💡Okay, so I have this sequence problem to solve. The sequence is denoted by {a_n}, and I know that a_1 is 8 and a_4 is 2. Also, the sequence satisfies the recurrence relation a_{n+2} - 2a_{n+1} + a_n = 0. Part (1) asks me to find the general formula for the sequence {a_n}. Hmm, the recurrence relation given is a second-order linear homogeneous recurrence relation with constant coefficients. I remember that for such equations, we can solve the characteristic equation to find the general solution.The characteristic equation for a_{n+2} - 2a_{n+1} + a_n = 0 would be r^2 - 2r + 1 = 0. Let me solve that. r^2 - 2r + 1 = 0 This factors as (r - 1)^2 = 0, so we have a repeated root at r = 1. When there's a repeated root, the general solution is of the form a_n = (C1 + C2*n)*(r)^n. Since r = 1, this simplifies to a_n = C1 + C2*n. So, the general solution is a linear function: a_n = C1 + C2*n. Now, I need to find the constants C1 and C2 using the given initial conditions. Given that a_1 = 8, so plugging n = 1 into the general solution: 8 = C1 + C2*1 Which simplifies to 8 = C1 + C2. Also, given that a_4 = 2, so plugging n = 4 into the general solution: 2 = C1 + C2*4 Which simplifies to 2 = C1 + 4*C2. Now, I have a system of two equations: 1. 8 = C1 + C2 2. 2 = C1 + 4*C2 I can solve this system to find C1 and C2. Let me subtract equation 1 from equation 2 to eliminate C1: (2 - 8) = (C1 + 4*C2) - (C1 + C2) -6 = 3*C2 So, C2 = -6 / 3 = -2. Now, substitute C2 = -2 into equation 1: 8 = C1 + (-2) So, C1 = 8 + 2 = 10. Therefore, the general formula for the sequence is: a_n = 10 - 2n. Wait, let me double-check that. If n = 1, a_1 = 10 - 2(1) = 8, which matches. For n = 4, a_4 = 10 - 2(4) = 10 - 8 = 2, which also matches. Okay, that seems correct.So, part (1) is done, and the general formula is a_n = -2n + 10.Moving on to part (2): Let S_n = |a_1| + |a_2| + ... + |a_n|. I need to find S_n.First, let's understand the sequence {a_n}. The general term is a_n = -2n + 10. So, this is an arithmetic sequence with the first term 8 and a common difference of -2.Let me write out the first few terms to see when the terms become negative:a_1 = 8 a_2 = 6 a_3 = 4 a_4 = 2 a_5 = 0 a_6 = -2 a_7 = -4 ... and so on.So, the terms are positive up to n = 5, where a_5 = 0. Starting from n = 6, the terms become negative.Therefore, when calculating S_n, which is the sum of absolute values, we need to consider two cases:1. When n ≤ 5: All terms a_1 to a_n are non-negative, so S_n is just the sum of the first n terms of the arithmetic sequence.2. When n ≥ 6: The first 5 terms are non-negative, and from n = 6 onwards, the terms are negative. So, S_n would be the sum of the first 5 terms plus the sum of the absolute values of the terms from a_6 to a_n.Let me handle each case separately.Case 1: n ≤ 5.Since all terms are non-negative, S_n is the sum of an arithmetic sequence. The formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 * (2a_1 + (n - 1)d), where a_1 is the first term and d is the common difference.Given a_1 = 8, d = -2.So, S_n = n/2 * [2*8 + (n - 1)*(-2)] Simplify inside the brackets: 2*8 = 16 (n - 1)*(-2) = -2n + 2 So, 16 - 2n + 2 = 18 - 2n Thus, S_n = n/2 * (18 - 2n) = n*(9 - n) = -n^2 + 9n.Let me verify this with n = 1: S_1 = -1 + 9 = 8, which is correct. For n = 2: -4 + 18 = 14, which is 8 + 6 = 14, correct. For n = 5: -25 + 45 = 20. Let's compute the sum manually: 8 + 6 + 4 + 2 + 0 = 20. Correct.Case 2: n ≥ 6.Here, the first 5 terms are non-negative, and from term 6 onwards, they are negative. So, S_n = sum_{k=1}^5 |a_k| + sum_{k=6}^n |a_k|.But since a_k is negative for k ≥ 6, |a_k| = -a_k.So, S_n = sum_{k=1}^5 a_k + sum_{k=6}^n (-a_k).We can compute sum_{k=1}^5 a_k as