💡Alright, so I've got these two geometry problems here, and I need to figure out how to prove the statements. Let me take them one at a time.Starting with part (a): There's a square with an area of 6, and inside this square, there are three polygons, each with an area of 3. I need to prove that among these three polygons, there are at least two whose overlapping area is at least 1. Hmm, okay.First, let me visualize this. The square has an area of 6, so each side must be sqrt(6) units long. The three polygons each have an area of 3, which is exactly half of the square's area. So each polygon covers half of the square. Now, since all three polygons are inside the same square, they must overlap somewhere.I remember something about the pigeonhole principle, which might be useful here. The idea is that if you have more pigeons than pigeonholes, at least two pigeons must share a hole. Maybe I can apply that concept here.Let me think: If each polygon has an area of 3, and the total area of the square is 6, then the combined area of the three polygons is 9. But the square itself is only 6, so there must be overlapping areas. The total overlapping area must account for the extra 3 units of area.But wait, how does that help me find the minimum overlapping area between two polygons? Maybe I need to consider the overlaps more carefully.Let me denote the three polygons as A, B, and C. Let’s say the area of overlap between A and B is S12, between B and C is S23, and between A and C is S13. Also, there might be an area where all three overlap, which I'll call S123.Using the principle of inclusion-exclusion, the total area covered by the three polygons is:Area(A) + Area(B) + Area(C) - S12 - S23 - S13 + S123.But since the total area of the square is 6, we have:6 = 3 + 3 + 3 - S12 - S23 - S13 + S123.Simplifying that, we get:6 = 9 - (S12 + S23 + S13) + S123.Rearranging terms:S12 + S23 + S13 = 3 + S123.Since S123 is the area where all three overlap, it can't be negative. So, S12 + S23 + S13 is at least 3.Now, if I have three overlaps (S12, S23, S13) and their sum is at least 3, then at least one of these overlaps must be at least 1. Because if all three were less than 1, their sum would be less than 3, which contradicts our earlier equation.Therefore, there must be at least two polygons whose overlapping area is at least 1. That seems to make sense.Moving on to part (b): Now, the square has an area of 5, and there are nine polygons each with an area of 1. I need to prove that among these nine polygons, there are at least two whose overlapping area is at least 1/9.Okay, so the square is smaller this time, area 5, and there are nine polygons each of area 1. So the total area covered by all nine polygons is 9, but the square is only 5, so there's a lot of overlapping.Again, maybe the pigeonhole principle applies here. But with nine polygons, the number of possible pairs is quite large. Let me calculate that: the number of unique pairs is C(9,2) which is 36.So, there are 36 possible pairs of polygons. Now, the total overlapping area across all these pairs must account for the excess area beyond the square's size.Let me denote M2 as the total overlapping area among all pairs. Then, using inclusion-exclusion again, the total area covered by all nine polygons is:9 - M2 + M3 - ... = 5.But this might get complicated with higher-order overlaps. Maybe I can simplify it by considering just the pairwise overlaps.If I ignore higher-order overlaps (like three or more polygons overlapping), then:Total area = 9 - M2.But the actual area is 5, so:5 >= 9 - M2.Which implies:M2 >= 4.So, the total overlapping area across all pairs is at least 4.Now, since there are 36 pairs, the average overlapping area per pair is at least 4/36, which simplifies to 1/9.Therefore, there must be at least one pair of polygons whose overlapping area is at least 1/9.Wait, that seems straightforward. But let me double-check.If the total overlapping area is 4, spread over 36 pairs, then on average, each pair overlaps by 1/9. So, by the pigeonhole principle, at least one pair must overlap by at least the average, which is 1/9.Yes, that makes sense.So, both parts (a) and (b) can be proven using the pigeonhole principle and inclusion-exclusion principles.I think I've got it. Let me summarize my thoughts.For part (a), by considering the total area and overlaps, we deduced that the sum of pairwise overlaps is at least 3, so at least one pair must overlap by at least 1.For part (b), by calculating the total overlapping area needed and dividing by the number of pairs, we found that at least one pair must overlap by at least 1/9.Yeah, that seems solid. I don't see any flaws in this reasoning.

