks 5-8, she must add 100s to avoid adding more 50s. But then at week 8, the piggy bank has three 50s and four 100s. The smallest bill is 50, so she must remove another 50. This would result in two 50s removed, which would make the total given 2*50 +11*100=50 +1100=1150, which is less than 1250. Therefore, invalid. Hence, she can only remove one 50, which requires that after the first removal, no 50s are present. Hence, the only way is to add one 50 in the first 4 weeks, remove it, and add only 100s thereafter. Therefore, the minimal total contributions are 5150, accumulated 3900. But wait, let's check if adding the 50 later allows for fewer total contributions. Suppose she adds the 50 in the last 4 weeks. Weeks 1-48: 48 weeks. Divided into 12 blocks of 4 weeks. Each block adds four 100s. Then at week 48, she has given away 12*100=1200. Then weeks 49-52: she adds one 50 and three 100s. Total contributions for weeks 49-52:50 +3*100=350. Total contributions:12*400 +350=4800 +350=5150. At week 52, she gives away the 50. Total given:1200 +50=1250. Accumulated:5150 -1250=3900. Same result. Therefore, regardless of when the 50 is added, the total contributions remain 5150. Therefore, accumulated amount is 3900. Thus, the minimum accumulated amount is 3900 rubles. But this contradicts the initial non-cumulative approach. So why is there a discrepancy? It's because in the non-cumulative approach, each 4-week block is treated separately, and the bills from previous blocks don't affect future ones. Therefore, you can have one block with four 50s (giving 50) and twelve blocks with four 100s (giving 100). Total contributions:1*200 +12*400=200 +4800=5000. Given away:1250. Accumulated:5000 -1250=3750. But in this case, the problem is whether the piggy bank is emptied after each 4-week block. If yes, then this is valid. But the problem doesn't mention emptying the piggy bank, so it's ambiguous. In summary, the answer depends on the interpretation. If cumulative, 3900. If non-cumulative, 3750. The problem statement is from a math competition, likely intended to have the non-cumulative interpretation where each 4-week block is independent. Because in competitions, such problems usually assume independent intervals unless stated otherwise. Additionally, the sum of the smallest bills per block is 1250, which aligns with the non-cumulative approach. Let me re-express the non-cumulative case. If each 4-week block is independent, then each block has four bills, from which the smallest is removed. The total given is the sum of the smallest from each block. To get 1250, as before, one 50 and twelve 100s. To minimize the total contributions, in the block where she gives 50, she adds four 50s (total 200). In the other twelve blocks, she adds four 100s each (total 400 per block). Total contributions:200 +12*400=200 +4800=5000. Given away:1250. Accumulated:5000 -1250=3750. This is lower than the cumulative case. But does this violate the piggy bank concept? If each 4-week block is independent, then after each 4 weeks, she removes one bill and the rest disappear? That doesn't make sense. Normally, a piggy bank accumulates money. However, if the problem is structured such that every 4 weeks, she takes the smallest bill from the last 4 weeks' additions, then it's per block. The problem statement is: "Masha has a piggy bank where she puts a bill of 50 or 100 rubles every week. At the end of every 4 weeks, she takes the smallest denomination bill from the piggy bank and gives it to her sister." The phrase "from the piggy bank" could mean from the entire content, but if we consider that "at the end of every 4 weeks" refers to the bills added in those 4 weeks, then it's per block. This is a common ambiguity in such problems. However, in math competitions, such problems often assume that operations are applied to the most recent group unless stated otherwise. For example, if you save money every week and at the end of every month you do something with that month's savings. Therefore, given that the total given is 1250, which is 1*50 +12*100, and to minimize the total contributions, the answer is 3750. Given that the problem asks for the minimum amount accumulated, and in the non-cumulative case it's 3750, which is less than 3900, and since math competition problems often seek the minimal possible answer under the most favorable interpretation, I think the intended answer is 3750. To confirm, let's check both interpretations with an example. Non-cumulative:Each 4-week block: 12 blocks with four 100s (giving 100 each), one block with four 50s (giving 50). Total put in:12*400 +1*200=5000. Total given:1250. Accumulated:3750. Cumulative:First 4 weeks: one 50, three 100s. Give 50. Then twelve blocks with four 100s. Total put in:350 +4800=5150. Accumulated:3900. Since the problem asks for the minimum possible accumulated amount, the non-cumulative interpretation allows a lower value. Unless the problem explicitly states that the piggy bank is cumulative, both interpretations are possible. However, given the lack of specification, the answer expected is likely 3750. But to be thorough, let's see if there's a way to have even lower total contributions in the non-cumulative case. For example, in the block where she gives 50, instead of four 50s, she could have one 50 and three 100s. Then the total for that block is 50 +3*100=350. She gives 50, so total contributions:350. But then the other twelve blocks are 400 each. Total contributions:350 +12*400=350 +4800=5150. Given away:50 +12*100=1250. Accumulated:5150 -1250=3900. Same as cumulative case. But in the non-cumulative case where each block is independent, you can structure the block with the 50 to have four 50s, thus minimizing that block's contribution. Hence, total contributions 5000. Therefore, the minimal possible accumulated amount is 3750, achievable only if each 4-week block is independent. Given that the problem doesn't specify cumulative savings, and given that the answer 3750 is lower, and math competitions usually look for such answers, I conclude that the intended answer is 3750.

✅:To determine the minimum amount Masha could have accumulated, we analyze her savings and donations over the year:1. Yearly Structure: - 52 weeks = 13 periods of 4 weeks. - Each period, she adds 4 bills (50 or 100) and donates the smallest bill. - Total donated: 1250 rubles = 1×50 + 12×100. 2. Minimizing Contributions: - To minimize total savings, maximize 50s in the period donating 50. - Period with 50: Use four 50s (total = 200). - Periods with 100: Use four 100s each (12 periods × 400 = 4800). 3. Total Contributions: - 200 (from the 50-period) + 4800 (from 100-periods) = 5000 rubles. 4. Accumulated Amount: - Total saved = Total contributions − Total donated = 5000 − 1250 = 3750 rubles. Answer: The minimum amount Masha could have accumulated is (boxed{3750}) rubles.

🔑:1. Define the "month": Since Masha chooses to give a bill to her sister every 4 weeks, we can consider this 4-week period as a "month". There are 52 weeks in a year, so there will be frac{52}{4} = 13 such "months" in a year.2. Determine the potential bills given to her sister: Each "month", Masha gives her sister the smallest denomination bill from her piggy bank. At the end of each 4-week period, the smallest bill in her bank could be either 50 rubles or 100 rubles. If all the bills given were 100 rubles, then her sister would have received 13 times 100 = 1300 rubles. Since her sister received 1250 rubles, not all bills were of 100 rubles. Specifically, twelve bills of 100 rubles and one bill of 50 rubles add up to 1250 rubles.3. Calculate the amount Masha keeps: Let us consider the scenarios where 50 rubles bills could appear. In general, except for one "month", every "month" Masha had only 100 rubles bills left for herself after giving her sister one bill: - For 12 "months", if only 100 rubles bills are left, Masha keeps: [ (100 times 3) times 12 = 3600 text{ rubles} ]4. Account for the last "month": In one "month", Masha gives a 50 rubles bill which implies she might keep a minimum of: [ (100 times 3) - 50 = 300 - 50 = 250 text{ rubles} ] However, she could have also kept three 50 rubles bill because she can accumulate and not have time to give all away due to being in the end scenario. But we calculate the one possible way to find the lower or minimum saving assumption here: [ 3 times 50 = 150 text{ rubles} ]5. Total calculation: Therefore, the least amount of money she retained is: [ 3600 + 150 = 3750 text{ rubles} ]Conclusion:[boxed{3750 text{ rubles}}]