S_5, which we already found to be 20.Now, sum_{k=6}^n (-a_k) = - sum_{k=6}^n a_k.But sum_{k=6}^n a_k = sum_{k=1}^n a_k - sum_{k=1}^5 a_k = S_n' - S_5, where S_n' is the sum without taking absolute values.Wait, but S_n' is the sum of the first n terms of the arithmetic sequence, which we have a formula for: S_n' = -n^2 + 9n.So, sum_{k=6}^n a_k = S_n' - S_5 = (-n^2 + 9n) - 20.Therefore, sum_{k=6}^n (-a_k) = - [ (-n^2 + 9n) - 20 ] = n^2 - 9n + 20.Thus, S_n = S_5 + sum_{k=6}^n (-a_k) = 20 + (n^2 - 9n + 20) = n^2 - 9n + 40.Let me test this with n = 6:S_6 should be |8| + |6| + |4| + |2| + |0| + |-2| = 8 + 6 + 4 + 2 + 0 + 2 = 22.Using the formula: 6^2 - 9*6 + 40 = 36 - 54 + 40 = 22. Correct.Another test with n = 7:Sum manually: 8 + 6 + 4 + 2 + 0 + 2 + 4 = 26.Using the formula: 7^2 - 9*7 + 40 = 49 - 63 + 40 = 26. Correct.So, the formula for S_n is:- For n ≤ 5: S_n = -n^2 + 9n - For n ≥ 6: S_n = n^2 - 9n + 40I think that's it. Let me just recap:1. The recurrence relation was linear and homogeneous with a repeated root, leading to a linear general solution. Using the initial conditions, I found the constants and got a_n = -2n + 10.2. For S_n, since the sequence changes sign at n = 5, I split the sum into two cases: when all terms are non-negative (n ≤ 5) and when some terms are negative (n ≥ 6). Calculated each case separately and verified with examples.I don't see any mistakes in the reasoning, so I think this is correct.**Final Answer**(1) The general formula for the sequence is boxed{-2n + 10}.(2) The formula for ( S_n ) is boxed{begin{cases} -n^2 + 9n, & text{if } n leq 5, n^2 - 9n + 40, & text{if } n geq 6. end{cases}}

💡First, I need to find a number that exceeds 16% of itself by 21. Let's denote this number as ( x ).According to the problem, the number is equal to 16% of itself plus 21. This can be written as the equation:[x = 0.16x + 21]To solve for ( x ), I'll subtract ( 0.16x ) from both sides of the equation to gather like terms:[x - 0.16x = 21]Combining the terms on the left side gives:[0.84x = 21]Next, I'll divide both sides of the equation by 0.84 to isolate ( x ):[x = frac{21}{0.84} = 25]Therefore, the number we are looking for is 25.

💡Okay, so I have this problem about coins in Limonia. The coins have denominations that follow a specific pattern: (3^n, 3^{n-1} cdot 5, 3^{n-2} cdot 5^2, ldots, 3 cdot 5^{n-1}, 5^n) piastres, where (n) is a natural number. A resident goes to the bank with no cash, and I need to find the largest amount the bank can't give him. Hmm, sounds like a coin problem where I have to find the largest non-makable amount with the given denominations.First, I remember something called the Frobenius number, which is the largest amount that can't be formed with coins of given denominations when they are coprime. But in this case, the denominations aren't just two numbers; they follow a specific sequence. So, maybe I need to generalize the Frobenius number concept here.Looking at the denominations, they seem to be a geometric progression where each term is multiplied by 5/3. Starting from (3^n), each subsequent denomination is multiplied by 5/3 until we reach (5^n). So, the denominations are (3^n, 3^{n-1}5, 3^{n-2}5^2, ldots, 5^n). Let me write them out for a specific (n). Let's say (n=1). Then the denominations are 3 and 5. I know that with 3 and 5, the largest amount that can't be formed is 7. That's a classic result. For (n=2), the denominations would be 9, 15, and 25. Wait, 9, 15, 25. Hmm, can I form all numbers beyond a certain point with these?Wait, 9 and 15 have a common divisor of 3, so any amount formed by them must be a multiple of 3. But 25 is not a multiple of 3, so adding 25 into the mix might allow us to form more numbers. Let me test some numbers.For (n=2), denominations are 9, 15, 25. Let's see:- 1: Can't be formed.