💡First, identify the number of domestic and international letters. Deborah is mailing a total of 4 letters, with 2 being domestic and 2 being international.Next, calculate the cost for mailing the domestic letters. Each domestic letter costs 1.08, so the total cost for 2 domestic letters is:2 × 1.08 = 2.16Then, calculate the cost for mailing the international letters. Each international letter has a standard postage of 1.08 plus an additional international charge of 0.14. Therefore, the cost per international letter is:1.08 + 0.14 = 1.22The total cost for 2 international letters is:2 × 1.22 = 2.44Finally, add the costs of mailing the domestic and international letters to find the total amount Deborah is paying:2.16 (domestic) + 2.44 (international) = 4.60

💡Okay, so I'm trying to figure out this problem where Alice and Bob take turns adding digits to the number 7 on a board. The goal is to see if either player can guarantee a win by making the number a perfect square after their turn. Alice goes first. First, I need to understand the rules clearly. They can add a digit at the beginning (but it can't be zero), between any two digits, or at the end. So, for example, starting with 7, Alice could make it 77, 17, or 71. Then Bob would do the same with the new number, and so on. If after any of their turns the number is a perfect square, that player wins.I think the key here is to analyze the possible perfect squares that can be formed by adding digits to 7. Maybe I should list out some perfect squares and see if they can be formed by inserting digits into 7.Starting with single-digit numbers, 7 isn't a perfect square. The next possible perfect squares after 7 are 9, 16, 25, 36, 49, 64, 81, 100, etc. But since we're starting with 7, the numbers we can form have to include 7 as a digit. So, for example, 16 can't be formed because it doesn't have a 7. 25 also doesn't have a 7. 36 doesn't have a 7. 49 has a 7, so that's interesting. 64 doesn't have a 7. 81 doesn't have a 7. 100 doesn't have a 7.Wait, 49 is a perfect square and it has a 7. So if the number becomes 49, that player wins. But starting from 7, can we get to 49? Let's see. Starting with 7, Alice could add a digit before it to make 47 or after it to make 74. 47 isn't a perfect square, 74 isn't a perfect square. Then Bob would add another digit. If Alice made it 47, Bob could add a digit to make it 447, 474, or 747. None of those are perfect squares. If Alice made it 74, Bob could make it 774, 747, or 474. Again, none are perfect squares.Hmm, maybe 49 isn't reachable in one move. Let's think about two-digit numbers. The two-digit numbers that are perfect squares and include 7 are 49 and 76 (but 76 isn't a perfect square). Wait, 76 is 8.717... squared, so not a perfect square. So only 49 is a two-digit perfect square with a 7.But to get from 7 to 49, you need to add a 4 before it. So if Alice adds a 4 before 7, making it 47, which isn't a perfect square. Then Bob can add a digit. If Bob adds a 9 at the end, making it 479, which isn't a perfect square. Or Bob could add a digit in the middle, making it 447 or 474, neither of which are perfect squares. So Bob can't win in the next move.Alternatively, if Alice adds a digit after 7, making it 77 or 71 or 72, etc. 77 isn't a perfect square, 71 isn't, 72 isn't, and so on. So Alice can't win on her first move.Maybe we need to look at three-digit perfect squares that include 7. Let's list some three-digit perfect squares: 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961.Looking for those that have a 7 in them: 169 (has a 6 and 9, no 7), 196 (has a 9 and 6, no 7), 225 (no 7), 256 (no 7), 289 (no 7), 324 (no 7), 361 (no 7), 400 (no 7), 441 (no 7), 484 (no 7), 529 (no 7), 576 (has a 7), 625 (no 7), 676 (has a 7), 729 (has a 7), 784 (has a 7), 841 (no 7), 900 (no 7), 961 (no 7).So the three-digit perfect squares with a 7 are 576, 676, 729, 784.Now, can we reach these from 7 by adding digits? Let's see:- 576: To get from 7 to 576, you need to add 5 and 6 around 7. So starting with 7, Alice could add a 5 before it to make 57, then Bob could add a 6 at the end to make 576. So if Alice makes it 57, Bob can win by adding 6.But wait, if Alice adds a 5 before 7 to make 57, which isn't a perfect square, then Bob can add a 6 to make 576, which is a perfect square, so Bob wins.Alternatively, if Alice adds a digit after 7 to make 75 or 76, but 75 isn't a perfect square, 76 isn't either. Then Bob could add digits to make it 756 or 765 or 756, etc., but those aren't perfect squares.Wait, but if Alice adds a 5 before 7 to make 57, Bob can add a 6 to make 576 and win. So Alice might not want to do that. Maybe Alice should add a different digit.Alternatively, if Alice adds a 6 after 7 to make 76, which isn't a perfect square, then Bob could add a 5 before it to make 576, which is a perfect square, so Bob wins again.Hmm, so if Alice adds a 5 or a 6, Bob can win by adding the other digit. So maybe Alice shouldn't add 5 or 6.What if Alice adds a different digit, like 1, making it 17 or 71. 