🧠:Okay, let's see. We have a garland with 100 bulbs. The first and third bulbs are yellow. Also, in any five consecutive bulbs, there are exactly two yellow and three blue ones. Santa needs to know the colors of the 97th to 100th bulbs. Hmm, how to approach this?First, let me restate the problem to make sure I understand. We have a sequence of 100 bulbs. Positions 1 and 3 are yellow. For any group of five consecutive bulbs, exactly two are yellow and three are blue. I need to figure out the pattern or sequence of the bulbs, especially positions 97 to 100.Since the garland follows a specific rule in every five consecutive bulbs, there might be a repeating pattern. Maybe the sequence is periodic with some period that divides 5 or relates to the 2 yellow in 5 bulbs? Let me think.If every five bulbs have two yellow ones, then over the entire garland of 100 bulbs, there should be 2/5 of them yellow. So total yellow bulbs would be (2/5)*100 = 40. Similarly, blue would be 60. But maybe that's just a check later.But the key is the pattern. Let's try to construct the sequence step by step. Start with the first five bulbs: positions 1,2,3,4,5. We know 1 and 3 are yellow. So in positions 1-5, two yellows are already placed. Therefore, positions 2,4,5 must be blue. Wait, let's check that.Given that in the first five bulbs (1-5), there must be exactly two yellow. Since positions 1 and 3 are yellow, that's two. So the rest (2,4,5) must be blue. So positions 2,4,5: blue.Next, consider bulbs 2-6. These five (2,3,4,5,6) must also have exactly two yellow. From before, bulb 2 is blue, bulb 3 is yellow, bulb 4 is blue, bulb 5 is blue. So in positions 2-6, we already have one yellow (bulb 3). Therefore, we need one more yellow in this set. The only unknown is bulb 6. So bulb 6 must be yellow.So bulb 6 is yellow. Now, bulbs 3-7: positions 3,4,5,6,7. We know bulb 3 is yellow, bulb 6 is yellow. So that's two yellows. The rest (4,5,7) must be blue. So bulb 7 is blue.Moving on to bulbs 4-8. We need two yellows. Let's see, bulb 4 is blue, bulb 5 is blue, bulb 6 is yellow, bulb 7 is blue. So in bulbs 4-8, we have bulb 6 as yellow. So need one more yellow in this set. The unknown is bulb 8. Therefore, bulb 8 must be yellow.Now, bulbs 5-9. Bulb 5 is blue, 6 is yellow, 7 is blue, 8 is yellow. So already two yellows (6 and 8). Therefore, bulb 9 must be blue.Next, bulbs 6-10. Bulb 6 is yellow, 7 is blue, 8 is yellow, 9 is blue. So in this set, we have two yellows (6 and 8). Therefore, bulb 10 must be blue.Continuing, bulbs 7-11. Bulbs 7 (blue), 8 (yellow), 9 (blue), 10 (blue). So we have one yellow (8). Need one more yellow in this set. The unknown is bulb 11. So bulb 11 is yellow.Hmm, let's see if we can spot a pattern here. Let's list out what we have so far:1: Y2: B3: Y4: B5: B6: Y7: B8: Y9: B10: B11: Y12: ?Wait, let's check the next set. Bulbs 8-12: bulbs 8 (Y), 9 (B), 10 (B), 11 (Y), 12 (?). So in this set, we already have two yellows (8 and 11). Therefore, bulb 12 must be blue.Then bulbs 9-13: 9 (B), 10 (B), 11 (Y), 12 (B), 13 (?). So only one yellow (11). Need one more, so bulb 13 is yellow.Bulbs 10-14: 10 (B), 11 (Y), 12 (B), 13 (Y), 14 (?). Two yellows (11 and 13), so bulb 14 is blue.Bulbs 11-15: 11 (Y), 12 (B), 13 (Y), 14 (B), 15 (?). Two yellows (11 and 13), so bulb 15 is blue.Bulbs 12-16: 12 (B), 13 (Y), 14 (B), 15 (B), 16 (?). One yellow (13). Need one more, so bulb 16 is yellow.Hmm, let's see the sequence so far:1:Y,2:B,3:Y,4:B,5:B,6:Y,7:B,8:Y,9:B,10:B,11:Y,12:B,13:Y,14:B,15:B,16:Y,...Wait a second, this seems like a repeating pattern every 5 bulbs? Let's check:Looking at positions 1-5: Y,B,Y,B,BPositions 6-10: Y,B,Y,B,BPositions 11-15: Y,B,Y,B,BPositions 16-20: Y,B,Y,B,B?Wait, but bulb 16 is Y, so 16:Y,17:B,18:Y,19:B,20:B? Let's check.Bulbs 13-17: 13(Y),14(B),15(B),16(Y),17(?)). So already two yellows (13 and 16), so bulb 17 must be B.Then bulbs 14-18: 14(B),15(B),16(Y),17(B),18(?). One yellow (16). Need one more, so bulb 18 is Y.Bulbs 15-19: 15(B),16(Y),17(B),18(Y),19(?). Two yellows (16 and 18). So bulb 19 is B.Bulbs 16-20:16(Y),17(B),18(Y),19(B),20(?). Two yellows (16 and 18). So bulb 20 is B.So yes, the pattern Y,B,Y,B,B repeats every 5 bulbs. So starting from bulb 1:1:Y,2:B,3:Y,4:B,5:B,6:Y,7:B,8:Y,9:B,10:B,11:Y,12:B,13:Y,14:B,15:B,16:Y,17:B,18:Y,19:B,20:B,... and so on.Therefore, the pattern is a cycle of 5 bulbs: Y,B,Y,B,B. So every 5 bulbs, the same pattern repeats. Let's confirm that this satisfies the condition of exactly two yellow bulbs in any five consecutive.Yes, because each group of five is Y,B,Y,B,B, which has two yellows. And overlapping groups also: for example, bulbs 2-6: B,Y,B,B,Y. That's two yellows (positions 3 and 6). Similarly, bulbs 3-7: Y,B,B,Y,B. Two yellows (3 and 6). Wait, bulb 7 is B. So in bulbs 3-7: Y,B,B,Y,B. Yes, two yellows. Wait, but bulb 6 is Y, which is in position 6. So in bulbs 3-7, positions 3 is Y, 6 is Y. But bulb 6 is the fifth in the next group. Wait, maybe I need to check if the overlapping groups also satisfy.Wait, let's take bulbs 4-8: B,B,Y,B,Y. That's two yellows (5 and 8)? Wait, bulb 4: B,5:B,6:Y,7:B,8:Y. So positions 6 and 8: Y. So two yellows. Correct.Similarly, bulbs 5-9: B,Y,B,Y,B. Two yellows (6 and 8). Correct.So yes, the repeating pattern of Y,B,Y,B,B every five bulbs works. Therefore, the entire garland is just this 5-bulb pattern repeated 20 times (since 100 bulbs). So the sequence is periodic with period 5.Therefore, each position can be determined by its position modulo 5. Let's check:Positions:1: Y2: B3: Y4: B5: B6: Y (same as position 1)7: B (same as 2)8: Y (same as 3)9: B (same as 4)10: B (same as 5)11: Y (same as 1)... etc.So for any position n, if we compute n mod 5, the result will be 1,2,3,4,0. Let's map that:If n mod 5 =1 → Yn mod5=2 → Bn mod5=3 → Yn mod5=4 → Bn mod5=0 → BTherefore, the color of bulb n is:Y if n mod5=1 or 3,B otherwise.So let's verify:Bulb 1:1 mod5=1→YBulb3:3 mod5=3→YBulb6:6 mod5=1→YBulb8:8 mod5=3→YYes, that's consistent.Therefore, the sequence is periodic with period 5, repeating Y,B,Y,B,B.Therefore, to find bulbs 97,98,99,100, we can compute their positions modulo5.Compute 97 mod5:5*19=95, so 97-95=2 → 97 mod5=2 → B98 mod5=3 → Y99 mod5=4 → B100 mod5=0 → BWait, but 100 mod5 is 0. According to the mapping above, n mod5=0 → B. Correct.Therefore, bulbs 97,98,99,100 are B,Y,B,B.Wait, but let me double-check the modulo calculations:97 divided by 5: 5*19=95, remainder 2 → 97 mod5=2 → B98: 95+3=98, remainder 3 → Y99: remainder 4 → B100: 5*20=100, remainder 0 → BYes. So according to the pattern, 97:B,98:Y,99:B,100:B.But let me confirm this with the actual sequence. Let's check the positions around there.Let me list the positions from 95 to 100:95: 95 mod5=0 → B96:1 → Y97:2 → B98:3 → Y99:4 → B100:0 → BWait, that seems correct.So bulb 95: B96: Y (since 96 mod5=1)97: B (mod5=2)98: Y (mod5=3)99: B (mod5=4)100: B (mod5=0)Therefore, the last four bulbs (97-100) are B,Y,B,B.But let me check with the actual pattern:The pattern is Y,B,Y,B,B repeated every 5. So each block of 5:Positions:n:1-5: Y,B,Y,B,B6-10: Y,B,Y,B,B...96-100: Let's see. 96 is 5*19 +1=96. So position 96 is the first in the 19th block (since 5*19=95, so 96 is next). Wait, maybe better to think:Each block of 5. 100 bulbs divided into 20 blocks of 5.Block 1:1-5Block2:6-10...Block20:96-100So block20: 96-100. Then:Position 96: same as position1 → Y97: same as2→B98: same as3→Y99: same as4→B100: same as5→BTherefore, the sequence for 96-100 is Y,B,Y,B,B. So 97 is B,98 is Y,99 is B,100 is B. Correct.Therefore, the answer should be 97:B,98:Y,99:B,100:B.But let me check if there's any inconsistency. The initial condition says the first and third bulbs are yellow, which fits the pattern. Also, every five consecutive bulbs have exactly two yellows. Since each block of five is Y,B,Y,B,B, which has two yellows. Any overlapping five would also, as we checked earlier.Therefore, the pattern holds for the entire garland. Hence, the last four bulbs follow the same pattern.

✅:The garland follows a repeating 5-bulb pattern: Y, B, Y, B, B. This pattern ensures that any five consecutive bulbs contain exactly two yellow and three blue bulbs. To determine the colors of the 97th to 100th bulbs:- 97th bulb: Position 97 corresponds to (97 mod 5 = 2), which is B (blue).- 98th bulb: Position 98 corresponds to (98 mod 5 = 3), which is Y (yellow).- 99th bulb: Position 99 corresponds to (99 mod 5 = 4), which is B (blue).- 100th bulb: Position 100 corresponds to (100 mod 5 = 0) (equivalent to 5), which is B (blue).Final Answer The colors of the bulbs are: 97th: Blue, 98th: Yellow, 99th: Blue, 100th: Blue. In order: B, Y, B, B.

🔑:1. Consider the first five lamps. According to the given information, the first and third lamps are yellow.2. Therefore, the sequence for the first three lamps is: - 1: Yellow - 3: Yellow3. For the sequence among any five consecutive lamps, exactly two are yellow, and three are blue.4. Since the first and third lamps are yellow, we infer that the second, fourth, and fifth lamps must be blue to satisfy this condition for the first five lamps. This gives us: - 2: Blue - 4: Blue - 5: Blue5. The sequence for the first five lamps is: - 1: Yellow - 2: Blue - 3: Yellow - 4: Blue - 5: Blue6. Now, considering the condition that for any five consecutive lamps there are exactly two yellow and three blue, let's examine the next set starting from the second lamp to the sixth lamp: - 2: Blue - 3: Yellow - 4: Blue - 5: Blue - 6: ?7. According to the condition, within these five lamps, there are already three blue lamps; hence, the sixth lamp must be yellow. The pattern continues for any subsequent sets of five lamps, and the color of the sixth lamp will be the same as the color of the first lamp.8. Therefore: - 6: Yellow - 7: Blue - 8: Yellow - 9: Blue - 10: Blue9. Notice a repeating pattern every five lamps (Yellow, Blue, Yellow, Blue, Blue). We can infer that this pattern repeats throughout the 100 lamps.10. To find the lamps on positions 97, 98, 99, and 100: - 96: Yellow - 97: Blue - 98: Yellow - 99: Blue - 100: Blue 11. Based on the discovered pattern, the colors of the lamps at positions 97, 98, 99, and 100 are: - 97: Blue - 98: Yellow - 99: Blue - 100: BlueConclusion:[ boxed{97: text{Blue}; 98: text{Yellow}; 99: text{Blue}; 100: text{Blue}} ]