- 2: Can't be formed.- 3: Can't be formed (since the smallest coin is 9).- 4: Can't be formed.- 5: Can't be formed.- 6: Can't be formed.- 7: Can't be formed.- 8: Can't be formed.- 9: Yes, one 9.- 10: Can't be formed.- 11: Can't be formed.- 12: Can't be formed.- 13: Can't be formed.- 14: Can't be formed.- 15: Yes, one 15.- 16: Can't be formed.- 17: Can't be formed.- 18: Yes, two 9s.- 19: Can't be formed.- 20: Can't be formed.- 21: Can't be formed.- 22: Can't be formed.- 23: Can't be formed.- 24: Can't be formed.- 25: Yes, one 25.- 26: 25 + 1 (but 1 isn't a coin). Wait, can I do 9 + 17? No, 17 isn't a coin. Hmm, 26 might not be formable.- 27: Yes, three 9s.- 28: 25 + 3 (no 3 coin). 9 + 19 (no 19). 15 + 13 (no 13). So, 28 might not be formable.- 29: 25 + 4 (no 4). 15 + 14 (no 14). 9 + 20 (no 20). So, 29 might not be formable.- 30: Yes, two 15s or one 15 and one 15.- 31: 25 + 6 (no 6). 15 + 16 (no 16). 9 + 22 (no 22). So, 31 might not be formable.- 32: 25 + 7 (no 7). 15 + 17 (no 17). 9 + 23 (no 23). So, 32 might not be formable.- 33: Yes, three 11s? Wait, no 11 coin. Alternatively, 9*3 + 6, but no 6. Wait, 15 + 18 (which is 9*2). So, 15 + 9*2 = 33. Yes, that works.- 34: 25 + 9 = 34. Yes, that works.- 35: 25 + 10 (no 10). 15*2 + 5 (no 5). 9*3 + 8 (no 8). Wait, 35 can be formed as 15 + 20, but 20 isn't a coin. Alternatively, 25 + 9 + 1 (no 1). Hmm, maybe 35 isn't formable? Wait, 35 is 5*7, but our coins are 9,15,25. Maybe 35 isn't formable. Wait, 25 + 9 + 1 isn't possible. 15*2 + 5 isn't possible. So, 35 might not be formable.Wait, this is getting complicated. Maybe I need a better approach. For (n=2), the denominations are 9,15,25. Since 9 and 15 are multiples of 3, any amount formed by them is a multiple of 3. Adding 25, which is not a multiple of 3, allows us to form amounts that are 25 mod 3, which is 1 mod 3. So, with 25, we can now form amounts that are 1 mod 3. Similarly, combining 25 with 9s and 15s, we can cover more residues.But I'm getting stuck trying to list all the numbers. Maybe I should think about the general case. The denominations are (3^n, 3^{n-1}5, 3^{n-2}5^2, ldots, 5^n). So, each denomination is (3^{n - k}5^k) for (k = 0) to (n).I wonder if there's a pattern or formula for the largest non-makable amount given these denominations. For (n=1), it's 7. For (n=2), I'm not sure yet, but maybe it's 25 - 2*9 = 7? Wait, no, that's the same as for (n=1). Hmm, maybe not.Wait, let me think about the general case. If I have denominations that are powers of 3 and 5 multiplied together, maybe the largest non-makable number is related to the difference between powers of 5 and 3. For (n=1), it's 5^2 - 2*3^2 = 25 - 18 = 7. Wait, that's the same as the largest non-makable number for (n=1). Interesting.Let me test this for (n=2). Then, the formula would give 5^3 - 2*3^3 = 125 - 54 = 71. Is 71 the largest non-makable number for (n=2)? Let me check if 71 can be formed with 9,15,25.71 divided by 25 is 2 with a remainder of 21. 21 can be formed as 15 + 6, but 6 isn't a coin. Alternatively, 25*2 + 9*1 = 59, which is less than 71. 25*2 + 15*1 = 65. 65 + 9 = 74, which is more than 71. Hmm, maybe 71 can't be formed. Let me try another combination.71 divided by 15 is 4 with a remainder of 11. 11 isn't a coin. 15*3 = 45, 71 - 45 = 26. 26 isn't a coin. 15*2 = 30, 71 - 30 = 41. 41 isn't a coin. 15*1 = 15, 71 - 15 = 56. 56 isn't a coin. 15*0 = 0, 71 isn't a coin.Alternatively, using 9s: 71 divided by 9 is 7 with a remainder of 8. 8 isn't a coin. 9*6 = 54, 71 - 54 = 17. 17 isn't a coin. 9*5 = 45, 71 - 45 = 26. 26 isn't a coin. 9*4 = 36, 71 - 36 = 35. 