💡Alright, so I've got this problem here: Let ( a ) be a root of the function ( f(x) = 2^x - log_{frac{1}{2}} x ). If ( x_0 > a ), then what can we say about the value of ( f(x_0) )? The options are A: ( f(x_0) = 0 ), B: ( f(x_0) < 0 ), C: ( f(x_0) > 0 ), and D: The sign of ( f(x_0) ) is uncertain.Okay, first, let me make sure I understand what's being asked. A root of a function is a value ( a ) where ( f(a) = 0 ). So, ( a ) is a solution to the equation ( 2^a - log_{frac{1}{2}} a = 0 ). Now, if we take another value ( x_0 ) that's greater than ( a ), we need to determine whether ( f(x_0) ) is positive, negative, zero, or uncertain.Hmm, I think this has to do with the behavior of the function ( f(x) ). Maybe if I can figure out whether the function is increasing or decreasing, I can determine what happens when ( x ) increases beyond ( a ).Let me rewrite the function to make it easier to analyze. The function is ( f(x) = 2^x - log_{frac{1}{2}} x ). I know that ( log_{frac{1}{2}} x ) can be rewritten using the change of base formula. Remember that ( log_{frac{1}{2}} x = frac{log_2 x}{log_2 frac{1}{2}} ). Since ( log_2 frac{1}{2} = -1 ), this simplifies to ( -log_2 x ). So, substituting back into the function, we have:( f(x) = 2^x - (-log_2 x) = 2^x + log_2 x ).Okay, that's a bit simpler. So now, ( f(x) = 2^x + log_2 x ). I need to analyze the behavior of this function.Let me think about the components of this function. The first term is ( 2^x ), which is an exponential function with base 2. I know that exponential functions with base greater than 1 are increasing functions. So, as ( x ) increases, ( 2^x ) increases.The second term is ( log_2 x ), which is a logarithmic function. Logarithmic functions with base greater than 1 are also increasing functions, but they increase much more slowly compared to exponential functions.So, both ( 2^x ) and ( log_2 x ) are increasing functions. Therefore, their sum, ( f(x) = 2^x + log_2 x ), should also be an increasing function. That makes sense because if both individual functions are increasing, adding them together would result in an overall increasing function.Wait, is that always true? Let me double-check. Suppose I have two increasing functions, ( g(x) ) and ( h(x) ). Then, their sum ( g(x) + h(x) ) is also increasing because the derivative of the sum is the sum of the derivatives, and if both derivatives are positive, their sum is positive. So, yes, ( f(x) ) is indeed an increasing function.Since ( f(x) ) is increasing, that means as ( x ) increases, ( f(x) ) increases as well. So, if ( x_0 > a ), then ( f(x_0) > f(a) ). But we know that ( a ) is a root of ( f(x) ), so ( f(a) = 0 ). Therefore, ( f(x_0) > 0 ).Wait, let me make sure I didn't make a mistake in rewriting the logarithm. The original function was ( log_{frac{1}{2}} x ), and I converted it to ( -log_2 x ). Let me verify that.Yes, because ( log_{frac{1}{2}} x = frac{ln x}{ln frac{1}{2}} = frac{ln x}{- ln 2} = - frac{ln x}{ln 2} = -log_2 x ). So that part is correct.Also, the function ( f(x) = 2^x + log_2 x ) is indeed increasing. Let's think about the derivative to confirm.The derivative of ( 2^x ) is ( ln 2 cdot 2^x ), which is always positive because ( ln 2 ) is positive and ( 2^x ) is always positive. The derivative of ( log_2 x ) is ( frac{1}{x ln 2} ), which is also positive for ( x > 0 ). Therefore, the derivative of ( f(x) ) is ( ln 2 cdot 2^x + frac{1}{x ln 2} ), which is always positive for ( x > 0 ). This confirms that ( f(x) ) is strictly increasing on its domain.Since ( f(x) ) is strictly increasing, once it crosses zero at ( x = a ), it will continue to increase beyond that point. So, for any ( x_0 > a ), ( f(x_0) ) must be greater than zero.Is there any possibility that ( f(x) ) could decrease after ( x = a )? Well, since the derivative is always positive, the function can't decrease anywhere. It's always going up as ( x ) increases.Just to visualize, let's consider the graph of ( f(x) ). The exponential term ( 2^x ) will dominate as ( x ) becomes large, so the function will shoot up rapidly. The logarithmic term ( log_2 x ) grows slowly but is always positive for ( x > 1 ) and negative for ( 0 < x < 1 ).Wait, so for ( x ) between 0 and 1, ( log_2 x ) is negative, but ( 2^x ) is still positive. So, ( f(x) ) might be negative or positive in that interval depending on the balance between ( 2^x ) and ( log_2 x ). But since ( f(x) ) is increasing, once it crosses zero at ( x = a ), it will stay positive for all ( x > a ).Let me think about specific values to test this. Suppose ( a ) is somewhere between 0 and 1. Let's say ( a = 0.5 ). Then ( f(0.5) = 2^{0.5} + log_2 0.5 ). ( 2^{0.5} ) is about 1.414, and ( log_2 0.5 ) is -1. So, ( f(0.5) = 1.414 - 1 = 0.414 ), which is positive. Wait, but if ( a ) is a root, then ( f(a) = 0 ). So, maybe ( a ) is less than 0.5?Wait, let's compute ( f(0.25) ). ( 2^{0.25} ) is about 1.189, and ( log_2 0.25 = -2 ). So, ( f(0.25) = 1.189 - 2 = -0.811 ). That's negative. So, somewhere between 0.25 and 0.5, ( f(x) ) crosses zero.So, if ( a ) is between 0.25 and 0.5, then for ( x_0 > a ), say ( x_0 = 0.3 ), which is greater than ( a ) if ( a ) is less than 0.3, then ( f(x_0) ) would be positive. Wait, but if ( a ) is, say, 0.3, then for ( x_0 = 0.4 ), which is greater than ( a ), ( f(x_0) ) is positive.But in my earlier calculation, ( f(0.5) ) was positive, and ( f(0.25) ) was negative, so the root ( a ) is somewhere between 0.25 and 0.5.But regardless of where ( a ) is, since ( f(x) ) is increasing, once ( x ) passes ( a ), ( f(x) ) becomes positive and stays positive.Wait, but what if ( a ) is greater than 1? Let me check ( f(1) ). ( 2^1 = 2 ), and ( log_2 1 = 0 ). So, ( f(1) = 2 + 0 = 2 ), which is positive. So, the root ( a ) must be less than 1 because ( f(1) ) is positive, and ( f(x) ) is increasing.So, ( a ) is somewhere between 0 and 1, specifically between 0.25 and 0.5 as we saw earlier.Therefore, for any ( x_0 > a ), ( f(x_0) ) is greater than zero.Is there any chance that ( f(x) ) could have multiple roots? Let's see. Since ( f(x) ) is strictly increasing, it can have at most one root. Because if it were to have another root, the function would have to decrease after increasing, which contradicts the fact that it's strictly increasing.So, ( a ) is the only root, and for all ( x > a ), ( f(x) > 0 ).Therefore, the correct answer should be C: ( f(x_0) > 0 ).I think that makes sense. I don't see any flaws in this reasoning. The key points are:1. Rewriting the logarithm to simplify the function.2. Recognizing that both terms in the function are increasing.3. Confirming that the derivative is always positive, making the function strictly increasing.4. Concluding that once ( x ) passes the root ( a ), the function remains positive.So, yeah, I'm pretty confident that the answer is C.

💡First, I recognize that if a positive integer ( n ) leaves a remainder of 7 when divided by 10, it can be expressed as ( n = 10k + 7 ), where ( k ) is an integer.Next, to find the remainder when ( n ) is divided by 5, I substitute ( n ) into the expression for division by 5:[n div 5 = (10k + 7) div 5]Since 10 is a multiple of 5, dividing ( 10k ) by 5 results in an integer with no remainder. However, dividing 7 by 5 gives a quotient of 1 with a remainder of 2. Therefore, the remainder when ( n ) is divided by 5 is 2.

💡Alright, so I have this number, 975312468, and I need to figure out if I can add a single digit somewhere in it—either at the beginning, end, or between any two digits—to make the new number divisible by 33. Hmm, okay. Let me think about how to approach this.First off, I remember that 33 is equal to 3 multiplied by 11. So, if a number is divisible by both 3 and 11, it will be divisible by 33. That gives me two things to check: whether the number is divisible by 3 and whether it's divisible by 11.Starting with divisibility by 3. I recall that a number is divisible by 3 if the sum of its digits is divisible by 3. Let me add up the digits of 975312468:9 + 7 + 5 + 3 + 1 + 2 + 4 + 6 + 8.Let me calculate that step by step:- 9 + 7 = 16- 16 + 5 = 21- 21 + 3 = 24- 24 + 1 = 25- 25 + 2 = 27- 27 + 4 = 31- 31 + 6 = 37- 37 + 8 = 45Okay, so the sum of the digits is 45. Now, 45 divided by 3 is 15, which means 45 is divisible by 3. So, the original number is already divisible by 3. That means that whatever digit I add, the new sum should still be divisible by 3. Since 45 is already divisible by 3, adding a digit that is also divisible by 3 (i.e., 0, 3, 6, or 9) will keep the sum divisible by 3. So, I can only add one of these digits: 0, 3, 6, or 9.Now, moving on to divisibility by 11. I remember that a number is divisible by 11 if the alternating sum of its digits is a multiple of 11 (including zero). The