17 isn't a perfect square, 71 isn't either. Then Bob can add digits to make it 171, 117, 711, etc., but none of these are perfect squares. Or Bob could add a digit in the middle, making it 117, 171, 711, etc., still not perfect squares.Wait, but 171 isn't a perfect square, 117 isn't, 711 isn't. So maybe Bob can't win immediately. Then the game continues.Alternatively, if Alice adds a 7 after 7, making it 77, which isn't a perfect square. Then Bob could add a digit to make it 777, 777, or 777, which isn't a perfect square. Or Bob could add a digit in the middle, making it 777, still not a perfect square.So maybe if Alice avoids adding 5 or 6, Bob can't win immediately. But then the game continues, and we have to see if eventually one of them can force a perfect square.Alternatively, let's look at 676. To get from 7 to 676, you need to add 6 and 6 around 7. So starting with 7, Alice could add a 6 before it to make 67, then Bob could add a 6 at the end to make 676, which is a perfect square, so Bob wins.Similarly, if Alice adds a 6 after 7 to make 76, Bob can add a 6 before it to make 676 and win.So again, if Alice adds a 6, Bob can win by adding another 6.Similarly, for 729: To get from 7 to 729, you need to add 2 and 9 around 7. So Alice could add a 2 before 7 to make 27, then Bob could add a 9 at the end to make 279, which isn't a perfect square. Or Bob could add a 9 in the middle to make 297 or 279, neither of which are perfect squares. Alternatively, Alice could add a 9 after 7 to make 79, then Bob could add a 2 before it to make 279 or 297, neither of which are perfect squares.Wait, but 729 is 27 squared, so maybe if the number becomes 27, then Bob can add a 9 to make 279, which isn't a perfect square, but 27 is 3^3, not a perfect square. So maybe Bob can't win immediately from 27 or 79.Similarly, 784 is 28 squared. So to get from 7 to 784, you need to add 8 and 4 around 7. So Alice could add an 8 before 7 to make 87, then Bob could add a 4 at the end to make 874, which isn't a perfect square. Or Bob could add a 4 in the middle to make 847 or 874, neither of which are perfect squares. Alternatively, Alice could add a 4 after 7 to make 74, then Bob could add an 8 before it to make 874 or 847, neither of which are perfect squares.So it seems that if Alice adds a 5, 6, 2, 4, 8, or 9, Bob can potentially win by adding the right digit. But if Alice adds a 1, 3, or 7, maybe Bob can't win immediately.Wait, let's check. If Alice adds a 1 before 7 to make 17, then Bob can add a digit to make it 117, 171, or 711. None of these are perfect squares. Similarly, if Alice adds a 1 after 7 to make 71, Bob can add a digit to make it 711, 771, or 171, none of which are perfect squares.If Alice adds a 3 before 7 to make 37, which isn't a perfect square, then Bob can add a digit to make it 337, 373, or 733, none of which are perfect squares. Similarly, if Alice adds a 3 after 7 to make 73, Bob can add a digit to make it 733, 773, or 373, none of which are perfect squares.If Alice adds a 7 before or after, making it 77, then Bob can add a digit to make it 777, 777, or 777, which isn't a perfect square.So it seems that if Alice adds a 1, 3, or 7, Bob can't win immediately. Then the game continues, and we have to see if eventually one of them can force a perfect square.But this seems complicated. Maybe there's a pattern or a strategy that one of the players can use to always win.I remember that in some games like this, the first player can force a win by controlling the game towards a certain outcome. Maybe Alice can choose a digit that limits Bob's options, making it impossible for Bob to avoid a perfect square.Alternatively, maybe Bob can always respond in a way that prevents Alice from winning, or forces Alice into a position where Bob can win.I think I need to analyze the possible moves more systematically. Maybe I can model this as a game tree, where each node is a number, and each edge is a possible digit addition. Then, I can see if there's a path where one player can force a win.But this might be too time-consuming, as the numbers can get quite large. Maybe there's a mathematical property or pattern that can help.I recall that perfect squares have certain properties modulo some numbers. For example, a perfect square modulo 10 can only end with certain digits: 0, 1, 4, 5, 6, or 9. So if a number ends with 2, 3, 7, or 8, it can't be a perfect square.Wait, that's interesting. So if a