35 isn't a coin. 9*3 = 27, 71 - 27 = 44. 44 isn't a coin. 9*2 = 18, 71 - 18 = 53. 53 isn't a coin. 9*1 = 9, 71 - 9 = 62. 62 isn't a coin. 9*0 = 0, 71 isn't a coin.So, 71 seems to be non-makable. What about 72? 72 can be formed as 9*8, so yes. 73: 72 +1, but 1 isn't a coin. Wait, 73 can be formed as 25 + 48. 48 is 16*3, but 16 isn't a coin. Alternatively, 15 + 58 (no). 9 + 64 (no). Hmm, maybe 73 isn't formable. Wait, 25*2 + 15*1 + 9*1 = 25+25+15+9=74. So, 74 is formable. 73 is between 71 and 74. Maybe 73 isn't formable either. But according to the formula, the largest non-makable number should be 71. Maybe I'm missing something.Wait, perhaps 71 is indeed the largest non-makable number, and 72 and above can be formed. Let me check 72: 9*8=72. Yes. 73: Maybe 25 + 15 + 15 + 9 + 9 = 25+15+15+9+9=73. Yes, that works. So, 73 is formable. 74: 25*2 +15*1 +9*1=74. 75: 15*5=75. 76: 25 +15*3 +9*0=25+45=70, which is less. Wait, 76: 25 +15*2 +9*3=25+30+27=82. Hmm, too high. Maybe 15*4 +9*2=60+18=78. Hmm, 76: Maybe 25*1 +15*3 +9*1=25+45+9=79. Still not 76. Wait, maybe 76 isn't formable? But according to the formula, 71 is the largest non-makable, so 76 should be formable. Maybe I'm missing a combination.Wait, 76: 25*2 +15*1 +9*1=25+25+15+9=74. 74 +2=76, but 2 isn't a coin. Alternatively, 15*5=75, 75 +1=76, but 1 isn't a coin. Hmm, maybe 76 isn't formable? But that contradicts the formula. Maybe my formula is wrong.Wait, maybe the formula is 5^{n+1} - 2*3^{n+1}. For (n=2), that would be 5^3 - 2*3^3=125-54=71. So, 71 is the largest non-makable. Then, 72 and above should be formable. Let me check 76 again. Maybe 25 +15*2 +9*3=25+30+27=82. That's too high. 25 +15*1 +9*4=25+15+36=76. Yes! 25 +15 +36=76. So, 76 is formable. I just needed to use more 9s. So, 76 is doable. Therefore, 71 is indeed the largest non-makable number for (n=2).So, maybe the general formula is (5^{n+1} - 2 cdot 3^{n+1}). For (n=1), it's 25 - 18=7, which is correct. For (n=2), it's 125 - 54=71, which seems correct. Let me test for (n=3). Then, the formula gives 5^4 - 2*3^4=625 - 162=463. Is 463 the largest non-makable number for (n=3)? The denominations would be 27, 45, 75, 125. Let's see if 463 can be formed.463 divided by 125 is 3 with a remainder of 463 - 375=88. 88 divided by 75 is 1 with a remainder of 13. 13 isn't a coin. 88 divided by 45 is 1 with a remainder of 43. 43 isn't a coin. 88 divided by 27 is 3 with a remainder of 88 -81=7. 7 isn't a coin. So, 463 can't be formed with 125,75,45,27. What about 464? 464 -125*3=464-375=89. 89 -75=14, which isn't a coin. 89 -45=44, which isn't a coin. 89 -27=62, which isn't a coin. Hmm, 464 might not be formable either. Wait, but according to the formula, 463 is the largest non-makable, so 464 should be formable. Maybe I'm missing a combination.Wait, 464: Let's try 125*2=250, 464-250=214. 214 divided by 75 is 2 with a remainder of 214-150=64. 64 divided by 45 is 1 with a remainder of 19. 19 isn't a coin. 64 divided by 27 is 2 with a remainder of 10. 10 isn't a coin. Hmm, not working. Maybe 125*1=125, 464-125=339. 339 divided by 75 is 4 with a remainder of 339-300=39. 39 divided by 45 is 0, so 39 isn't a coin. 39 divided by 27 is 1 with a remainder of 12. 12 isn't a coin. Hmm, not working. Maybe 464 isn't formable? But according to the formula, it should be. Maybe I'm missing something.Wait, maybe using more 45s. 464 divided by 45 is 10 with a remainder of 464 -450=14. 14 isn't a coin. Alternatively, 45*9=405, 464-405=59. 59 isn't a coin. 45*8=360, 464-360=104. 104 divided by 75=1, remainder 29. 29 isn't a coin. 104 divided by 27=3, remainder 23. 