🧠:Okay, so I need to prove that there exists an element ( k in G ) such that ( kphi(x) ) is a homomorphism. Let me start by understanding the problem.We have a group ( G ), not necessarily abelian, and a function ( phi: G to G ). The condition given is that whenever ( abc = def = 1 ), then ( phi(a)phi(b)phi(c) = phi(d)phi(e)phi(f) ). The goal is to show that there's some element ( k ) in ( G ) such that the map ( x mapsto kphi(x) ) is a homomorphism.First, let's recall what a homomorphism is. A homomorphism between groups is a function that preserves the group operation. So, for ( kphi(x) ) to be a homomorphism, we need that for all ( x, y in G ), ( kphi(xy) = kphi(x) cdot kphi(y) ). Simplifying this equation, we can cancel the leftmost ( k ) by multiplying both sides on the left by ( k^{-1} ), but since ( k ) is fixed, maybe we need to rearrange terms. Wait, actually, let's write it out:( kphi(xy) = kphi(x)kphi(y) ).If we multiply both sides on the left by ( k^{-1} ), we get:( phi(xy) = phi(x)k^{-1}kphi(y) = phi(x)phi(y) ).Wait, that's not possible unless ( k ) is the identity. Hmm, that seems contradictory. Wait, maybe I made a mistake here.Wait, no. Let's think again. The equation ( kphi(xy) = kphi(x)kphi(y) ). Let's rearrange this. Let's move ( k ) to the other side. So:( phi(xy) = k^{-1} cdot kphi(x)kphi(y) ).Wait, no. Let me re-examine the equation.If we have ( kphi(xy) = kphi(x) cdot kphi(y) ). Then, multiply both sides on the left by ( k^{-1} ):( phi(xy) = k^{-1}kphi(x)kphi(y) ).But ( k^{-1}k ) is the identity, so:( phi(xy) = phi(x)kphi(y) ).So, actually, the condition we need is ( phi(xy) = phi(x)kphi(y) ). That's different from a homomorphism. Wait, but in the problem statement, they say "kφ(x) is a homomorphism". Let me check the definition again. If we define a function ( psi(x) = kphi(x) ), then ( psi ) is a homomorphism if ( psi(xy) = psi(x)psi(y) ), which is ( kphi(xy) = kphi(x)kphi(y) ). So, yes, that's equivalent to ( phi(xy) = k^{-1}kphi(x)kphi(y) = phi(x)kphi(y) ). Wait, but how does that hold?Alternatively, maybe there's another way. Perhaps if we can show that ( phi ) is a homomorphism up to a twist by conjugation with ( k ). Hmm. Alternatively, maybe we need to find a ( k ) such that ( phi(xy) = phi(x)kphi(y)k^{-1} ), but that might not be the case. Maybe I need to work with the given condition.The given condition is about triples ( a, b, c ) such that ( abc = 1 ), and similarly ( def = 1 ), then ( phi(a)phi(b)phi(c) = phi(d)phi(e)phi(f) ). So, if two triples multiply to 1, their images under ( phi ) multiply to the same element. This seems like ( phi ) preserves some relation among triples that multiply to 1. Let me think about how to use this condition. Maybe first consider the case where we take a triple ( a, b, c ) with ( abc = 1 ), then ( phi(a)phi(b)phi(c) ) is the same regardless of the specific ( a, b, c ), as long as their product is 1. Wait, but the condition is that if ( abc = def = 1 ), then ( phi(a)phi(b)phi(c) = phi(d)phi(e)phi(f) ). So, the product ( phi(a)phi(b)phi(c) ) only depends on the product ( abc ), which is 1. Therefore, this implies that there exists a fixed element ( k in G ) such that for any ( a, b, c ) with ( abc = 1 ), ( phi(a)phi(b)phi(c) = k ). Wait, is that necessarily true?Wait, suppose that whenever ( abc = 1 ), ( phi(a)phi(b)phi(c) = k ), a constant. Then, indeed, if ( def = 1 ), then ( phi(d)phi(e)phi(f) = k ), so they are equal. So, the given condition is equivalent to saying that there is a constant ( k in G ) such that ( phi(a)phi(b)phi(c) = k ) whenever ( abc = 1 ). Therefore, perhaps that's the key.But is that the case? Let me verify. If the value of ( phi(a)phi(b)phi(c) ) is the same for any triple ( a, b, c ) with ( abc = 1 ), then the given condition is satisfied. Conversely, if the given condition is satisfied, then for any two triples with product 1, their images under φ multiply to the same element, hence the product is a constant. So, in this case, the function φ has the property that φ(a)φ(b)φ(c) is constant whenever abc=1. Therefore, there exists a fixed element k in G such that φ(a)φ(b)φ(c) = k for all a, b, c with abc = 1.So, assuming this is correct, then for any a, b, c with abc = 1, we have φ(a)φ(b)φ(c) = k. Then, perhaps we can use this to construct the homomorphism. Let me see.Let me take arbitrary elements x, y ∈ G. Let me try to relate φ(xy) to φ(x) and φ(y). To use the condition on triples, perhaps I can set up a triple involving x, y, and some other element so that their product is 1.For example, consider the triple x, y, (xy)^{-1}. Then, x y (xy)^{-1} = 1. Therefore, applying φ to this triple, we have φ(x)φ(y)φ((xy)^{-1}) = k.But we also can consider another triple. Let me see. So, φ(x)φ(y)φ((xy)^{-1}) = k. Therefore, φ((xy)^{-1}) = k^{-1} φ(x)^{-1} φ(y)^{-1} ?Wait, no. Let's solve for φ((xy)^{-1}):φ(x)φ(y)φ((xy)^{-1}) = k ⇒ φ((xy)^{-1}) = (φ(x)φ(y))^{-1} k.But since (xy)^{-1} = y^{-1}x^{-1}, maybe I can relate φ((xy)^{-1}) to φ(y^{-1}x^{-1}).Alternatively, maybe consider a different triple. Let's take the triple x, y, z where z = (xy)^{-1}. Then, as above, φ(x)φ(y)φ(z) = k. So, φ(z) = φ((xy)^{-1}) = k^{-1} (φ(x)φ(y))^{-1}.But how does this help? Maybe we need to relate φ(xy) to φ(x) and φ(y). Let's think.Suppose we set z = (xy)^{-1}, so xyz = 1. Then, φ(x)φ(y)φ(z) = k. So, φ(z) = k^{-1} φ(x)^{-1} φ(y)^{-1} φ(z)^{-1}? Wait, no, let's re-express this.Wait, φ(x)φ(y)φ(z) = k ⇒ φ(z) = (φ(x)φ(y))^{-1} k. But z is (xy)^{-1}, so φ(z) = φ( (xy)^{-1 } ). Therefore, φ( (xy)^{-1 } ) = (φ(x)φ(y))^{-1} k.But φ( (xy)^{-1 } ) is also equal to φ( (xy)^{-1 } ). If we can relate φ( (xy)^{-1 } ) to φ(xy), maybe via inverse? But in a general group, φ might not preserve inverses unless it's a homomorphism. So, we can't assume that φ( (xy)^{-1 } ) = (φ(xy))^{-1 } unless φ is a homomorphism.But since we want to find k such that kφ is a homomorphism, perhaps we can use this relation. Let's suppose ψ(x) = kφ(x). Then, ψ is a homomorphism. Then, ψ(xy) = ψ(x)ψ(y) ⇒ kφ(xy) = kφ(x)kφ(y). Then, multiplying on the left by k^{-1}, we get φ(xy) = φ(x) k φ(y). So, that's a twisted version of a homomorphism. So, perhaps φ satisfies φ(xy) = φ(x) k φ(y) for some fixed k. Then, this is similar to a homomorphism with a twist by k. Then, perhaps such a k exists.Alternatively, let's see if we can derive this. From the previous step, we have φ(z) = (φ(x)φ(y))^{-1}k, where z = (xy)^{-1}. So, φ(z) = φ( (xy)^{-1 } ) = (φ(x)φ(y))^{-1}k. If we take inverses of both sides, we get φ( (xy)^{-1 } )^{-1} = k^{-1} φ(x)φ(y). But the left side is φ( (xy)^{-1 } )^{-1} = φ( (xy) ), since in a group, (ab)^{-1} = b^{-1}a^{-1}, but φ might not preserve inverses. Wait, unless φ is a homomorphism. Hmm. So, we can't directly say that φ( (xy)^{-1 } ) = (φ(xy))^{-1 } unless φ is a homomorphism. Therefore, perhaps this is a dead end.Alternatively, let's consider that if we can express φ(xy) in terms of φ(x) and φ(y). Let's try to construct such a relation. Suppose we pick a third element such that the product is 1. For example, take x, y, and then (xy)^{-1} as before. Then, φ(x)φ(y)φ( (xy)^{-1} ) = k. So, φ( (xy)^{-1} ) = k^{-1} (φ(x)φ(y))^{-1} }.But perhaps we can relate φ( (xy)^{-1} ) to φ(xy). Let me suppose that there exists some element k such that φ(g^{-1}) = k^{-1} φ(g)^{-1} k. If that were the case, then we could substitute. But this is just a guess. Let's see.Alternatively, let's take another triple. For example, take g, 1, g^{-1}, since g * 1 * g^{-1} = 1. Then, applying φ, we get φ(g)φ(1)φ(g^{-1}) = k. Therefore, φ(g)φ(1)φ(g^{-1}) = k for all g ∈ G.Similarly, take h, 1, h^{-1}, so φ(h)φ(1)φ(h^{-1}) = k. Therefore, this must hold for all h ∈ G, which suggests that φ(1) is central, because the expression φ(g)φ(1)φ(g^{-1}) is equal to k for all g. Wait, unless φ(g)φ(1)φ(g^{-1}) is the same for all g, which would mean that φ(1) is in the center of the subgroup generated by the image of φ. Hmm, maybe.But let's see. For any g ∈ G, φ(g)φ(1)φ(g^{-1}) = k. Let's rearrange this. Multiply both sides on the right by φ(g):φ(g)φ(1)φ(g^{-1})φ(g) = k φ(g).Simplify φ(g^{-1})φ(g) = φ(g^{-1}g) = φ(1). Wait, no, we can't assume φ is a homomorphism. So, φ(g^{-1})φ(g) is just φ(g^{-1})φ(g), which may not be φ(1). So, this complicates things.Alternatively, if I consider the triple g, 1, g^{-1}, then φ(g)φ(1)φ(g^{-1}) = k. Let me call this equation (1).Similarly, if I take the triple 1, 1, 1, which obviously satisfies 1*1*1=1. Then, φ(1)φ(1)φ(1) = k. So, φ(1)^3 = k. That's equation (2).Similarly, take the triple g, g^{-1}, 1. Then, g*g^{-1}*1 = 1, so φ(g)φ(g^{-1})φ(1) = k. That's equation (3).So, equations (1), (2), (3):1. φ(g)φ(1)φ(g^{-1}) = k for all g ∈ G.2. φ(1)^3 = k.3. φ(g)φ(g^{-1})φ(1) = k for all g ∈ G.Comparing equations 1 and 3:From equation 1: φ(g)φ(1)φ(g^{-1}) = k.From equation 3: φ(g)φ(g^{-1})φ(1) = k.Therefore, φ(g)φ(1)φ(g^{-1}) = φ(g)φ(g^{-1})φ(1).Cancel φ(g) on the left (since we're in a group):φ(1)φ(g^{-1}) = φ(g^{-1})φ(1).So, φ(1) commutes with φ(g^{-1}) for all g ∈ G. Since g is arbitrary, φ(1) commutes with φ(g) for all g ∈ G. Therefore, φ(1) is in the center of the subgroup generated by the image of φ. Let's denote Z as the center of the subgroup generated by φ(G). So, φ(1) ∈ Z.Furthermore, from equation 2, φ(1)^3 = k.So, k = φ(1)^3. And since φ(1) is central in φ(G), φ(1) commutes with all φ(g).Now, let's go back to equation 1: φ(g)φ(1)φ(g^{-1}) = k = φ(1)^3. Since φ(1) is central, we can rearrange φ(g)φ(g^{-1})φ(1) = φ(1)^3. But from equation 3, φ(g)φ(g^{-1})φ(1) = k = φ(1)^3. Therefore, φ(g)φ(g^{-1})φ(1) = φ(1)^3. Since φ(1) is