alternating sum means I subtract the second digit from the first, add the third, subtract the fourth, and so on.Let me apply this rule to the original number, 975312468. I'll write out the digits with their positions:Position: 1 2 3 4 5 6 7 8 9Digits: 9 7 5 3 1 2 4 6 8Now, starting from the right (position 9), I'll alternate adding and subtracting:8 (position 9) - 6 (position 8) + 4 (position 7) - 2 (position 6) + 1 (position 5) - 3 (position 4) + 5 (position 3) - 7 (position 2) + 9 (position 1)Let me compute that step by step:- 8 - 6 = 2- 2 + 4 = 6- 6 - 2 = 4- 4 + 1 = 5- 5 - 3 = 2- 2 + 5 = 7- 7 - 7 = 0- 0 + 9 = 9So, the alternating sum is 9. Since 9 is not a multiple of 11, the original number is not divisible by 11. Therefore, adding a digit will change this alternating sum, and I need to find a position and a digit (from 0, 3, 6, 9) such that the new alternating sum is a multiple of 11.Let me think about how adding a digit affects the alternating sum. Depending on where I add the digit, it will either be added or subtracted in the alternating sum. For example, if I add a digit at the beginning (position 1), it will be added in the alternating sum. If I add it at position 2, it will be subtracted, and so on.Let me consider each possible position where I can add a digit and see what effect it has on the alternating sum.1. Adding a digit at the beginning (position 1): - The new digit will be added in the alternating sum. - Let's denote the new digit as 'd'. - The new alternating sum will be d + (8 - 6 + 4 - 2 + 1 - 3 + 5 - 7 + 9) = d + 9. - We need d + 9 to be a multiple of 11. - Possible values for d: 0, 3, 6, 9. - Let's check: - d = 0: 0 + 9 = 9 → Not a multiple of 11. - d = 3: 3 + 9 = 12 → Not a multiple of 11. - d = 6: 6 + 9 = 15 → Not a multiple of 11. - d = 9: 9 + 9 = 18 → Not a multiple of 11. - So, adding a digit at the beginning doesn't work.2. Adding a digit at position 2: - The new digit will be subtracted in the alternating sum. - New alternating sum: 9 - d + (8 - 6 + 4 - 2 + 1 - 3 + 5 - 7 + 9) = 9 - d + 9 = 18 - d. - We need 18 - d to be a multiple of 11. - Possible values for d: 0, 3, 6, 9. - Let's check: - d = 0: 18 - 0 = 18 → Not a multiple of 11. - d = 3: 18 - 3 = 15 → Not a multiple of 11. - d = 6: 18 - 6 = 12 → Not a multiple of 11. - d = 9: 18 - 9 = 9 → Not a multiple of 11. - So, adding a digit at position 2 doesn't work.3. Adding a digit at position 3: - The new digit will be added in the alternating sum. - New alternating sum: 9 - 7 + d + (8 - 6 + 4 - 2 + 1 - 3 + 5 - 7 + 9) = 9 - 7 + d + 9 = 11 + d. - We need 11 + d to be a multiple of 11. - Possible values for d: 0, 3, 6, 9. - Let's check: - d = 0: 11 + 0 = 11 → Yes, multiple of 11. - d = 3: 11 + 3 = 14 → Not a multiple of 11. - d = 6: 11 + 6 = 17 → Not a multiple of 11. - d = 9: 11 + 9 = 20 → Not a multiple of 11. - So, adding a digit '0' at position 3 works.Wait, but adding '0' at position 3 would make the number 9705312468. Is that correct? Let me verify.Original number: 9 7 5 3 1 2 4 6 8Adding '0' at position 3: 9 7 0 5 3 1 2 4 6 8Now, let's compute the alternating sum for this new number:Starting from the right:8 (position 10) - 6 (position 9) + 4 (position 8) - 2 (position 7) + 1 (position 6) - 3 (position 5) + 5 (position 4) - 0 (position 3) + 7 (position 2) - 9 (position 1)Calculating step by step:- 8 - 6 = 2- 2 + 4 = 6- 6 - 2 = 4- 4 + 1 = 5- 5 - 3 = 2- 2 + 5 = 7- 7 - 0 = 7- 7 + 7 = 14- 14 - 9 = 5Wait, that's not 11. Did I make a mistake?Let me recount the positions correctly. When adding a digit at position 3, the positions shift:New number: 9 (1), 7 (2), 0 (3), 5 (4), 3 (5), 1 (6), 2 (7), 4 (8), 6 (9), 8 (10)So, starting from position 10:8 (10) - 6 (9) + 4 (8) - 2 (7) + 1 (6) - 3 (5) + 5 (4) - 0 (3) + 7 (2) - 9 (1)Calculating:- 8 - 6 = 2- 2 + 4 = 6- 6 - 2 = 4- 4 + 1 = 5- 5 - 3 = 2- 2 + 5 = 7- 7 - 0 = 7- 7 + 7 = 14- 14 - 9 = 5Hmm, still 5, not 11. That's confusing. I thought adding '0' at position 3 would make the alternating sum 11, but it's not. Maybe I miscalculated earlier.Wait, let's go back. When adding a digit at position 3, the alternating sum becomes 11 + d. If d = 0, it should be 11. But when I actually compute it, it's 5. There must be a mistake in how I'm calculating the alternating sum.Perhaps I need to consider that adding a digit changes the positions of the subsequent digits, which affects the alternating sum differently. Maybe