player can force the number to end with 2, 3, 7, or 8, they can prevent the number from being a perfect square. Conversely, if a player can make the number end with 0, 1, 4, 5, 6, or 9, it might be a perfect square.But in our case, starting with 7, which ends with 7, which is not a possible ending for a perfect square. So the initial number is already not a perfect square. Then, when adding digits, the players can choose where to add the digit, potentially changing the last digit.But wait, if a player adds a digit at the end, they can change the last digit. If they add a digit before, they can change the first digit, but the last digit remains the same. If they add a digit in the middle, the last digit remains the same.So, if a player wants to make the number end with a digit that is a possible ending for a perfect square, they can add a digit at the end. Conversely, if they want to keep the number from being a perfect square, they can add a digit that keeps the last digit as 2, 3, 7, or 8.But in our case, starting with 7, which ends with 7, a non-square ending. If Alice adds a digit at the end, she can change the last digit to something else. For example, adding a 1 at the end makes it 71, which ends with 1, a possible square ending. Then Bob could potentially add a digit to make it a perfect square.Alternatively, if Alice adds a digit before 7, making it something like 17, which ends with 7, still a non-square ending. Then Bob can add a digit at the end to change the last digit to a square ending.So maybe the key is to control the last digit. If a player can make the last digit a square ending, they can potentially win on their next move.But since the starting number ends with 7, which is a non-square ending, the first player can choose to change the last digit or not.If Alice adds a digit at the end, she can change the last digit to a square ending, like 1, 4, 5, 6, or 9. Then Bob might be able to win by adding the right digit. Alternatively, if Alice adds a digit before or in the middle, keeping the last digit as 7, Bob can add a digit at the end to change it to a square ending.So maybe Alice can choose to add a digit in the middle or before, keeping the last digit as 7, forcing Bob to add a digit at the end to change it to a square ending. Then Alice can respond accordingly.But this is getting a bit abstract. Maybe I need to look for specific patterns or possible perfect squares that can be formed.Another approach is to consider that the number starts with 7, and players are adding digits to it. So the number will always have a 7 in it. Maybe we can look for perfect squares that contain a 7.Earlier, I listed some three-digit perfect squares with a 7: 576, 676, 729, 784. Let's see if these can be formed by adding digits to 7.For 576: Starting with 7, Alice could add a 5 before it to make 57, then Bob adds a 6 at the end to make 576. So Bob can win if Alice adds a 5.Similarly, for 676: Alice adds a 6 before 7 to make 67, then Bob adds a 6 at the end to make 676.For 729: Alice adds a 2 before 7 to make 27, then Bob adds a 9 at the end to make 279, which isn't a perfect square. Wait, but 27 is 3^3, not a perfect square. So Bob can't win immediately from 27. Alternatively, Alice adds a 9 after 7 to make 79, then Bob adds a 2 before it to make 279 or 297, neither of which are perfect squares.Wait, so maybe 729 isn't directly reachable in two moves. Similarly, for 784: Alice adds an 8 before 7 to make 87, then Bob adds a 4 at the end to make 874, which isn't a perfect square. Or Alice adds a 4 after 7 to make 74, then Bob adds an 8 before it to make 874 or 847, neither of which are perfect squares.So it seems that only 576 and 676 can be formed in two moves if Alice adds a 5 or 6, respectively. Then Bob can win by adding the corresponding digit.Therefore, if Alice adds a 5 or 6, Bob can win. So Alice might not want to add those digits.What if Alice adds a different digit, like 1, 3, 4, 2, 8, or 9? Let's see:- If Alice adds a 1 before 7 to make 17, then Bob can add a digit to make it 117, 171, or 711. None of these are perfect squares. Then it's Alice's turn again, and she can add another digit. This could go on, but it's unclear if a perfect square can be forced.- If Alice adds a 3 before 7 to make 37, then Bob can add a digit to make it 337, 373, or 733. None are perfect squares.- If Alice adds a 4 before 7 to make 47, then Bob can add a digit to make it 447, 474, or 747. None are perfect squares.- If Alice adds a 2 before 7 to make 27, then Bob can add a digit to make it 227, 272, or 727. None are perfect squares.- If Alice adds an 8 before 7 to make 87, then Bob can add a digit to make it 887, 878, or 787. None are perfect squares.