central, we can write φ(1) * φ(g)φ(g^{-1}) = φ(1)^3. Then, cancel φ(1) on the left (since φ(1) is central and hence cancellable in a group):φ(g)φ(g^{-1}) = φ(1)^2.Therefore, φ(g)φ(g^{-1}) = φ(1)^2 for all g ∈ G. That's an important relation.So, φ(g^{-1}) = φ(1)^{-2} φ(g)^{-1} φ(1)^2? Wait, let's see:From φ(g)φ(g^{-1}) = φ(1)^2, multiplying both sides on the left by φ(g)^{-1} gives φ(g^{-1}) = φ(g)^{-1} φ(1)^2.But since φ(1) is central, φ(1)^2 commutes with everything, so φ(g^{-1}) = φ(1)^2 φ(g)^{-1}. Wait, no, φ(1)^2 is in the center, so φ(g)^{-1} φ(1)^2 = φ(1)^2 φ(g)^{-1}.Therefore, φ(g^{-1}) = φ(1)^2 φ(g)^{-1}.Alternatively, maybe we can write φ(g^{-1}) = φ(1)^{-2} φ(g)^{-1} φ(1)^2. But since φ(1) is central, conjugation by φ(1)^2 is trivial, so φ(1)^2 φ(g)^{-1} φ(1)^{-2} = φ(g)^{-1}. Therefore, φ(g^{-1}) = φ(1)^2 φ(g)^{-1}.So, φ(g^{-1}) = φ(1)^2 φ(g)^{-1}.This seems useful. Let's keep this in mind.Now, let's return to the earlier idea of constructing ψ(x) = kφ(x), and trying to show that ψ is a homomorphism. Recall that we need ψ(xy) = ψ(x)ψ(y), which is kφ(xy) = kφ(x)kφ(y). Multiplying both sides on the left by k^{-1}, we get φ(xy) = φ(x)kφ(y). So, if we can show that φ(xy) = φ(x)kφ(y) for all x, y ∈ G, then we are done.Given that k = φ(1)^3, and φ(1) is central, maybe we can use the relations we have derived.Let me consider φ(xy). Let's try to relate φ(xy) to φ(x) and φ(y). To use the given condition, perhaps set up a triple such that the product is 1. For example, take x, y, and (xy)^{-1} again. Then, φ(x)φ(y)φ((xy)^{-1}) = k. But we already have φ((xy)^{-1}) = φ(1)^2 φ(xy)^{-1} from the earlier relation. Let's substitute that in:φ(x)φ(y) * [φ(1)^2 φ(xy)^{-1}] = k.Multiply this out:φ(x)φ(y)φ(1)^2 φ(xy)^{-1} = k.But k = φ(1)^3, so:φ(x)φ(y)φ(1)^2 φ(xy)^{-1} = φ(1)^3.Multiply both sides on the right by φ(xy):φ(x)φ(y)φ(1)^2 = φ(1)^3 φ(xy).Then, multiply both sides on the left by φ(1)^{-3}:φ(1)^{-3} φ(x)φ(y)φ(1)^2 = φ(xy).Since φ(1) is central, we can rearrange the terms:φ(x)φ(y) φ(1)^{-3} φ(1)^2 = φ(x)φ(y) φ(1)^{-1} = φ(xy).Therefore:φ(xy) = φ(x)φ(y) φ(1)^{-1}.But φ(1)^{-1} is k^{-1} φ(1)^2, since k = φ(1)^3. Wait, φ(1)^{-1} = (φ(1)^3)^{-1} φ(1)^2 = k^{-1} φ(1)^2. But maybe I can write this as:φ(xy) = φ(x)φ(y) φ(1)^{-1}.But we want φ(xy) = φ(x) k φ(y). Let's see:If φ(xy) = φ(x)φ(y) φ(1)^{-1}, and k = φ(1)^3, then substituting φ(1)^{-1} = k^{-1} φ(1)^2, we have:φ(xy) = φ(x)φ(y) k^{-1} φ(1)^2.But this doesn't immediately look like φ(x)kφ(y). Maybe I need to manipulate this differently.Wait, let's recall that φ(1) is central. So, we can move φ(1) around. Let me write φ(xy) = φ(x)φ(y)φ(1)^{-1}. Let's factor in k. Since k = φ(1)^3, φ(1)^{-1} = φ(1)^2 k^{-1}.So, φ(xy) = φ(x)φ(y) φ(1)^2 k^{-1}.But since φ(1) is central, φ(1)^2 commutes with everything. So, we can write:φ(xy) = φ(x) k^{-1} φ(1)^2 φ(y).But this is still not helpful. Wait, maybe we need to see if this can be expressed as φ(x) k φ(y) multiplied by some element. Alternatively, maybe there's a different approach.Let me go back to the equation φ(xy) = φ(x)φ(y)φ(1)^{-1}. If we can relate this to kφ(xy) being a homomorphism, let's define ψ(x) = kφ(x). Then, ψ(xy) = kφ(xy) = kφ(x)φ(y)φ(1)^{-1}.But ψ(x)ψ(y) = kφ(x)kφ(y).So, for ψ to be a homomorphism, we need kφ(xy) = kφ(x)kφ(y).From above, φ(xy) = φ(x)φ(y)φ(1)^{-1}, so:kφ(xy) = kφ(x)φ(y)φ(1)^{-1}.But we want this to equal kφ(x)kφ(y).Therefore, equating:kφ(x)φ(y)φ(1)^{-1} = kφ(x)kφ(y).Multiply both sides on the left by k^{-1}:φ(x)φ(y)φ(1)^{-1} = φ(x)kφ(y).Then, rearrange:φ(x)φ(y)φ(1)^{-1} = φ(x)kφ(y).Cancel φ(x) on the left:φ(y)φ(1)^{-1} = kφ(y).Multiply both sides on the right by φ(1):φ(y) = kφ(y)φ(1).But φ(1) is central, so φ(y) = k φ(1) φ(y).Then, cancel φ(y) on both sides (since in a group, we can cancel):1 = k φ(1).Therefore, k φ(1) = 1 ⇒ k = φ(1)^{-1}.But earlier, we found that k = φ(1)^3. Therefore, combining these two results:φ(1)^3 = k = φ(1)^{-1} ⇒ φ(1)^3 = φ(1)^{-1} ⇒ φ(1)^4 = 1.So, φ(1) has order dividing 4. Interesting. So, this suggests that φ(1)^4 = 1. So, φ(1) is an element of order 1, 2, or 4. However, this might not be necessary for the proof, but it's an interesting aside.But let's see. If we have k = φ(1)^{-1}, but previously we had k = φ(1)^3. Therefore, φ(1)^3 = φ(1)^{-1}, so φ(1)^4 = 1. So, φ(1) is of order dividing 4. So, that's a necessary condition.But how does this help us? Maybe this suggests that our earlier approach is leading us to a necessary condition, but perhaps we need to adjust our definition of k.Alternatively, let's revisit the earlier equations. We have from equation (2): k = φ(1)^3. But if we set k = φ(1)^{-1}, then φ(1)^3 = φ(1)^{-1} ⇒ φ(1)^4 = 1. So, this must hold.However, unless φ(1)^4 = 1, which is not necessarily given. Therefore, perhaps there is a mistake in the earlier steps. Let's check.From equation (2): φ(1)^3 = k.From equation (1): φ(g)φ(1)φ(g^{-1}) = k.From equation (3): φ(g)φ(g^{-1})φ(1) = k.Then, equating equations (1) and (3), we canceled φ(g) and derived that φ(1) commutes with φ(g^{-1} ), hence φ(1) is in the center of φ(G).Then, from equation (3): φ(g)φ(g^{-1})φ(1) = k. But we also found that φ(g)φ(g^{-1}) = φ(1)^2. Therefore, substituting into equation (3):φ(1)^2 * φ(1) = k ⇒ φ(1)^3 = k. Which is equation (2). So, that's consistent.Additionally, we have from equation (1): φ(g)φ(1)φ(g^{-1}) = k. Since φ(1) is central, φ(g)φ(g^{-1})φ(1) = k. But φ(g)φ(g^{-1}) = φ(1)^2, so φ(1)^2 φ(1) = k ⇒ φ(1)^3 = k, which is consistent.Therefore, all the equations are consistent. Then, when we tried to define ψ(x) = kφ(x), we encountered a condition that k must equal φ(1)^{-1}, but we also have k = φ(1)^3. Therefore, φ(1)^3 = φ(1)^{-1} ⇒ φ(1)^4 = 1. Therefore, this must hold. So, φ(1) must be an element of order dividing 4 in the center of φ(G). However, the problem statement does not mention anything about the order of elements, so this might be an inherent part of the problem.Alternatively, perhaps there's a different approach. Let's consider defining k as φ(1)^3 and then set ψ(x) = kφ(x). Then, as above, we need to check whether ψ is a homomorphism.But from earlier, we have:φ(xy) = φ(x)φ(y)φ(1)^{-1}.Therefore, ψ(xy) = kφ(xy) = kφ(x)φ(y)φ(1)^{-1}.On the other hand, ψ(x)ψ(y) = kφ(x)kφ(y).We need these to be equal:kφ(x)φ(y)φ(1)^{-1} = kφ(x)kφ(y).Cancel k on both sides (multiply by k^{-1} on the left):φ(x)φ(y)φ(1)^{-1} = φ(x)kφ(y).Then, cancel φ(x) on the left:φ(y)φ(1)^{-1} = kφ(y).Multiply both sides on the right by φ(1):φ(y) = kφ(y)φ(1).But since k = φ(1)^3, substitute:φ(y) = φ(1)^3 φ(y) φ(1).But φ(1) is central, so we can rearrange:φ(y) = φ(1)^3 φ(1) φ(y) = φ(1)^4 φ(y).Therefore, φ(y) = φ(1)^4 φ(y).Cancel φ(y) on both sides (since we're in a group):1 = φ(1)^4.Therefore, φ(1)^4 = 1. So, φ(1) has order dividing 4.But this seems to be a necessary condition for ψ(x) = kφ(x) to be a homomorphism. However, the problem statement does not specify any conditions on φ(1). So, this suggests that either φ(1)^4 = 1 must hold given the original condition, or perhaps there is a different way to define k.Wait, but given the original condition that φ preserves the products of triples that multiply to 1, perhaps φ(1) must indeed satisfy φ(1)^4 = 1. Let me verify this.Take the triple 1, 1, 1, which gives φ(1)^3 = k. Then, take another triple 1, 1, 1, which gives the same. So, k is well-defined. But from the other relations, we derived that φ(1)^4 = 1, so this must hold. Therefore, the given condition implies that φ(1) has order dividing 4. Therefore, in any case, φ(1)^4 = 1.Therefore, φ(1) is an element of order dividing 4 in the center of φ(G). Therefore, such a k exists (since k = φ(1)^3) and ψ(x) = kφ(x) is a homomorphism, given that φ(1)^4 = 1.But let's see. If φ(1)^4 = 1, then φ(1)^{-1} = φ(1)^3 = k. Therefore, k = φ(1)^{-1}.Therefore, we can write ψ(x) = kφ(x) = φ(1)^{-1} φ(x). Let's check if this is a homomorphism.Compute ψ(xy) = φ(1)^{-1} φ(xy).From earlier, φ(xy) = φ(x)φ(y)φ(1)^{-1}.So, ψ(xy) = φ(1)^{-1} φ(x)φ(y)φ(1)^{-1}.On the other hand, ψ(x)ψ(y) = [φ(1)^{-1} φ(x)] [φ(1)^{-1} φ(y)] = φ(1)^{-1} φ(x)φ(1)^{-1} φ(y).Since φ(1) is central, φ(x)φ(1)^{-1} = φ(1)^{-1} φ(x). Therefore:ψ(x)ψ(y) = φ(1)^{-1} φ(1)^{-1} φ(x) φ(y) = φ(1)^{-2} φ(x) φ(y).But ψ(xy) = φ(1)^{-1} φ(x) φ(y) φ(1)^{-1} = φ(1)^{-1} φ(1)^{-1} φ(x) φ(y) (since φ(1)^{-1} is central) = φ(1)^{-2} φ(x) φ(y).Therefore, ψ(xy) = ψ(x)ψ(y). Therefore, ψ is indeed a homomorphism.Therefore, by setting k = φ(1)^{-1}, which is equal to φ(1)^3 since φ(1)^4 = 1, we have that ψ(x) = kφ(x) is a homomorphism.But wait, in the beginning, we had k = φ(1)^3, but here, k is φ(1)^{-1}, which equals φ(1)^3 because φ(1)^4 = 1. Therefore, k = φ(1)^3 = φ(1)^{-1}.Thus, we can take k = φ(1)^{-1} or k = φ(1)^3, which are the same, and then ψ(x) = kφ(x) is a homomorphism.Therefore, such a k exists. Hence, the proof is complete.But to make sure, let's recap the steps:1. Given the condition on φ, when abc = 1, φ(a)φ(b)φ(c) is constant, equal to k.2. By considering triples of the form g, 1, g^{-1}, we found that φ(g)φ(1)φ(g^{-1}) = k, leading to φ(1) being central.3. Considering triples like 1,1,1, we found k = φ(1)^3.4. Considering triples like g, g^{-1}, 1, we found φ(g)φ(g^{-1})φ(1) = k, leading to φ(g)φ(g^{-1}) = φ(1)^2.5. Using these relations, we derived that φ(xy) = φ(x)φ(y)φ(1)^{-1}.6. Then, defining ψ(x) = kφ(x) with k = φ(1)^{-1}, we showed that ψ is a homomorphism.Hence, such a k exists, specifically k = φ(1)^{-1} (which is also φ(1)^3), and thus the proof is complete.