my initial approach was too simplistic.Let me try a different method. Instead of trying to adjust the alternating sum based on where I add the digit, I'll consider all possible positions and digits, compute the new alternating sum, and see if it's a multiple of 11.There are 10 possible positions to add a digit (before the first digit, between each pair of digits, and after the last digit). For each position, I'll try adding 0, 3, 6, or 9 and check if the new number is divisible by 11.This might take a while, but let's proceed step by step.1. Adding a digit at the beginning (position 1): - New number: d 9 7 5 3 1 2 4 6 8 - Alternating sum: d - 9 + 7 - 5 + 3 - 1 + 2 - 4 + 6 - 8 - Simplify: d - 9 + 7 = d - 2 - d - 2 + (-5) = d - 7 - d - 7 + 3 = d - 4 - d - 4 - 1 = d - 5 - d - 5 + 2 = d - 3 - d - 3 - 4 = d - 7 - d - 7 + 6 = d - 1 - d - 1 - 8 = d - 9 - So, alternating sum = d - 9 - We need d - 9 to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 0 - 9 = -9 → Not a multiple of 11. - d = 3: 3 - 9 = -6 → Not a multiple of 11. - d = 6: 6 - 9 = -3 → Not a multiple of 11. - d = 9: 9 - 9 = 0 → Yes, multiple of 11. - So, adding '9' at the beginning works.Let me verify this. Adding '9' at the beginning makes the number 9975312468.Compute the alternating sum:Starting from the right:8 - 6 + 4 - 2 + 1 - 3 + 5 - 7 + 9 - 9Calculating:- 8 - 6 = 2- 2 + 4 = 6- 6 - 2 = 4- 4 + 1 = 5- 5 - 3 = 2- 2 + 5 = 7- 7 - 7 = 0- 0 + 9 = 9- 9 - 9 = 0Yes, the alternating sum is 0, which is a multiple of 11. So, 9975312468 is divisible by 11 and already divisible by 3, hence by 33.Okay, that works. But let's continue checking other positions to see if there are other possibilities.2. Adding a digit at position 2: - New number: 9 d 7 5 3 1 2 4 6 8 - Alternating sum: 9 - d + 7 - 5 + 3 - 1 + 2 - 4 + 6 - 8 - Simplify: 9 - d + 7 = 16 - d - 16 - d - 5 = 11 - d - 11 - d + 3 = 14 - d - 14 - d - 1 = 13 - d - 13 - d + 2 = 15 - d - 15 - d - 4 = 11 - d - 11 - d + 6 = 17 - d - 17 - d - 8 = 9 - d - So, alternating sum = 9 - d - We need 9 - d to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 9 - 0 = 9 → Not a multiple of 11. - d = 3: 9 - 3 = 6 → Not a multiple of 11. - d = 6: 9 - 6 = 3 → Not a multiple of 11. - d = 9: 9 - 9 = 0 → Yes, multiple of 11. - So, adding '9' at position 2 works.Let me verify. Adding '9' at position 2 makes the number 9975312468, which is the same as adding '9' at the beginning. Wait, no, actually, it's different. Adding '9' at position 2 would make it 9 9 7 5 3 1 2 4 6 8, which is 9975312468, same as before. So, it's the same number, just added at a different position.But actually, in this case, adding '9' at position 2 also results in the same number, so it's redundant. So, only one unique number here.3. Adding a digit at position 3: - New number: 9 7 d 5 3 1 2 4 6 8 - Alternating sum: 9 - 7 + d - 5 + 3 - 1 + 2 - 4 + 6 - 8 - Simplify: 9 - 7 = 2 - 2 + d = 2 + d - 2 + d - 5 = d - 3 - d - 3 + 3 = d - d - 1 = d - 1 - d - 1 + 2 = d + 1 - d + 1 - 4 = d - 3 - d - 3 + 6 = d + 3 - d + 3 - 8 = d - 5 - So, alternating sum = d - 5 - We need d - 5 to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 0 - 5 = -5 → Not a multiple of 11. - d = 3: 3 - 5 = -2 → Not a multiple of 11. - d = 6: 6 - 5 = 1 → Not a multiple of 11. - d = 9: 9 - 5 = 4 → Not a multiple of 11. - So, adding a digit at position 3 doesn't work.4. Adding a digit at position 4: - New number: 9 7 5 d 3 1 2 4 6 8 - Alternating sum: 9 - 7 + 5 - d + 3 - 1 + 2 - 4 + 6 - 8 - Simplify: 9 - 7 = 2 - 2 + 5 = 7 - 7 - d = 7 - d - 7 - d + 3 = 10 - d - 10 - d - 1 = 9 - d - 9 - d + 2 = 11 - d - 11 - d - 4 = 7 - d - 7 - d + 6 = 13 - d - 13 - d - 8 = 5 - d - So, alternating sum = 5 - d - We need 5 - d to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 5 - 0 = 5 → Not a multiple of 11. - d = 3: 5 - 3 = 2 → Not a multiple of 11. - d = 6: 5 - 6 = -1 → Not a multiple of 11. - d = 9: 5 - 9 = -4 → Not a multiple of 11. - So, adding a digit at position 4 doesn't work.5. Adding a digit at position 5: - New number: 9 7 5 3 d 1 2 4 6 8 - Alternating sum: 9 - 7 + 5 - 3 + d - 1 + 2 - 4 + 6 - 8 - Simplify: 9 - 7 = 2 - 2 + 5 = 7 - 7 - 3 = 4 - 4 + d = 4 + d - 4 + d - 1 = 3 + d - 3 + d + 2 = 5 + d - 5 + d - 4 = 1 + d - 1 + d + 6 = 7 + d - 7 + d - 8 = d - 1 - So, alternating sum = d - 1 - We need d - 1 to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 0 - 1 = -1 → Not a multiple of 11. - d = 3: 