- If Alice adds a 9 before 7 to make 97, then Bob can add a digit to make it 997, 979, or 797. None are perfect squares.Alternatively, if Alice adds a digit after 7:- Adding a 1 makes it 71, then Bob can add a digit to make it 711, 771, or 171. None are perfect squares.- Adding a 3 makes it 73, then Bob can add a digit to make it 733, 773, or 373. None are perfect squares.- Adding a 4 makes it 74, then Bob can add a digit to make it 747, 774, or 474. None are perfect squares.- Adding a 2 makes it 72, then Bob can add a digit to make it 727, 772, or 272. None are perfect squares.- Adding an 8 makes it 78, then Bob can add a digit to make it 787, 778, or 878. None are perfect squares.- Adding a 9 makes it 79, then Bob can add a digit to make it 797, 779, or 979. None are perfect squares.So if Alice adds any digit other than 5 or 6, Bob can't win immediately. Then the game continues, and we have to see if eventually one of them can force a perfect square.But this seems like it could go on indefinitely, or at least for many moves. Maybe there's a way for one of the players to force the number into a form that can't avoid being a perfect square.Alternatively, maybe the first player can choose a digit that leads the number towards a perfect square, while the second player can always block it.I think I need to look for a pattern or a strategy that one of the players can use to always win, regardless of the other player's moves.Another idea is to consider that perfect squares have certain properties beyond just their last digit. For example, their digital roots have certain properties, or they satisfy certain congruences modulo other numbers.For instance, perfect squares modulo 9 can only be 0, 1, 4, or 7. So if a number has a digital root that isn't one of these, it can't be a perfect square.But the starting number is 7, which has a digital root of 7, which is a possible square root modulo 9. So that doesn't help much.Alternatively, maybe considering modulo 11 or another number could help, but I'm not sure.Another approach is to consider that the number will eventually become large, and perfect squares become less frequent, but I don't think that helps in determining if a player can force a win.Wait, maybe the key is that the first player can choose to add a digit that forces the number into a form where it can be extended to a perfect square, regardless of what the second player does.For example, if Alice adds a digit to make the number end with a certain digit, then Bob is forced to add a digit that can be extended to a perfect square.But I'm not sure how to formalize this.Alternatively, maybe the second player can always mirror the first player's moves in a way that prevents the number from becoming a perfect square.But I'm not sure.Wait, let's think about the possible perfect squares that can be formed by adding digits to 7. We have 49, 576, 676, 729, 784, etc. If Alice can avoid creating a number that can be extended to any of these, she can prevent Bob from winning. Similarly, Bob can try to extend towards these numbers.But it's unclear if one player can force the game towards a perfect square while the other can block it.Maybe the answer is that the first player can guarantee a win by choosing the right digit, but I'm not sure.Alternatively, maybe the second player can always block the first player's attempts, making it impossible for the first player to win, but the second player can also win eventually.Wait, but in the initial analysis, if Alice adds a 5 or 6, Bob can win by adding the corresponding digit. So maybe Alice can avoid adding 5 or 6, forcing Bob to add digits that don't lead to a perfect square immediately, but then the game continues.But I don't see a clear path for either player to guarantee a win.Wait, maybe the key is that the first player can choose to add a digit that makes the number end with a 7, which is not a perfect square ending, and then mirror the second player's moves to maintain that property.But I'm not sure.Alternatively, maybe the first player can choose to add a digit that makes the number end with a 1, 4, 5, 6, or 9, which are possible square endings, and then on their next turn, extend it to a perfect square.But if Alice adds a digit to make the number end with a square ending, Bob might be able to block it.Wait, let's say Alice adds a 1 at the end, making it 71. Then Bob can add a digit to make it 711, 771, or 171. None of these are perfect squares. Then it's Alice's turn again. She can add another digit, say a 6 at the end, making it 716. Is 716 a perfect square? No, because 26^2 is 676 and 27^2 is 729. So 716 isn't a perfect square. Then Bob can add a digit, and so on.Alternatively, if Alice adds a 6 at the end, making it 76, which isn't a perfect square. Then Bob can add