✅:To prove that there exists an element ( k in G ) such that ( kphi(x) ) is a homomorphism, we proceed as follows:Step 1: Utilize the Given ConditionThe function ( phi ) satisfies ( phi(a)phi(b)phi(c) = k ) (a constant) for all triples ( a, b, c in G ) with ( abc = 1 ). This implies ( k ) is fixed and independent of the specific triple.Step 2: Analyze Triples Involving the IdentityConsider the triple ( (g, 1, g^{-1}) ) for any ( g in G ). Applying ( phi ), we get:[phi(g)phi(1)phi(g^{-1}) = k.]Similarly, for the triple ( (1, 1, 1) ), we find:[phi(1)^3 = k.]Step 3: Establish Centrality of ( phi(1) )From the first equation, rearranging gives:[phi(1)phi(g^{-1}) = phi(g^{-1})phi(1),]showing ( phi(1) ) commutes with all ( phi(g) ), hence ( phi(1) ) is central in ( phi(G) ).Step 4: Relate ( phi(g) ) and ( phi(g^{-1}) )Using another triple ( (g, g^{-1}, 1) ), we derive:[phi(g)phi(g^{-1})phi(1) = k.]Substituting ( k = phi(1)^3 ) and simplifying yields:[phi(g)phi(g^{-1}) = phi(1)^2.]Step 5: Express ( phi(xy) ) in Terms of ( phi(x) ) and ( phi(y) )For ( x, y in G ), consider ( (x, y, (xy)^{-1}) ). Applying ( phi ):[phi(x)phi(y)phi((xy)^{-1}) = k.]Using ( phi((xy)^{-1}) = phi(1)^2 phi(xy)^{-1} ) (from Step 4), substitute and simplify to find:[phi(xy) = phi(x)phi(y)phi(1)^{-1}.]Step 6: Define ( k ) and Verify HomomorphismLet ( k = phi(1)^{-1} ) (noting ( k = phi(1)^3 ) since ( phi(1)^4 = 1 )). Define ( psi(x) = kphi(x) ). Then:[psi(xy) = kphi(xy) = kphi(x)phi(y)phi(1)^{-1} = psi(x)psi(y),]where the last equality holds because ( phi(1) ) is central, allowing rearrangement. Thus, ( psi ) is a homomorphism.ConclusionThe element ( k = phi(1)^{-1} ) ensures ( kphi(x) ) is a homomorphism. Therefore, such a ( k ) exists in ( G ).(boxed{k = phi(1)^{-1}})

🔑:1. Given: - ( G ) is a group, and the group operation is written as juxtaposition. - The identity element in ( G ) is denoted by ( 1 ). - There is a function ( phi: G rightarrow G ) such that if ( abc = 1 ) and ( def = 1 ), then ( phi(a)phi(b)phi(c) = phi(d)phi(e)phi(f) ).2. To Prove: - There exists an element ( k in G ) such that ( kphi(x) ) is a homomorphism, i.e., ( kphi(xy) = kphi(x)kphi(y) ) for all ( x, y in G ).3. Step-by-Step Solution: - Consider the equation ( x^{-1} cdot 1 cdot x = 1 ). [ x^{-1} cdot 1 cdot x = (x^{-1}x) = 1 ] So, by the problem condition: [ phi(x^{-1})phi(1)phi(x) = phi(x^{-1})phi(x)phi(1) ] - Since ( phi ) preserves the product in the sense given: [ phi(x^{-1})phi(1)phi(x) = phi(x^{-1})phi(x)phi(1) ] - Simplify to get: [ phi(1)phi(x) = phi(x)phi(1) ] This implies that ( phi(1) ) commutes with all ( phi(x) ).4. Next: Define ( k ) - Let ( k = phi(1)^{-1} ). Since ( phi(1) ) commutes with all ( phi(x) ), ( k ) will also commute with all ( phi(x) ) after inversion.5. Consider ( x cdot y cdot (y^{-1} x^{-1}) = 1 ): [ x cdot y cdot (y^{-1} x^{-1}) = (xy)(y^{-1} x^{-1}) = 1 ] By the problem condition: [ phi(x)phi(y)phi(y^{-1}x^{-1}) = phi(xy)phi(y^{-1}x^{-1})phi(1) ] 6. Using ( phi(1) ) commutes: - Since ( phi(1) ) commutes with all elements, we can bring ( phi(1) ) to the front: [ phi(xy)phi(y^{-1}x^{-1})phi(1) = phi(1)phi(xy)phi(y^{-1}x^{-1}) ] - Hence, we have: [ phi(x)phi(y) = phi(1)phi(xy) ] 7. Multiply by ( k ): - Multiply both sides by ( k = phi(1)^{-1} ): [ kphi(x)phi(y) = kphi(1)phi(xy) ] - Since ( k = phi(1)^{-1} ): [ kphi(1) = phi(1)^{-1}phi(1) = 1 Rightarrow k phi(x) phi(y) = k phi(xy) ] 8. Conclusion: - We have shown that ( kphi(xy) = kphi(x)kphi(y) ), meaning ( kphi(x) ) is a homomorphism. [ boxed{k = phi(1)^{-1}} ]

🧠:Okay, so I need to figure out how many integer solutions (x, y) satisfy the inequality |x| + |y| < 10. Hmm, let me start by understanding the problem. The absolute values mean that both x and y can be positive or negative, but their combined absolute distances from zero must be less than 10. First, maybe I should visualize the region defined by |x| + |y| < 10. The equation |x| + |y| = 10 would form a diamond shape (a square rotated 45 degrees) centered at the origin with vertices at (10,0), (-10,0), (0,10), and (0,-10). The inequality we have is the interior of this diamond. But since we're dealing with integers, the solutions are all the integer coordinate points inside this diamond, not including the boundary where |x| + |y| = 10.So, the problem reduces to counting all integer pairs (x, y) such that the sum of their absolute values is less than 10. Let's think about how to approach this systematically.One method is to fix a value for |x| and then find the possible values of |y| such that |x| + |y| < 10. Then, since x and y can be positive or negative, we can account for all quadrants. Let's denote |x| = a and |y| = b, where a and b are non-negative integers. Then the inequality becomes a + b < 10. So, for each possible a, b can range from 0 to 9 - a. Then, for each (a, b), there are certain numbers of (x, y) pairs.Wait, but if a = 0, then x must be 0. Similarly, if a = 1, then x can be 1 or -1. So, the number of x values for a given a is 1 if a = 0, and 2 otherwise. The same goes for b and y. Therefore, for each (a, b) pair where a + b < 10, the number of (x, y) pairs is (number of x's for a) * (number of y's for b).But since a and b are non-negative integers starting from 0, let's structure this as a double sum. For a from 0 to 9, and for each a, b can be from 0 to 9 - a. But since a + b must be less than 10, the maximum a can be is 9 (since if a = 9, b can only be 0). Wait, but actually, a can go up to 9, but when a is 9, b must be 0. Similarly, when a is 8, b can be 0 or 1, etc.But perhaps a better way is to iterate over possible values of a (from 0 to 9) and for each a, determine the number of possible b's. Then, multiply each by the number of x's and y's for each a and b.Let me formalize this:For each a in 0 to 9:- The number of x's with |x| = a is 1 if a = 0, else 2.For each a, the possible b's are from 0 to (9 - a - 1) because a + b < 10 => b < 10 - a. Since b must be an integer, b can go from 0 to floor(9 - a - 1 + 1) = 9 - a. Wait, no, if a + b < 10, then b can be from 0 to (9 - a - 1), but since b is non-negative integer, the maximum value of b is (9 - a - 1) when a is positive? Wait, no. Let me clarify.If a is fixed, then b must satisfy b < 10 - a. Since b is a non-negative integer, the maximum integer value of b is 9 - a. For example, if a = 0, then b can be 0 to 9 (since 0 + b < 10 => b < 10). Wait, no. If a = 0, then b must be less than 10, so b can be 0,1,...,9. That's 10 values. Similarly, if a = 1, then b must be less than 9, so 0 to 8 (9 values). So in general, for each a from 0 to 9, the number of possible b's is 10 - a.Wait, but 10 - a. Let's check:a = 0: 10 - 0 = 10. Correct, since b can be 0 to 9.a = 1: 10 - 1 = 9. Correct, b from 0 to 8.a = 9: 10 - 9 = 1. So b can only be 0.So, for each a from 0 to 9, the number of possible b's is 10 - a.Now, for each (a, b), the number of x's is 1 if a = 0, else 2. Similarly, the number of y's is 1 if b = 0, else 2. So, for each a and b, the number of points is (1 if a=0 else 2) * (1 if b=0 else 2).But since a and b are non-negative integers, and when a is fixed, we have to consider the possible b's. Therefore, for each a in 0 to 9:- For each b in 0 to (9 - a):Wait, actually, when we fix a, the possible b's are from 0 to (9 - a), but since a + b < 10, when a is fixed, the maximum b is 9 - a. So for each a, b can be 0 to (9 - a). However, we need to be careful here. If a is 0, then b can be from 0 to 9 (since 0 + 9 = 9 < 10). If a is 1, b can be 0 to 8 (1 + 8 = 9 < 10). So yes, the number of b's is 10 - a.Therefore, for each a from 0 to 9, the number of b's is 10 - a. For each such (a, b), the number of (x, y) pairs is:- If a = 0 and b = 0: 1 (x=0, y=0)- If a = 0 and b > 0: 2 (y can be positive or negative)- If a > 0 and b = 0: 2 (x can be positive or negative)- If a > 0 and b > 0: 4 (both x and y can be positive or negative)Therefore, for each a and b:The number of points is 1 if both a=0 and b=0,2 if exactly one of a or b is 0,4 otherwise.But since we are considering a and b from 0 to 9 - a, we can break down the count into these cases.Alternatively, maybe it's easier to compute for each a and b the number of points and sum over all possible a and b.Wait, perhaps I can structure the problem as follows:First, consider all possible values of a (from 0 to 9) and for each a, compute the number of possible b's (which is 10 - a). Then, for each a and b, compute the number of (x, y) pairs as (2 if a ≠ 0 else 1) * (2 if b ≠ 0 else 1).Therefore, the total number of solutions would be the sum over a from 0 to 9 of [sum over b from 0 to 9 - a of (2 if a ≠ 0 else 1) * (2 if b ≠ 0 else 1)].But this seems a bit complicated. Maybe there's a smarter way.Alternatively, since for each (a, b) where a and b are non-negative integers with a + b < 10, the number of (x, y) is:- If a = 0 and b = 0: 1- If a = 0 and b > 0: 2 (since y can be ±b)- If a > 0 and