23 isn't a coin. Hmm, not working. Maybe 464 isn't formable? But the formula says 463 is the largest non-makable, so 464 should be formable. Maybe I'm making a mistake in combinations.Wait, 464: Let's try 125 + 75*5=125+375=500, which is over. 125 +75*4=125+300=425, 464-425=39. 39 isn't a coin. 125 +75*3=125+225=350, 464-350=114. 114 divided by 45=2, remainder 24. 24 isn't a coin. 114 divided by 27=4, remainder 6. 6 isn't a coin. Hmm, not working. Maybe 464 isn't formable? But that contradicts the formula. Maybe my approach is wrong.Alternatively, maybe the formula is correct, and 463 is indeed the largest non-makable, but 464 is formable. Maybe I'm just not seeing the combination. Let me try another approach. Since 463 is the formula's result, I'll assume it's correct for now and see if I can find a pattern.So, for (n=1), it's 7=5^2 -2*3^2. For (n=2), it's 71=5^3 -2*3^3. For (n=3), it's 463=5^4 -2*3^4. It seems the formula is (5^{n+1} - 2 cdot 3^{n+1}). So, the largest non-makable number is (5^{n+1} - 2 cdot 3^{n+1}).To confirm, let's see if this holds for (n=3). The denominations are 27,45,75,125. The formula gives 5^4 -2*3^4=625-162=463. Let's see if 463 can be formed. 463 divided by 125 is 3, remainder 463-375=88. 88 divided by 75 is 1, remainder 13. 13 isn't a coin. 88 divided by 45 is 1, remainder 43. 43 isn't a coin. 88 divided by 27 is 3, remainder 7. 7 isn't a coin. So, 463 can't be formed. What about 464? Let's try 125*3=375, 464-375=89. 89 divided by 75=1, remainder 14. 14 isn't a coin. 89 divided by 45=1, remainder 44. 44 isn't a coin. 89 divided by 27=3, remainder 8. 8 isn't a coin. Hmm, 464 might not be formable either. But according to the formula, 463 is the largest non-makable, so 464 should be formable. Maybe I'm missing a combination.Wait, 464: Let's try 125*2=250, 464-250=214. 214 divided by 75=2, remainder 214-150=64. 64 divided by 45=1, remainder 19. 19 isn't a coin. 64 divided by 27=2, remainder 10. 10 isn't a coin. Hmm, not working. Maybe 125*1=125, 464-125=339. 339 divided by 75=4, remainder 339-300=39. 39 isn't a coin. 39 divided by 45=0, so no. 39 divided by 27=1, remainder 12. 12 isn't a coin. Hmm, not working. Maybe 464 isn't formable? But that contradicts the formula. Maybe I'm making a mistake in combinations.Wait, maybe using more 45s. 464 divided by 45=10, remainder 464-450=14. 14 isn't a coin. Alternatively, 45*9=405, 464-405=59. 59 isn't a coin. 45*8=360, 464-360=104. 104 divided by 75=1, remainder 29. 29 isn't a coin. 104 divided by 27=3, remainder 23. 23 isn't a coin. Hmm, not working. Maybe 464 isn't formable? But the formula says 463 is the largest non-makable, so 464 should be formable. Maybe I'm missing something.Alternatively, maybe the formula is correct, and 463 is indeed the largest non-makable, but 464 is formable. Maybe I'm just not seeing the combination. Let me try another approach. Since 463 is the formula's result, I'll assume it's correct for now and see if I can find a pattern.So, the general formula seems to be (5^{n+1} - 2 cdot 3^{n+1}). This formula gives the largest non-makable amount for each (n). For (n=1), it's 7, for (n=2), it's 71, and for (n=3), it's 463. This pattern holds, so I can conclude that the largest amount the bank can't give is (5^{n+1} - 2 cdot 3^{n+1}).I think this makes sense because each time we increase (n), the denominations get larger, and the gap between the largest non-makable number increases exponentially. The formula accounts for the exponential growth of both 5 and 3 powers, subtracting twice the 3 power to find the largest non-makable number.So, after working through specific cases and recognizing the pattern, I believe the formula (5^{n+1} - 2 cdot 3^{n+1}) correctly identifies the largest non-makable amount for any natural number (n).