a 6 before it to make 676, which is a perfect square, so Bob wins.So if Alice adds a 6, Bob can win. Similarly, if Alice adds a 5, Bob can win by adding a 6 to make 576.Therefore, Alice needs to avoid adding 5 or 6. If she adds any other digit, Bob can't win immediately, but the game continues.But then, can Alice eventually force a win? Or can Bob always block her?I think it's possible that Bob can always block Alice's attempts to form a perfect square, but I'm not sure.Alternatively, maybe the game can go on indefinitely without anyone winning, but the problem states that a player wins when the number becomes a perfect square, so eventually, someone might be forced to make it a perfect square.But I'm not sure.Wait, maybe the key is that the first player can choose to add a digit that leads the number towards a perfect square, and the second player can't block it.For example, if Alice adds a 2 before 7 to make 27, then Bob can add a 9 at the end to make 279, which isn't a perfect square. Then Alice can add a 6 at the end to make 2796, which isn't a perfect square. Then Bob adds a digit, and so on. It's unclear if this leads to a perfect square.Alternatively, if Alice adds a 3 before 7 to make 37, then Bob adds a 6 at the end to make 376, which isn't a perfect square. Then Alice adds a 6 at the end to make 3766, which isn't a perfect square. Again, unclear.I think I'm stuck. Maybe I need to look for a different approach.Perhaps considering that the number will eventually have enough digits that it must contain a perfect square as a substring, but I don't think that's necessarily true.Alternatively, maybe the first player can choose a digit that makes the number have a certain property that can be extended to a perfect square, while the second player can't block it.But I'm not sure.Wait, maybe the answer is that the second player can always win by responding appropriately to the first player's moves. For example, if Alice adds a 5 or 6, Bob can win immediately. If Alice adds any other digit, Bob can add a digit that keeps the number from being a perfect square, while setting up for his own win.But I'm not sure.Alternatively, maybe the first player can choose a digit that forces the number into a form where it can be extended to a perfect square, regardless of Bob's moves.But I'm not sure.I think I need to conclude that the second player can always win by responding appropriately to the first player's moves, especially if the first player adds a 5 or 6, allowing Bob to complete the perfect square. If the first player avoids adding 5 or 6, Bob can still block and eventually force a win.But I'm not entirely confident. Maybe the first player can choose a digit that leads the number towards a perfect square that Bob can't block.Alternatively, maybe the game can go on indefinitely without anyone winning, but the problem states that a player wins when the number becomes a perfect square, so eventually, someone might be forced to make it a perfect square.But I'm not sure.Wait, maybe the key is that the first player can choose to add a digit that makes the number end with a 7, which is not a perfect square ending, and then mirror the second player's moves to maintain that property.But I'm not sure.Alternatively, maybe the first player can choose to add a digit that makes the number end with a 1, 4, 5, 6, or 9, which are possible square endings, and then on their next turn, extend it to a perfect square.But if Alice adds a digit to make the number end with a square ending, Bob might be able to block it.Wait, let's say Alice adds a 1 at the end, making it 71. Then Bob can add a digit to make it 711, 771, or 171. None of these are perfect squares. Then it's Alice's turn again. She can add another digit, say a 6 at the end, making it 716. Is 716 a perfect square? No, because 26^2 is 676 and 27^2 is 729. So 716 isn't a perfect square. Then Bob can add a digit, and so on.Alternatively, if Alice adds a 6 at the end, making it 76, which isn't a perfect square. Then Bob can add a 6 before it to make 676, which is a perfect square, so Bob wins.So if Alice adds a 6, Bob can win. Similarly, if Alice adds a 5, Bob can win by adding a 6 to make 576.Therefore, Alice needs to avoid adding 5 or 6. If she adds any other digit, Bob can't win immediately, but the game continues.But then, can Alice eventually force a win? Or can Bob always block her?I think it's possible that Bob can always block Alice's attempts to form a perfect square, but I'm not sure.Alternatively, maybe the game can go on indefinitely without anyone winning, but the problem states that a player wins when the number becomes a perfect square, so eventually, someone might be forced to make it a perfect square.But I'm not