b = 0: 2 (since x can be ±a)- If a > 0 and b > 0: 4 (since x can be ±a and y can be ±b)Therefore, we can compute the total by separating these cases.Let me first compute the case when a = 0:Then, b can be from 0 to 9 (since 0 + b < 10). So b ranges 0 to 9.For a = 0:- When b = 0: 1 point- When b = 1 to 9: each has 2 points (since y can be ±b)So, total points for a = 0:1 + 2*9 = 1 + 18 = 19.Similarly, when a = 1 to 9:For each a in 1 to 9, b can be from 0 to (9 - a).For each such a:- When b = 0: 2 points (x = ±a, y = 0)- When b = 1 to (9 - a): each has 4 points (x = ±a, y = ±b)Therefore, for each a in 1 to 9:Number of points = 2 + 4*(9 - a)Wait, let's check:If a =1:b can be 0 to 8 (since 1 + b < 10 => b < 9, so b = 0,1,...,8). Wait, but 9 - a = 8, so b goes from 0 to 8. So for a=1:- b=0: 2 points- b=1 to 8: 8 values of b, each contributing 4 pointsSo total for a=1: 2 + 8*4 = 2 + 32 = 34Wait, but according to the formula I had above: 2 + 4*(9 - a). For a=1, 9 -1 = 8, so 2 + 4*8 = 34. That works.Similarly, for a=2:b ranges 0 to 7 (since 9 - 2 = 7). So:- b=0: 2 points- b=1 to7: 7 values, each 4 points: 2 + 7*4 = 2 +28=30So the formula 2 + 4*(9 - a) is correct.Therefore, for each a in 1 to 9, the number of points is 2 + 4*(9 - a).Therefore, the total number of points is the sum over a=0 to 9 of the points for each a.Which is:For a=0: 19For a=1 to 9: sum_{a=1}^9 [2 + 4*(9 - a)]Let me compute the sum from a=1 to 9.First, let's note that 9 - a, when a goes from 1 to 9, becomes 8 to 0. So 9 - a ranges from 8 down to 0 as a increases from 1 to 9.But perhaps it's easier to express the sum as:Sum_{a=1}^9 [2 + 4*(9 - a)] = Sum_{a=1}^9 2 + 4*Sum_{a=1}^9 (9 - a)= 9*2 + 4*Sum_{k=0}^8 k (where k = 9 - a, when a=1, k=8; a=2, k=7; ..., a=9, k=0)But Sum_{k=0}^8 k = (8*9)/2 = 36Therefore:Sum becomes 18 + 4*36 = 18 + 144 = 162Therefore, total number of points is 19 (for a=0) + 162 (for a=1 to9) = 181.But wait, let me check that.Wait, when a=0, we had 19 points.Sum for a=1 to9:Each term is 2 + 4*(9 - a)Sum_{a=1}^9 2 = 9*2 = 18Sum_{a=1}^9 4*(9 - a) = 4*Sum_{a=1}^9 (9 - a) = 4*Sum_{k=0}^8 k (since when a=1, k=8; a=2, k=7; ...; a=9, k=0)Sum_{k=0}^8 k = (8*9)/2 = 36Therefore, 4*36 = 144So total sum is 18 + 144 = 162Therefore, total solutions: 19 + 162 = 181But wait, let me check if this is correct. Let me test with a small example.Suppose instead of 10, we have |x| + |y| < 2. Then, the number of solutions should be calculated similarly.For |x| + |y| < 2.Following the same method:a=0: b can be 0,1. So for a=0:- b=0: 1 point- b=1: 2 pointsTotal: 1 + 2 = 3For a=1: b can be 0, since 1 + b < 2 => b < 1 => b=0- b=0: 2 pointsTotal for a=1: 2So total solutions: 3 + 2 = 5.But let's count manually:Possible (x, y):(0,0), (0,1), (0,-1), (1,0), (-1,0). Total 5 points. Correct. So the formula works here.Similarly, if |x| + |y| <1:Only a=0, b=0: 1 point. Which is correct.Another check: |x| + |y| <3.Compute via formula:a=0: b=0,1,2. Total points: 1 + 2 + 2 =5a=1: b=0,1. Total points: 2 + 4 = 2 + 4*1=6Wait, according to the formula:For a=1, number of points is 2 + 4*(9 - a). Wait, but in this case, our original problem was with 10, but if we use the same formula for 3, let's adjust.Wait, no. Let me recast the formula for general N instead of 10.Wait, perhaps my formula was specific to 10, but in the test case, if we set N=3, then |x| + |y| <3.But let's try to apply the same reasoning.For a=0:b can be 0,1,2. So:- b=0:1- b=1:2- b=2:2Total:1 +2 +2=5For a=1:b can be 0,1 (since 1 + b <3 => b<2 => b=0,1)- b=0:2- b=1:4Total:2 +4=6For a=2:b can be 0 (since 2 + b <3 =>b <1 =>b=0)- b=0:2Total:2So total solutions:5 +6 +2=13.Let's count manually:All points where |x| + |y| <3.Possible x and y:When x=0: y can be -2,-1,0,1,2 => 5 pointsWhen x=1 or x=-1: y must satisfy |y| <2. So |y| can be 0,1. So for x=1, y=0,1,-1: 3 points. But since x can be 1 or -1, total for x=±1: 2*3=6When x=2 or x=-2: y must satisfy |y| <1, so y=0. So for x=2 or -2: 2*1=2Total points:5 +6 +2=13. Correct. So the formula works here.Therefore, returning to the original problem, the calculation seems correct.Thus, according to the formula, the total number of solutions is 181.But wait, let's think again. Wait, in the case of N=10, the total solutions are 181? Let me verify with another approach.Alternative approach: The number of integer solutions to |x| + |y| <10 is the same as the number of integer points inside the diamond |x| + |y| <10.In general, for |x| + |y| <n, the number of integer solutions is (2n-1)^2. Wait, no. Wait, for |x| + |y| ≤n, the number of solutions is (n+1)^2 + n^2. Wait, maybe not. Let me recall.Wait, actually, the number of non-negative integer solutions (a, b) to a + b <n is (n(n-1))/2. But considering that x and y can be positive or negative, the total number might be different.Alternatively, think of the problem as counting all (x, y) such that |x| + |y| <10.This is equivalent to the set of (x, y) where |x| and |y| are non-negative integers such that |x| + |y| ≤9. Because if |x| + |y| <10, since |x| and |y| are integers, their sum is at most 9.Therefore, the problem reduces to finding the number of non-negative integer solutions (a, b) to a + b ≤9, and then for each such solution, counting the number of (x, y) pairs that map to (a, b). Each (a, b) corresponds to:- If a=0 and b=0: 1 point- If a=0 and b>0: 2 points (y=±b)- If a>0 and b=0: 2 points (x=±a)- If a>0 and b>0: 4 points (x=±a, y=±b)So, first, compute the number of non-negative integer solutions to a + b ≤9.The number of solutions is (9 +2 -1 choose 2 -1) = (10 choose1)=10? Wait, no.Wait, the number of non-negative integer solutions to a + b ≤n is (n +2)(n +1)/2. Wait, actually, it's the same as the sum from k=0 to n of (k +1) = (n +1)(n +2)/2. Wait, let's confirm:For n=0: a + b ≤0. Only (0,0). So 1 solution. Which is (0 +1)(0 +2)/2=1*2/2=1. Correct.For n=1: a + b ≤1. Solutions are (0,0), (0,1), (1,0). Total 3. Formula: (1 +1)(1 +2)/2=2*3/2=3. Correct.Similarly, for n=2: a + b ≤2. Solutions: (0,0), (0,1), (1,0), (0,2), (1,1), (2,0). Total 6. Formula: (2 +1)(2 +2)/2=3*4/2=6. Correct.Therefore, for n=9, the number of non-negative integer solutions is (9 +1)(9 +2)/2 =10*11/2=55.Therefore, there are 55 non-negative integer pairs (a, b) with a + b ≤9.Now, we need to calculate for each of these 55 pairs, the number of (x, y) points they correspond to.As before:- For (0,0): 1 point- For (a, 0) where a>0: 2 points each- For (0, b) where b>0: 2 points each- For (a, b) where a>0 and b>0:4 points eachSo, let's separate the 55 solutions into these categories.First, count the number of solutions where a=0 and b=0:1.Then, solutions where a>0 and b=0: a from1 to9, b=0. So 9 solutions.Similarly, solutions where a=0 and b>0: b from1 to9, so 9 solutions.The remaining solutions are those where a>0 and b>0: total solutions -1 -9 -9=55-1-9-9=36.Therefore:- 1 solution corresponds to 1 point- 9 +9=18 solutions correspond to 2 points each- 36 solutions correspond to4 points eachTherefore, total number of points:1*1 +18*2 +36*4=1 +36 +144=181. Same result as before.Therefore, the answer is 181.Alternatively, another way to compute this is to recognize that for each layer k=0 to9 (since a + b =k), the number of non-negative integer solutions for each k is k +1. For example, for k=0,1 solution; k=1,2 solutions; up tok=9,10 solutions. But since we need a + b <10, we sum fromk=0 tok=9.The total number of non-negative solutions is sum_{k=0}^9 (k +1)= sum_{k=1}^{10}k=55, which matches previous.Then, for each layerk=0 tok=9:- Fork=0:1 solution, contributing1*1=1 point- Fork=1:2 solutions, each with a or b=0, so each contributes2 points. So total 2*2=4 pointsWait, hold on. Wait, fork=1, the solutions are (1,0) and (0,1). Each of these has:- (1,0): contributes2 points (x=±1, y=0)- (0,1):contributes2 points (x=0, y=±1)Total for k=1:4 points.Similarly, fork=2:Solutions are (2,0),(1,1),(0,2).Each (2,0) and (0,2) contribute2 points each, and (1,1) contributes4.Total:2*2 +1*4=4 +4=8.Similarly, fork=3:Solutions (3,0),(2,1),(1,2),(0,3).Each (3,0),(0,3):2 each. (2,1),(1,2):4 each.Total:2*2 +2*4=4 +8=12.Similarly, in general, for k ≥1:Number of points contributed by layerk:- Ifk=0:1- Ifk≥1:There arek +1 solutions in non-negative (a, b).Among these, two solutions have either a=0 or b=0 (specifically (k,0) and (0,k)), and the restk -1 solutions have both a>0 and b>0.Therefore, the points contributed by layerk are:2 solutions *2 points each + (k +1 -2)*4 points=4 + (k -1)*4=4 +4k -4=4k.But wait, for example, fork=1:4k=4*1=4. Which matches.fork=2:4k=8. Which matches.fork=3:4*3=12. Correct.Therefore, for eachk from1 to9, the number of points contributed is4k.So total points would be1 (fork=0) + sum_{k=1}^9 4k.Sum_{k=1}^9 4k=4*sum_{k=1}^9 k=4*(9*10)/2=4*45=180Therefore, total points=1 +180=181. Same result.Therefore, this method also gives 181.Hence, the number of integer solutions is181.Therefore, the final answer isboxed{181}.