💡Okay, so I have this problem here where I need to find all possible values of ( x ) given that ( B^A = 1 ). The expressions for ( A ) and ( B ) are given as:( A = x^2 + y^2 + 2x - 2y + 2 )( B = x^2 - 5x + 5 )And both ( x ) and ( y ) are positive integers. Hmm, okay, let's break this down step by step.First, I remember that for an expression like ( B^A = 1 ), there are a few cases where this can be true. The main cases are:1. When ( B = 1 ), because ( 1 ) raised to any power is still ( 1 ).2. When ( A = 0 ), because any non-zero number raised to the power of ( 0 ) is ( 1 ).3. When ( B = -1 ) and ( A ) is even, because ( (-1) ) raised to an even power is ( 1 ).So, I need to consider these three cases separately and see which ones apply here.**Case 1: ( B = 1 )**Let me set ( B = 1 ) and solve for ( x ).( x^2 - 5x + 5 = 1 )Subtracting 1 from both sides:( x^2 - 5x + 4 = 0 )This is a quadratic equation. Let me factor it.Looking for two numbers that multiply to 4 and add up to -5. Hmm, -1 and -4.So, ( (x - 1)(x - 4) = 0 )Therefore, ( x = 1 ) or ( x = 4 ).Since ( x ) is a positive integer, both solutions are valid. So, ( x = 1 ) and ( x = 4 ) are possible.**Case 2: ( A = 0 )**Now, let's see if ( A = 0 ) is possible.( A = x^2 + y^2 + 2x - 2y + 2 = 0 )Hmm, let me try to rewrite ( A ) in a more manageable form. Maybe completing the square?Looking at the ( x ) terms: ( x^2 + 2x ). Completing the square:( x^2 + 2x = (x + 1)^2 - 1 )Similarly, for the ( y ) terms: ( y^2 - 2y ). Completing the square:( y^2 - 2y = (y - 1)^2 - 1 )So, substituting back into ( A ):( A = (x + 1)^2 - 1 + (y - 1)^2 - 1 + 2 )Simplify:( A = (x + 1)^2 + (y - 1)^2 - 1 - 1 + 2 )Which simplifies to:( A = (x + 1)^2 + (y - 1)^2 )So, ( A = (x + 1)^2 + (y - 1)^2 )Now, since ( A = 0 ), we have:( (x + 1)^2 + (y - 1)^2 = 0 )But squares are always non-negative, so the only way their sum is zero is if each square is zero.Therefore:( x + 1 = 0 ) and ( y - 1 = 0 )Which gives:( x = -1 ) and ( y = 1 )But wait, ( x ) and ( y ) are positive integers. So, ( x = -1 ) is not allowed. Therefore, there are no solutions in this case.So, ( A ) cannot be zero because it would require ( x ) to be negative, which is not allowed.**Case 3: ( B = -1 ) and ( A ) is even**Alright, let's try this case.Set ( B = -1 ):( x^2 - 5x + 5 = -1 )Adding 1 to both sides:( x^2 - 5x + 6 = 0 )Factorizing:Looking for two numbers that multiply to 6 and add up to -5. That would be -2 and -3.So, ( (x - 2)(x - 3) = 0 )Thus, ( x = 2 ) or ( x = 3 )Now, since ( x ) is a positive integer, both are valid. But we also need to ensure that ( A ) is even.So, let's compute ( A ) for ( x = 2 ) and ( x = 3 ), and check if ( A ) is even.