3 - 1 = 2 → Not a multiple of 11. - d = 6: 6 - 1 = 5 → Not a multiple of 11. - d = 9: 9 - 1 = 8 → Not a multiple of 11. - So, adding a digit at position 5 doesn't work.6. Adding a digit at position 6: - New number: 9 7 5 3 1 d 2 4 6 8 - Alternating sum: 9 - 7 + 5 - 3 + 1 - d + 2 - 4 + 6 - 8 - Simplify: 9 - 7 = 2 - 2 + 5 = 7 - 7 - 3 = 4 - 4 + 1 = 5 - 5 - d = 5 - d - 5 - d + 2 = 7 - d - 7 - d - 4 = 3 - d - 3 - d + 6 = 9 - d - 9 - d - 8 = 1 - d - So, alternating sum = 1 - d - We need 1 - d to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 1 - 0 = 1 → Not a multiple of 11. - d = 3: 1 - 3 = -2 → Not a multiple of 11. - d = 6: 1 - 6 = -5 → Not a multiple of 11. - d = 9: 1 - 9 = -8 → Not a multiple of 11. - So, adding a digit at position 6 doesn't work.7. Adding a digit at position 7: - New number: 9 7 5 3 1 2 d 4 6 8 - Alternating sum: 9 - 7 + 5 - 3 + 1 - 2 + d - 4 + 6 - 8 - Simplify: 9 - 7 = 2 - 2 + 5 = 7 - 7 - 3 = 4 - 4 + 1 = 5 - 5 - 2 = 3 - 3 + d = 3 + d - 3 + d - 4 = d - 1 - d - 1 + 6 = d + 5 - d + 5 - 8 = d - 3 - So, alternating sum = d - 3 - We need d - 3 to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 0 - 3 = -3 → Not a multiple of 11. - d = 3: 3 - 3 = 0 → Yes, multiple of 11. - d = 6: 6 - 3 = 3 → Not a multiple of 11. - d = 9: 9 - 3 = 6 → Not a multiple of 11. - So, adding '3' at position 7 works.Let me verify. Adding '3' at position 7 makes the number 9753123468.Compute the alternating sum:Starting from the right:8 - 6 + 4 - 3 + 2 - 1 + 3 - 5 + 7 - 9Calculating:- 8 - 6 = 2- 2 + 4 = 6- 6 - 3 = 3- 3 + 2 = 5- 5 - 1 = 4- 4 + 3 = 7- 7 - 5 = 2- 2 + 7 = 9- 9 - 9 = 0Yes, the alternating sum is 0, which is a multiple of 11. So, 9753123468 is divisible by 11 and already divisible by 3, hence by 33.8. Adding a digit at position 8: - New number: 9 7 5 3 1 2 4 d 6 8 - Alternating sum: 9 - 7 + 5 - 3 + 1 - 2 + 4 - d + 6 - 8 - Simplify: 9 - 7 = 2 - 2 + 5 = 7 - 7 - 3 = 4 - 4 + 1 = 5 - 5 - 2 = 3 - 3 + 4 = 7 - 7 - d = 7 - d - 7 - d + 6 = 13 - d - 13 - d - 8 = 5 - d - So, alternating sum = 5 - d - We need 5 - d to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 5 - 0 = 5 → Not a multiple of 11. - d = 3: 5 - 3 = 2 → Not a multiple of 11. - d = 6: 5 - 6 = -1 → Not a multiple of 11. - d = 9: 5 - 9 = -4 → Not a multiple of 11. - So, adding a digit at position 8 doesn't work.9. Adding a digit at position 9: - New number: 9 7 5 3 1 2 4 6 d 8 - Alternating sum: 9 - 7 + 5 - 3 + 1 - 2 + 4 - 6 + d - 8 - Simplify: 9 - 7 = 2 - 2 + 5 = 7 - 7 - 3 = 4 - 4 + 1 = 5 - 5 - 2 = 3 - 3 + 4 = 7 - 7 - 6 = 1 - 1 + d = 1 + d - 1 + d - 8 = d - 7 - So, alternating sum = d - 7 - We need d - 7 to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 0 - 7 = -7 → Not a multiple of 11. - d = 3: 3 - 7 = -4 → Not a multiple of 11. - d = 6: 6 - 7 = -1 → Not a multiple of 11. - d = 9: 9 - 7 = 2 → Not a multiple of 11. - So, adding a digit at position 9 doesn't work.10. Adding a digit at the end (position 10): - New number: 9 7 5 3 1 2 4 6 8 d - Alternating sum: 9 - 7 + 5 - 3 + 1 - 2 + 4 - 6 + 8 - d - Simplify: 9 - 7 = 2 - 2 + 5 = 7 - 7 - 3 = 4 - 4 + 1 = 5 - 5 - 2 = 3 - 3 + 4 = 7 - 7 - 6 = 1 - 1 + 8 = 9 - 9 - d = 9 - d - We need 9 - d to be a multiple of 11. - Possible d: 0, 3, 6, 9 - d = 0: 9 - 0 = 9 → Not a multiple of 11. - d = 3: 9 - 3 = 6 → Not a multiple of 11. - d = 6: 9 - 6 = 3 → Not a multiple of 11. - d = 9: 9 - 9 = 0 → Yes, multiple of 11. - So, adding '9' at the end works.Let me verify. Adding '9' at the end makes the number 9753124689.Compute the alternating sum:Starting from the right:9 - 8 + 6 - 4 + 2 - 1 + 3 - 5 + 7 - 9Calculating:- 9 - 8 = 1- 1 + 6 = 7- 7 - 4 = 3- 3 + 2 = 5- 5 - 1 = 4- 4 + 3 = 7- 7 - 5 = 2- 2 + 7 = 9- 9 - 9 = 0Yes, the alternating sum is 0, which is a multiple of 11. So, 9753124689 is divisible by 11 and already divisible by 3, hence by 33.So, in summary, I found three possible ways to add a digit to make the number divisible by 33:1. Adding '9' at the beginning: 99753124682. Adding '3' at position 7: 97531234683. Adding '9' at the end: 9753124689Each of these new numbers has an alternating sum of 0, which is a multiple of 11, and the sum of their digits is 45 + 9 = 54, 45 + 3 = 48, and 45 + 9 = 54 respectively, all of which are divisible by 3. Therefore, all three numbers are divisible by 33.I think that's all the possibilities. I checked every position and the required digits, and these are the only ones that work.