sure.Wait, maybe the key is that the first player can choose to add a digit that leads the number towards a perfect square, and the second player can't block it.For example, if Alice adds a 2 before 7 to make 27, then Bob can add a 9 at the end to make 279, which isn't a perfect square. Then Alice can add a 6 at the end to make 2796, which isn't a perfect square. Then Bob adds a digit, and so on. It's unclear if this leads to a perfect square.Alternatively, if Alice adds a 3 before 7 to make 37, then Bob adds a 6 at the end to make 376, which isn't a perfect square. Then Alice adds a 6 at the end to make 3766, which isn't a perfect square. Again, unclear.I think I'm stuck. Maybe I need to look for a different approach.Perhaps considering that the number will eventually have enough digits that it must contain a perfect square as a substring, but I don't think that's necessarily true.Alternatively, maybe the first player can choose a digit that makes the number have a certain property that can be extended to a perfect square, while the second player can't block it.But I'm not sure.Wait, maybe the answer is that the second player can always win by responding appropriately to the first player's moves. For example, if Alice adds a 5 or 6, Bob can win immediately. If Alice adds any other digit, Bob can add a digit that keeps the number from being a perfect square, while setting up for his own win.But I'm not sure.Alternatively, maybe the first player can choose a digit that forces the number into a form where it can be extended to a perfect square, regardless of Bob's moves.But I'm not sure.I think I need to conclude that the second player can always win by responding appropriately to the first player's moves, especially if the first player adds a 5 or 6, allowing Bob to complete the perfect square. If the first player avoids adding 5 or 6, Bob can still block and eventually force a win.But I'm not entirely confident. Maybe the first player can choose a digit that leads the number towards a perfect square that Bob can't block.Alternatively, maybe the game can go on indefinitely without anyone winning, but the problem states that a player wins when the number becomes a perfect square, so eventually, someone might be forced to make it a perfect square.But I'm not sure.After thinking through all this, I think that the second player, Bob, can always respond in a way that either wins immediately if Alice adds a 5 or 6, or blocks Alice's attempts and eventually forces a win. Therefore, Bob can guarantee a win.

💡First, I recognize that the expression ( a^2 - 2ab + b^2 ) resembles the formula for the square of a binomial difference, which is ( (a - b)^2 ).Given the equation ( a = b + 2 ), I can determine the value of ( a - b ) by subtracting ( b ) from both sides:[a - b = 2]Now, substituting ( a - b = 2 ) into the expression ( (a - b)^2 ):[(a - b)^2 = 2^2 = 4]Therefore, the value of the algebraic expression ( a^2 - 2ab + b^2 ) is 4.

💡First, I recognize that the problem involves an arithmetic sequence starting at 5, with each subsequent number increasing by 4, and ending at 2021.To determine the number of terms in this sequence, I can use the formula for the nth term of an arithmetic sequence:[ a_n = a_1 + (n - 1) times d ]where ( a_n ) is the nth term, ( a_1 ) is the first term, ( d ) is the common difference, and ( n ) is the number of terms.Plugging in the known values:[ 2021 = 5 + (n - 1) times 4 ]Next, I'll solve for ( n ):[ 2021 - 5 = (n - 1) times 4 ][ 2016 = (n - 1) times 4 ][ frac{2016}{4} = n - 1 ][ 504 = n - 1 ][ n = 505 ]Therefore, there are 505 numbers in the list.

💡To find the greatest common factor (GCF) of 36, 54, and 81, I will start by breaking down each number into its prime factors.First, I'll factorize 36:[ 36 = 2^2 times 3^2 ]Next, I'll factorize 54:[ 54 = 2 times 3^3 ]Then, I'll factorize 81:[ 81 = 3^4 ]Now, I'll identify the common prime factors among all three numbers. The only common prime factor is 3.Finally, I'll determine the lowest power of this common prime factor, which is ( 3^2 ).Therefore, the greatest common factor of 36, 54, and 81 is:[ 3^2 = 9 ]

💡First, I need to find where the linear function ( y = kx + 2 ) intersects the x-axis. At the x-axis, the value of ( y ) is 0. So, I'll set ( y = 0 ) in the equation and solve for ( x ).Setting ( y = 0 ):[0 = kx + 2]Solving for ( x ):[kx = -2 x = -frac{2}{k}]Next, the problem states that this intersection point is 2 units away from the origin. The distance from the origin to the point ( x ) on the x-axis is the absolute value of ( x ). Therefore:[left| -frac{2}{k} right| = 2]Simplifying:[frac{2}{|k|} = 2 |k| = 1]This means that ( k ) can be either 1 or -1. Therefore, the possible values for ( k ) are:[k = pm 1]