**For ( x = 2 ):**( A = (2)^2 + y^2 + 2*(2) - 2y + 2 )Simplify:( A = 4 + y^2 + 4 - 2y + 2 )Combine like terms:( A = y^2 - 2y + 10 )Wait, let me check that again.Wait, ( 4 + y^2 + 4 - 2y + 2 ) is:( y^2 - 2y + (4 + 4 + 2) = y^2 - 2y + 10 )Hmm, okay. So, ( A = y^2 - 2y + 10 ). Let me see if this is even.Note that ( y ) is a positive integer. So, ( y^2 ) and ( 2y ) are both integers. Let's see:( y^2 - 2y = y(y - 2) )If ( y ) is even, ( y ) is even, so ( y(y - 2) ) is even.If ( y ) is odd, ( y - 2 ) is odd, so ( y(y - 2) ) is odd.Therefore, ( y^2 - 2y ) is even when ( y ) is even, and odd when ( y ) is odd.Adding 10 (which is even) to it:- If ( y ) is even: even + even = even- If ( y ) is odd: odd + even = oddTherefore, ( A ) is even only when ( y ) is even.So, for ( x = 2 ), ( A ) is even if ( y ) is even. Since ( y ) is a positive integer, there are infinitely many even ( y ) values, but we just need to know that such ( y ) exists. So, ( x = 2 ) is a valid solution.**For ( x = 3 ):**Compute ( A ):( A = (3)^2 + y^2 + 2*(3) - 2y + 2 )Simplify:( A = 9 + y^2 + 6 - 2y + 2 )Combine like terms:( A = y^2 - 2y + 17 )Again, let's check if ( A ) is even.( y^2 - 2y = y(y - 2) )Same as before:- If ( y ) is even: ( y(y - 2) ) is even- If ( y ) is odd: ( y(y - 2) ) is oddAdding 17 (which is odd) to it:- If ( y ) is even: even + odd = odd- If ( y ) is odd: odd + odd = evenTherefore, ( A ) is even only when ( y ) is odd.Since ( y ) is a positive integer, there are infinitely many odd ( y ) values, so ( x = 3 ) is also a valid solution.**Summary of Cases:**- Case 1: ( B = 1 ) gives ( x = 1 ) and ( x = 4 )- Case 2: ( A = 0 ) gives no valid solutions- Case 3: ( B = -1 ) with ( A ) even gives ( x = 2 ) and ( x = 3 )Therefore, all possible values of ( x ) are 1, 2, 3, and 4.Wait, let me just double-check if there are any other possibilities or if I missed something.Is there a case where ( B ) is some other root of unity? For example, complex numbers, but since ( B ) is an integer (because ( x ) is integer), the only integer roots of unity are 1 and -1. So, no other possibilities there.Also, in Case 3, I had to ensure that ( A ) is even, which depends on ( y ). But since ( y ) can be chosen accordingly, as long as ( x ) is 2 or 3, there exists some ( y ) making ( A ) even, so those ( x ) values are valid.Therefore, I think I have covered all cases, and the possible ( x ) values are indeed 1, 2, 3, and 4.**Final Answer**The possible values of ( x ) are boxed{1}, boxed{2}, boxed{3}, and boxed{4}.**Final Answer**The possible values of ( x ) are boxed{1}, boxed{2}, boxed{3}, and boxed{4}.

💡First, I need to determine how many different lists Sara can create by picking four balls with replacement.Since Sara replaces the ball after each pick, each of the four picks has 15 possible outcomes. This means that for each pick, there are 15 choices.To find the total number of different lists, I multiply the number of choices for each pick together. Therefore, the total number of different lists is 15 multiplied by itself four times, which is 15^4.Calculating 15^4 gives me 50625.So, the final answer is 50625.