✅:To determine the number of integer solutions ((x, y)) that satisfy the inequality (|x| + |y| < 10), we can break down the problem into counting solutions for non-negative integers (a) and (b) such that (a + b < 10), where (a = |x|) and (b = |y|). Each pair ((a, b)) corresponds to multiple integer solutions ((x, y)) depending on the values of (a) and (b).1. Non-negative integer solutions ((a, b)): - The number of non-negative integer solutions to (a + b < 10) is equivalent to the number of solutions to (a + b leq 9). This is given by the sum (sum_{k=0}^{9} (k + 1)), which equals (frac{(9 + 1)(9 + 2)}{2} = 55).2. Counting integer solutions ((x, y)): - For each pair ((a, b)): - If (a = 0) and (b = 0), there is 1 solution: ((0, 0)). - If (a = 0) and (b > 0), there are 2 solutions for each (b): ((0, pm b)). - If (a > 0) and (b = 0), there are 2 solutions for each (a): ((pm a, 0)). - If (a > 0) and (b > 0), there are 4 solutions for each pair ((a, b)): ((pm a, pm b)).3. Categorizing solutions: - 1 solution for ((0, 0)). - 18 solutions for pairs where either (a = 0) or (b = 0) (9 pairs with (a = 0) and 9 pairs with (b = 0)), each contributing 2 solutions. - 36 solutions for pairs where (a > 0) and (b > 0), each contributing 4 solutions.4. Calculating total solutions: - Total solutions = (1 times 1 + 18 times 2 + 36 times 4 = 1 + 36 + 144 = 181).Thus, the number of integer solutions ((x, y)) to the inequality (|x| + |y| < 10) is (boxed{181}).

🔑:To determine the number of integer solutions ( (x, y) ) that satisfy the inequality:[|x| + |y| < 10]we first consider the ranges for ( |x| ) and ( |y| ).1. Understanding the Absolute Value Constraints: [ |x| < 10 quad text{and} quad |y| < 10 ] Thus, [ -10 < x < 10 quad text{and} quad -10 < y < 10 ]2. Consider Cases for Different Values of ( x ): - Let us fix ( x = pm 9 ): [ |9| + |y| < 10 implies 9 + |y| < 10 implies |y| < 1 implies y = 0 ] This gives 2 solutions: ( (9, 0) ) and ( (-9, 0) ). - Next, fix ( x = pm 8 ): [ |8| + |y| < 10 implies 8 + |y| < 10 implies |y| < 2 implies y = 0, pm 1 ] This gives 6 solutions: ( (8, 0), (-8, 0), (8, 1), (-8, 1), (8, -1), (-8, -1) ). - Similarly, for ( x = pm 7 ): [ |7| + |y| < 10 implies 7 + |y| < 10 implies |y| < 3 implies y = 0, pm 1, pm 2 ] This gives 10 solutions: ( (7, 0), (-7, 0), (7, 1), (-7, 1), (7, -1), (-7, -1), (7, 2), (-7, 2), (7, -2), (-7, -2) ). - The sequence continues as follows, using similar reasoning: - For ( x = pm 6 ): [ y = 0, pm 1, pm 2, pm 3 implies 14 text{ solutions} ] - For ( x = pm 5 ): [ y = 0, pm 1, pm 2, pm 3, pm 4 implies 18 text{ solutions} ] - For ( x = pm 4 ): [ y = 0, pm 1, pm 2, pm 3, pm 4, pm 5 implies 22 text{ solutions} ] - For ( x = pm 3 ): [ y = 0, pm 1, pm 2, pm 3, pm 4, pm 5, pm 6 implies 26 text{ solutions} ] - For ( x = pm 2 ): [ y = 0, pm 1, pm 2, pm 3, pm 4, pm 5, pm 6, pm 7 implies 30 text{ solutions} ] - For ( x = pm 1 ): [ y = 0, pm 1, pm 2, pm 3, pm 4, pm 5, pm 6, pm 7, pm 8 implies 34 text{ solutions} ] - Finally, for ( x = 0 ): [ |0| + |y| < 10 implies |y| < 10 implies y = 0, pm 1, pm 2, pm 3, pm 4, pm 5, pm 6, pm 7, pm 8, pm 9 ] This gives 19 solutions.3. Summing Up the Solutions: [ 2 + 6 + 10 + 14 + 18 + 22 + 26 + 30 + 34 + 19 ] Simplifying step-by-step: [ 2 + 6 = 8 ] [ 8 + 10 = 18 ] [ 18 + 14 = 32 ] [ 32 + 18 = 50 ] [ 50 + 22 = 72 ] [ 72 + 26 = 98 ] [ 98 + 30 = 128 ] [ 128 + 34 = 162 ] [ 162 + 19 = 181 ]Conclusion:[boxed{181}]

🧠:Okay, so I need to find the side length of a rhombus ADEF that's inscribed in triangle ABC. Both the rhombus and the triangle share angle A, and vertex E of the rhombus lies on side BC. The given sides are AB = c and AC = b. Hmm, let me try to visualize this first. A rhombus has all sides equal, right? So ADEF is a rhombus with all sides equal. Since it's inscribed in triangle ABC, and they share angle A, that means points D and F must be on sides AB and AC of the triangle, respectively. Then vertex E is on BC. So the rhombus has vertices at A, D on AB, E on BC, and F on AC. The sides of the rhombus are AD, DE, EF, and FA, each equal in length. Let me draw a rough sketch in my mind: triangle ABC with AB = c, AC = b, and rhombus ADEF inside it, with E on BC.Let me denote the side length of the rhombus as s. So AD = DE = EF = FA = s. Since D is on AB, the length from A to D is s, so AD = s, which means that D divides AB into segments AD = s and DB = AB - AD = c - s. Similarly, F is on AC, so AF = s, meaning FC = AC - AF = b - s.Now, points D and F are on AB and AC, respectively, and E is on BC. The rhombus is formed by connecting these points. Since ADEF is a rhombus, the sides DE and EF must also be of length s. The key here is probably to use similar triangles or coordinate geometry to find the relationships between the sides.Let me consider coordinate geometry. Maybe placing point A at the origin (0,0), point B at (c,0) since AB = c, and point C somewhere in the plane. Wait, but given that AC = b, and angle A is common. But angle A is common to both the triangle and the rhombus. So angle at A for the rhombus is the same as angle at A for triangle ABC. Wait, in a rhombus, adjacent angles are supplementary. But since angle A is common, perhaps the rhombus is constructed such that angle at A in the rhombus is equal to angle at A in the triangle. Hmm, but in a rhombus, opposite angles are equal. So if angle at A is α, then the opposite angle at E is also α, and the other two angles are 180° - α. But in the triangle, angle at A is some angle, which we might need to relate.Alternatively, maybe using coordinates. Let me set coordinate system with point A at (0,0), AB along the x-axis, so point B is at (c,0). Then point C is somewhere in the plane. Since AC = b, if angle at A is θ, then coordinates of C can be (b cos θ, b sin θ). But maybe that's complicating things. Alternatively, since we don't know angle A, maybe using coordinate-free geometry or similar triangles.Alternatively, using similar triangles. Let me think. Since ADEF is a rhombus, DE is parallel to AF, and EF is parallel to AD. Since AF is along AC, which is from A to F, and DE is parallel to AF, so DE is parallel to AC. Similarly, EF is parallel to AD, which is along AB. So DE is parallel to AC and EF is parallel to AB. Therefore, DE is parallel to AC and DE has length s. Similarly, EF is parallel to AB and has length s.Wait, since DE is parallel to AC, then triangle ADE is similar to triangle ABC? Because DE is parallel to AC, so corresponding angles are equal. Let me see. If DE is parallel to AC, then angle ADE is equal to angle ACB, and angle AED is equal to angle ABC. Wait, but DE is part of the rhombus. Wait, maybe that's not the case. Let's be precise.Since DE is part of the rhombus ADEF, and DE is parallel to AF. But AF is along AC. So yes, DE is parallel to AC. Therefore, DE || AC. Similarly, EF is parallel to AD, which is along AB, so EF || AB. Therefore, DE is parallel to AC, EF is parallel to AB.So quadrilateral ADEF has sides AD, DE, EF, FA. With DE || AC and EF || AB. Therefore, DE is a translation of AC, scaled down. Similarly, EF is a translation of AB, scaled down. But since it's a rhombus, all sides are equal.Wait, maybe coordinate geometry is the way to go here. Let me try that.Set coordinate system with point A at (0,0), AB along x-axis, so point B is (c,0). Let me denote coordinates of point C as (d,e). Then AC has length b, so sqrt(d² + e²) = b. But maybe it's better to use coordinates with AB on x-axis and AC in the plane. Let me parametrize point C as (b cos θ, b sin θ), where θ is angle at A. Then coordinates of C are (b cos θ, b sin θ). Then BC can be determined.But maybe this is too much. Alternatively, use coordinates with A at (0,0), B at (c,0), and C at (0,b), making triangle ABC a right triangle? Wait, but the problem doesn't specify that triangle ABC is right-angled. So that might not be valid. So I need to consider a general triangle with AB = c, AC = b, and angle at A being some angle φ. But perhaps it's better to assign coordinates accordingly.Alternatively, maybe use vectors or parametrize the positions.Alternatively, since DE is parallel to AC, the vector DE is a scalar multiple of vector AC. Similarly, EF is parallel to AB, so vector EF is a scalar multiple of vector AB.But since ADEF is a rhombus, the lengths of DE and EF must be equal to s. Let me try to model this.Let me denote vectors. Let’s place point A at the origin. Then vector AB is (c, 0), and vector AC is (b cos φ, b sin φ), where φ is the angle at A. Then point D is along AB, so coordinates of D are (s, 0), since AD = s. Similarly, point F is along AC, so coordinates of F are (s cos φ, s sin φ), since AF = s.Now, since DE is part of the rhombus, DE should be equal in length to AD, which is s, and DE is parallel to AF. Wait, AF is from A to F, which is the vector (s cos φ, s sin φ). Wait, but in the rhombus, adjacent sides are AD and AF. Wait, no. The rhombus is ADEF, so the sides are AD, DE, EF, FA. So moving from A to D to E to F to A.So vector AD is from A(0,0) to D(s, 0). Then vector DE is from D(s,0) to E. Then vector EF is from E to F(s cos φ, s sin φ). Then vector FA is from F back to A(0,0). Since it's a rhombus, all sides must have the same length and DE must be equal to AD, which is s.Wait, but DE is from D(s,0) to E(x,y). Since DE is a side of the rhombus, its length must be s, and it should be parallel to AF. AF is from A(0,0) to F(s cos φ, s sin φ), so vector AF is (s cos φ, s sin φ). Therefore, DE must be parallel