💡Okay, so I have this problem where I'm given that a and b are positive numbers, and they satisfy the equation ab = 2a + b. Then, it says that a + 2b is always greater than or equal to m² - 8m, and I need to find the range of the real number m. The options are given, so I need to figure out which one is correct.First, let me try to understand the given equation: ab = 2a + b. Maybe I can manipulate this equation to express one variable in terms of the other. Let me see... If I rearrange the equation, I can write ab - 2a - b = 0. Hmm, maybe factor something out. Let's see, if I factor out a from the first two terms, I get a(b - 2) - b = 0. That doesn't seem too helpful yet.Wait, maybe I can express this as ab - 2a - b = 0, and then add 2 to both sides to make it factorable. Let me try that: ab - 2a - b + 2 = 2. Now, let's see if I can factor the left side. Maybe group terms: (ab - 2a) - (b - 2) = 2. Factoring a from the first group and 1 from the second: a(b - 2) - 1(b - 2) = 2. Oh, nice! So, (a - 1)(b - 2) = 2. That's a useful form.So, (a - 1)(b - 2) = 2. Since a and b are positive, a - 1 and b - 2 must also be positive? Wait, not necessarily. Because if a is less than 1, then a - 1 would be negative, but b would have to be greater than 2 to make the product positive. Similarly, if a is greater than 1, then b - 2 is positive. Hmm, but since a and b are positive, let's see:If a > 1, then b - 2 must be positive, so b > 2.If a < 1, then b - 2 must be negative, so b < 2.But since a and b are positive, let's consider both cases.Case 1: a > 1, so b > 2.Case 2: a < 1, so b < 2.But maybe it's easier to express one variable in terms of the other. Let's solve for b in terms of a.Starting from ab = 2a + b.Let me bring all terms to one side: ab - b = 2a.Factor out b: b(a - 1) = 2a.So, b = (2a)/(a - 1). Since a > 0 and b > 0, the denominator a - 1 must be positive because if a - 1 were negative, then b would be negative, which isn't allowed. So, a - 1 > 0, which implies a > 1.Therefore, a must be greater than 1, and b is given by b = (2a)/(a - 1).So, now, I can express a + 2b in terms of a.Let me compute a + 2b:a + 2b = a + 2*(2a)/(a - 1) = a + (4a)/(a - 1).Let me combine these terms:First, write a as a*(a - 1)/(a - 1) to have a common denominator:a + (4a)/(a - 1) = [a(a - 1) + 4a]/(a - 1) = [a² - a + 4a]/(a - 1) = [a² + 3a]/(a - 1).So, a + 2b = (a² + 3a)/(a - 1).Now, I need to find the minimum value of this expression because the problem states that a + 2b is always greater than or equal to m² - 8m. So, if I can find the minimum value of a + 2b, then I can set that minimum value greater than or equal to m² - 8m and solve for m.To find the minimum of (a² + 3a)/(a - 1) for a > 1, I can use calculus or maybe some algebraic manipulation. Let me try calculus first.Let me denote f(a) = (a² + 3a)/(a - 1). To find its minimum, compute its derivative and set it equal to zero.First, find f'(a):Using the quotient rule: f'(a) = [(2a + 3)(a - 1) - (a² + 3a)(1)] / (a - 1)².Let me compute the numerator:(2a + 3)(a - 1) = 2a² - 2a + 3a - 3 = 2a² + a - 3.Subtract (a² + 3a): 2a² + a - 3 - a² - 3a = a² - 2a - 3.So, f'(a) = (a² - 2a - 3)/(a - 1)².Set f'(a) = 0:(a² - 2a - 3) = 0.Solve for a:a² - 2a - 3 = 0.Using quadratic formula: a = [2 ± sqrt(4 + 12)] / 2 = [2 ± sqrt(16)] / 2 = [2 ± 4]/2.So, a = (2 + 4)/2 = 3, or a = (2 - 4)/2 = -1.But since a > 1, we discard a = -1. So, critical point at a = 3.Now, check if this is a minimum. Let's compute the second derivative or test intervals.Alternatively, since f(a) approaches infinity as a approaches 1 from the right, and as a approaches infinity, f(a) behaves like a² / a = a, which also approaches infinity. So, the critical point at a = 3 is likely a minimum.Compute f(3):f(3) = (9 + 9)/(3 - 1) = 18/2 = 9.So, the minimum value of a + 2b is 9.Therefore, since a + 2b is always greater than or equal to 9, and the problem states that a + 2b >= m² - 8m always holds true, we have:9 >= m² - 8m.So, m² - 8m <= 9.Bring all terms to one side:m² - 8m - 9 <= 0.Solve the quadratic inequality:First, find the roots of m² - 8m - 9 = 0.Using quadratic formula: m = [8 ± sqrt(64 + 36)] / 2 = [8 ± sqrt(100)] / 2 = [8 ± 10]/2.So, m = (8 + 10)/2 = 18/2 = 9, or m = (8 - 10)/2 = (-2)/2 = -1.So, the quadratic can be factored as (m - 9)(m + 1) <= 0.To solve this inequality, consider the intervals determined by the roots: m < -1, -1 < m < 9, and m > 9.Test each interval:1. For m < -1, say m = -2: (-2 - 9)(-2 + 1) = (-11)(-1) = 11 > 0. Doesn't satisfy the inequality.2. For -1 < m < 9, say m = 0: (0 - 9)(0 + 1) = (-9)(1) = -9 < 0. Satisfies the inequality.3. For m > 9, say m = 10: (10 - 9)(10 + 1) = (1)(11) = 11 > 0. Doesn't satisfy the inequality.Now, check the endpoints:At m = -1: (-1 - 9)(-1 + 1) = (-10)(0) = 0. So, equality holds.At m = 9: (9 - 9)(9 + 1) = (0)(10) = 0. So, equality holds.Therefore, the solution to the inequality is -1 <= m <= 9.Looking back at the options:A: 4 - 2√6 ≤ m ≤ 4 + 2√6B: m ≥ 4 + 2√6 or m ≤ 4 - 2√6C: -1 ≤ m ≤ 9D: m ≥ 9 or m ≤ -1So, the correct answer is C: -1 ≤ m ≤ 9.Wait, just to make sure I didn't make a mistake, let me recap:Given ab = 2a + b, I expressed b in terms of a as b = 2a/(a - 1), since a > 1. Then, a + 2b became (a² + 3a)/(a - 1). Took derivative, found critical point at a = 3, computed f(3) = 9, so minimum value is 9. Therefore, 9 >= m² - 8m, leading to m² - 8m - 9 <= 0, which solves to -1 <= m <= 9. That seems correct.Alternatively, maybe I can solve this without calculus, using the AM-GM inequality or something. Let me try that approach to verify.From the given equation ab = 2a + b, as before, I can write 1 = (2)/(b) + (1)/(a). Let me denote x = 1/a and y = 1/b. Then, 1 = 2y + x.We need to find the minimum of a + 2b, which is 1/x + 2*(1/y).So, minimize 1/x + 2/y subject to 2y + x = 1.Let me use Lagrange multipliers or substitution.From the constraint, x = 1 - 2y.So, substitute into the expression: 1/(1 - 2y) + 2/y.Let me denote this as f(y) = 1/(1 - 2y) + 2/y.Find the minimum of f(y) for y > 0 and 1 - 2y > 0, so y < 1/2.So, y is in (0, 1/2).Compute derivative f'(y):f'(y) = [0 + 2/(1 - 2y)^2] + [-2/y²].Set derivative equal to zero:2/(1 - 2y)^2 - 2/y² = 0.So, 2/(1 - 2y)^2 = 2/y².Divide both sides by 2: 1/(1 - 2y)^2 = 1/y².Take reciprocals: (1 - 2y)^2 = y².Take square roots: |1 - 2y| = |y|.Since y > 0 and y < 1/2, 1 - 2y is positive, so 1 - 2y = y.Thus, 1 = 3y => y = 1/3.So, y = 1/3, then x = 1 - 2*(1/3) = 1 - 2/3 = 1/3.Therefore, x = 1/3, y = 1/3.Thus, a = 1/x = 3, b = 1/y = 3.So, a + 2b = 3 + 6 = 9, same as before.Therefore, the minimum is indeed 9, so the same conclusion holds.Therefore, the range of m is -1 <= m <= 9, which is option C.**Final Answer**boxed{C}

💡First, I need to determine the number of terms in each polynomial. The first polynomial, (x + y + z), has 3 distinct terms. The second polynomial, (u + v + w + x + y), has 5 distinct terms.Next, I should consider how multiplying these polynomials affects the number of terms. When multiplying two polynomials, each term in the first polynomial is multiplied by each term in the second polynomial. This means that every term in (x + y + z) will be paired with every term in (u + v + w + x + y).To find the total number of distinct terms in the expansion, I multiply the number of terms in the first polynomial by the number of terms in the second polynomial. Therefore, (3 times 5 = 15).This calculation shows that there are 15 distinct terms in the expansion of ((x + y + z)(u + v + w + x + y)).