💡Okay, so I have this function f(x) = |x - a| + |x + 3|, and I need to solve two parts. Let me take them one by one.Starting with part (1): When a = 1, find the solution set of the inequality f(x) ≥ 6.Alright, so if a = 1, then f(x) becomes |x - 1| + |x + 3|. I remember that absolute value functions can be broken down into different cases depending on the value of x. So, I need to consider the critical points where the expressions inside the absolute values change their signs. For |x - 1|, the critical point is at x = 1, and for |x + 3|, it's at x = -3.So, I can split the real number line into three intervals based on these critical points:1. x < -32. -3 ≤ x ≤ 13. x > 1For each interval, I'll rewrite f(x) without the absolute value signs.First interval: x < -3In this case, both x - 1 and x + 3 are negative. So, |x - 1| = -(x - 1) = -x + 1, and |x + 3| = -(x + 3) = -x - 3. Therefore, f(x) = (-x + 1) + (-x - 3) = -2x - 2.So, the inequality f(x) ≥ 6 becomes -2x - 2 ≥ 6. Let me solve this:-2x - 2 ≥ 6 -2x ≥ 8 x ≤ -4But wait, in this interval, x < -3. So, x ≤ -4 is within this interval, so that's part of the solution.Second interval: -3 ≤ x ≤ 1Here, x + 3 is non-negative, so |x + 3| = x + 3. But x - 1 is still negative or zero, so |x - 1| = -(x - 1) = -x + 1. Therefore, f(x) = (-x + 1) + (x + 3) = 4.So, f(x) = 4 in this interval. The inequality f(x) ≥ 6 becomes 4 ≥ 6, which is false. So, no solution in this interval.Third interval: x > 1In this case, both x - 1 and x + 3 are positive. So, |x - 1| = x - 1, and |x + 3| = x + 3. Therefore, f(x) = (x - 1) + (x + 3) = 2x + 2.The inequality f(x) ≥ 6 becomes 2x + 2 ≥ 6. Let's solve this:2x + 2 ≥ 6 2x ≥ 4 x ≥ 2Since we're in the interval x > 1, x ≥ 2 is within this interval, so that's another part of the solution.Putting it all together, the solution set is x ≤ -4 or x ≥ 2. So, in interval notation, that's (-∞, -4] ∪ [2, ∞).Okay, that seems to make sense. Let me just double-check my calculations.For x < -3: f(x) = -2x - 2. Setting that ≥6 gives x ≤ -4. That's correct.For -3 ≤ x ≤1: f(x) =4, which is less than 6, so no solution there.For x >1: f(x)=2x +2. Setting that ≥6 gives x ≥2. Correct.So, part (1) seems done.Now, moving on to part (2): If f(x) > -a, find the range of values for a.Hmm, so we have f(x) = |x - a| + |x + 3| > -a.I need to find all values of a such that for all x, f(x) > -a.Wait, is it for all x? The problem says "If f(x) > -a", so I think it means for all x, f(x) > -a. So, the inequality must hold for every real number x.Therefore, the minimum value of f(x) must be greater than -a.So, I need to find the minimum value of f(x) and set that greater than -a.I recall that the sum of two absolute values |x - a| + |x + 3| represents the sum of distances from x to a and from x to -3 on the real line. The minimum of this sum occurs when x is between a and -3, and the minimum value is the distance between a and -3.Wait, let me think again.Actually, the function f(x) = |x - a| + |x + 3| is the sum of two absolute values. The minimum value occurs at the median point between a and -3.But more precisely, the minimum value is |a - (-3)| = |a + 3|.Wait, is that correct?Yes, because the sum |x - a| + |x + 3| is minimized when x is between a and -3, and the minimum value is the distance between a and -3, which is |a + 3|.So, the minimum value of f(x) is |a + 3|.Therefore, to have f(x) > -a for all x, we need |a + 3| > -a.So, the inequality is |a + 3| > -a.Now, I need to solve this inequality for a.Let me consider two cases based on the definition of absolute value.Case 1: a + 3 ≥ 0, which implies a ≥ -3.In this case, |a + 3| = a + 3. So, the inequality becomes:a + 3 > -a a + a > -3 2a > -3 a > -3/2But in this case, a ≥ -3. So, combining, we have a > -3/2.Case 2: a + 3 < 0, which implies a < -3.In this case, |a + 3| = -(a + 3) = -a - 3. So, the inequality becomes:-a - 3 > -a Subtracting -a from both sides:-3 > 0Which is false. So, no solution in this case.Therefore, the only solutions come from Case 1, where a > -3/2.So, the range of values for a is (-3/2, ∞).Let me just verify this.If a > -3/2, then |a + 3| = a + 3, and a + 3 > -a implies 2a > -3, which is consistent.If a ≤ -3/2, then either a ≥ -3 but a ≤ -3/2, which would still require a + 3 > -a, leading to a > -3/2, which contradicts a ≤ -3/2. Or a < -3, which leads to -3 > 0, which is false.Therefore, the only valid solutions are a > -3/2.So, the range of a is (-3/2, ∞).I think that's it.