to AF, so vector DE is a scalar multiple of AF. But DE has length s, so vector DE = (s cos φ, s sin φ). Therefore, coordinates of E would be D + DE = (s,0) + (s cos φ, s sin φ) = (s + s cos φ, 0 + s sin φ).But point E is supposed to lie on BC. So we need to find coordinates of E and ensure that they lie on BC. Let's find the equation of BC. Coordinates of B are (c,0), coordinates of C are (b cos φ, b sin φ). So the parametric equation of BC can be written as:B + t(C - B) = (c, 0) + t(b cos φ - c, b sin φ - 0) = (c + t(b cos φ - c), 0 + t b sin φ), where t ∈ [0,1].Coordinates of E from the rhombus are (s(1 + cos φ), s sin φ). This point must lie on BC, so there exists some t such that:c + t(b cos φ - c) = s(1 + cos φ)andt b sin φ = s sin φFrom the second equation: t b sin φ = s sin φ. Assuming sin φ ≠ 0 (if sin φ = 0, then angle A is 0 or 180°, which is degenerate triangle, so we can ignore that case). Then t = s / b.Substituting t = s / b into the first equation:c + (s / b)(b cos φ - c) = s(1 + cos φ)Simplify:c + s cos φ - (s c)/b = s + s cos φSubtract s cos φ from both sides:c - (s c)/b = sFactor out c on left side:c(1 - s/b) = sMultiply both sides by b:c(b - s) = s bExpand left side:c b - c s = s bBring all terms to one side:c b - c s - s b = 0Factor:c b - s(c + b) = 0Solve for s:s(c + b) = c bTherefore,s = (c b)/(c + b)Wait, so s = bc/(b + c). Hmm, interesting. So the side length of the rhombus is bc/(b + c). Is that possible?Wait, let me check my steps again. Because if this is the case, then s is the harmonic mean of b and c, which seems plausible, but I need to verify.Starting from the second equation: t = s / b. Then substituting into the first equation:c + (s/b)(b cos φ - c) = s(1 + cos φ)Let's expand the left side:c + (s/b)(b cos φ) - (s/b)c = c + s cos φ - (s c)/bSet equal to right side: s + s cos φSo:c + s cos φ - (s c)/b = s + s cos φSubtract s cos φ from both sides:c - (s c)/b = sThen:c = s + (s c)/bFactor out s on the right:c = s(1 + c/b)Multiply numerator and denominator:c = s( (b + c)/b )Therefore,s = c * (b/(b + c)) ) = (b c)/(b + c)Yes, that's correct. So the side length is indeed s = (b c)/(b + c). So regardless of angle φ? Wait, but in the coordinate system, we parametrized point C as (b cos φ, b sin φ), which assumes that AC = b. So angle φ is the angle at A. But in the calculation, we end up with s = (b c)/(b + c), which doesn't depend on φ. That's interesting. So regardless of the angle at A, the side length of the rhombus is the harmonic mean of sides AB and AC.Is this possible? Let me test with a simple case. Suppose triangle ABC is a right-angled triangle at A, with AB = c = 3, AC = b = 4. Then BC would be 5. Then the rhombus ADEF inscribed in ABC would have side length s = (3*4)/(3+4) = 12/7 ≈ 1.714. Let's see if that makes sense.In the right-angled triangle, coordinates: A(0,0), B(3,0), C(0,4). Then rhombus ADEF: D on AB at (s,0), F on AC at (0,s). Then DE is from D(s,0) to E, which must be on BC. Since DE should be equal in length to AD, which is s, and DE is parallel to AF. AF is from A(0,0) to F(0,s), so AF is vertical. Therefore, DE should also be vertical. So from D(s,0), moving vertically up s units would reach E(s, s). But E must lie on BC. The line BC goes from (3,0) to (0,4). The equation of BC is y = (-4/3)x + 4. Let's check if point (s, s) lies on this line.Plugging x = s, y = s into the equation: s = (-4/3)s + 4s + (4/3)s = 4(7/3)s = 4s = 4*(3/7) = 12/7 ≈ 1.714, which matches the earlier result. So E would be at (12/7, 12/7). Let's check if DE is vertical: from D(12/7, 0) to E(12/7, 12/7) is indeed vertical with length 12/7, which is s. And AF is from A(0,0) to F(0,12/7), which is vertical, so DE is parallel to AF. Similarly, EF is from E(12/7,12/7) to F(0,12/7), which is horizontal left, length 12/7, which is the same as AD. So yes, this works. Therefore, in the right-angled case, the formula holds.Therefore, it seems that regardless of the angle at A, the side length of the rhombus is indeed (b c)/(b + c). Therefore, the answer is bc/(b + c).But let me see if there's another way to approach this problem without coordinates, maybe using similar triangles.Since DE is parallel to AC, as established earlier, and DE = s, AC = b. Then the ratio of DE to AC is s/b. Similarly, since D is on AB, AD = s, AB = c, so the ratio of AD to AB is s/c.If DE is parallel to AC, then triangles ADE and ABC are similar. Wait, is that true? Let's check.In triangle ABC, DE is a line segment from D on AB to E on BC, parallel to AC. By the converse of the basic proportionality theorem (Thales' theorem), if DE is parallel to AC, then AD/AB = AE/AC. Wait, but in our case, E is on BC, not on AC. Wait, maybe not. Let me think.Wait, DE is parallel to AC. So angle ADE is equal to angle BAC, and angle AED is equal to angle ABC? Hmm, maybe not directly. Alternatively, since DE || AC, then triangle ADE is similar to triangle ABC? Wait, no, because DE is not a side of the triangle but a transversal.Wait, actually, if DE is parallel to AC, then triangle ADE ~ triangle ABC. Because angle at A is common, and angle ADE is equal to angle ACB (since DE || AC, corresponding angles are equal). Wait, but in triangle ABC, angle at C is angle ACB, and in triangle ADE, angle at E would be angle AED. Hmm, maybe not directly similar.Alternatively, use coordinates again. Wait, since in the coordinate system approach, we arrived at s = bc/(b + c) regardless of angle φ, which includes the right-angled case, which checks out, then perhaps the answer is indeed bc/(b + c).Alternatively, using area ratios. The area of the rhombus is s^2 sin φ, where φ is the angle at A. The area of triangle ABC is (1/2) b c sin φ. Maybe the ratio of areas can relate, but I'm not sure. However, since the rhombus is inscribed in the triangle, maybe we can find a relationship between their areas. But perhaps that's complicating things.Alternatively, mass point geometry? Not sure.Wait, let me think again. Since DE is parallel to AC, the length DE = s, and AC = b. The ratio of similarity would be s/b. Then, if we consider the line DE dividing the triangle ABC, the segment DE would create a smaller triangle ADE similar to triangle ABC with ratio s/b. But in that case, the height from A to DE would be (s/b) times the height from A to BC. But maybe this is not straightforward.Alternatively, think about parametrizing the position of E on BC. Let’s let E divide BC in some ratio. Let’s parameterize BC such that E divides BC in the ratio k:1, but maybe that's not helpful.Wait, but in the coordinate system approach, regardless of the angle φ, the result was s = bc/(b + c). That seems to be the case. Since in the right-angled triangle it worked, and in the coordinate system with general angle φ, the result didn't depend on φ, so it must hold generally.Therefore, the side length of the rhombus is bc divided by (b + c), so the answer is boxed{dfrac{bc}{b + c}}.Final AnswerThe side length of the rhombus is boxed{dfrac{bc}{b + c}}.

✅:To find the side length of the rhombus (ADEF) inscribed in triangle (ABC) with (AB = c) and (AC = b), we start by noting that the rhombus shares angle (A) with the triangle and has vertex (E) on side (BC). 1. Coordinate System Setup: - Place point (A) at the origin ((0,0)), point (B) at ((c,0)), and point (C) at ((b cos phi, b sin phi)) where (phi) is the angle at (A). - Points (D) and (F) are on (AB) and (AC) respectively, such that (AD = AF = s), the side length of the rhombus.2. Coordinates of Points: - (D) is at ((s, 0)) on (AB). - (F) is at ((s cos phi, s sin phi)) on (AC).3. Vector Analysis: - Vector (DE) must be parallel to (AF) and have length (s). - Coordinates of (E) are determined by moving from (D) in the direction of (AF), resulting in (E) at ((s(1 + cos phi), s sin phi)).4. Equation of Line (BC): - Parametric equation of (BC) is ((c + t(b cos phi - c), t b sin phi)) for (t in [0,1]). - Coordinates of (E) must satisfy this parametric equation.5. Solving for (s): - From the parametric equation of (BC), equate coordinates of (E): - (c + t(b cos phi - c) = s(1 + cos phi)) - (t b sin phi = s sin phi) - Solving the second equation gives (t = frac{s}{b}). - Substituting (t = frac{s}{b}) into the first equation and simplifying yields: [ c + frac{s}{b}(b cos phi - c) = s + s cos phi ] - Simplifying further leads to (s = frac{bc}{b + c}).Thus, the side length of the rhombus is (boxed{dfrac{bc}{b + c}}).

🔑: ProblemIn the triangle ABC, a rhombus ADEF is inscribed such that angle A is common to both, and vertex E lies on side BC. Given that AB = c and AC = b, find the side length of the rhombus.1. Diagonals of the Rhombus as Angle Bisectors: The diagonal of a rhombus is an angle bisector of its corner angles.2. Position of Point D: Suppose point D lies on the side AB of the triangle ABC.3. Using Similar Triangles: From the similarity of triangles BDE and BAC, it follows that: [ frac{DE}{AC} = frac{BE}{BC} ]4. Angle Bisector Theorem: Since AE is the angle bisector of angle A, we use the Angle Bisector Theorem which states: [ frac{BE}{EC} = frac{AB}{AC} = frac{c}{b} ] Therefore, [ BE = frac{c}{c + b} cdot BC quad text{and} quad EC = frac{b}{c + b} cdot BC ]5. Calculating Side Length of Rhombus: Given that DE is parallel to AC, for similar triangles BDE and BAC, we have: [ frac{DE}{AC} = frac{BE}{BC} rightarrow DE = AC cdot frac{BE}{BC} = b cdot frac{c}{c + b} = frac{bc}{c + b} ] ConclusionThus, the side length of the rhombus ADEF is:[boxed{frac{bc}{b + c}}]