answer：First, I recognize that the quadratic expression (4x^2 - 20x - 32) needs to be rewritten in the form ((px + q)^2 + r). Expanding ((px + q)^2 + r) gives (p^2x^2 + 2pqx + q^2 + r).To match the coefficients of (x^2), I set (p^2 = 4), which means (p = 2) or (p = -2). I'll start with (p = 2).Next, I equate the coefficients of (x): (2pq = -20). Substituting (p = 2) into this equation gives (2 cdot 2 cdot q = -20), leading to (q = -5).Finally, I match the constant terms: (q^2 + r = -32). Plugging in (q = -5) results in (25 + r = -32), so (r = -57).The product (pq) is (2 times (-5) = -10).

answer：First, I need to determine how much Gondor earns from repairing phones and laptops separately.For phone repairs:- On Monday, he repaired 3 phones.- On Tuesday, he repaired 5 phones.- Total phones repaired: 3 + 5 = 8 phones.- Earnings from phones: 8 phones × 10 per phone = 80.For laptop repairs:- On Wednesday, he repaired 2 laptops.- On Thursday, he repaired 4 laptops.- Total laptops repaired: 2 + 4 = 6 laptops.- Earnings from laptops: 6 laptops × 20 per laptop = 120.Finally, I add the earnings from both phone and laptop repairs to find the total earnings:- Total earnings: 80 (from phones) + 120 (from laptops) = 200.

answer：Okay, so I have this problem here that involves the Chinese Remainder Theorem. It says that among the numbers from 2 to 2017, I need to find those that leave a remainder of 1 when divided by both 3 and 5. Then, these numbers form a sequence, and I have to figure out how many terms are in this sequence. Hmm, let me try to break this down step by step.First, let me understand the problem. I need numbers that when divided by 3 give a remainder of 1, and when divided by 5 also give a remainder of 1. So, these numbers satisfy two conditions:1. When divided by 3, the remainder is 1. So, mathematically, this can be written as: [ x equiv 1 pmod{3} ]2. When divided by 5, the remainder is also 1. So, this can be written as: [ x equiv 1 pmod{5} ]Now, I remember that the Chinese Remainder Theorem helps solve systems of congruences like this. It tells us that if the moduli (the numbers we're dividing by, which are 3 and 5 here) are coprime, then there's a unique solution modulo the product of the moduli. Since 3 and 5 are coprime (their greatest common divisor is 1), the theorem applies.So, the solution to the system:[x equiv 1 pmod{3}][x equiv 1 pmod{5}]is a number that is congruent to 1 modulo the least common multiple (LCM) of 3 and 5. Since 3 and 5 are coprime, their LCM is just 3*5=15. Therefore, the solution is:[x equiv 1 pmod{15}]This means that all numbers which are 1 more than a multiple of 15 will satisfy both conditions. So, the numbers we're looking for are of the form:[x = 15k + 1]where ( k ) is a non-negative integer.Now, the problem specifies that we're looking at numbers from 2 to 2017. So, I need to find all numbers ( x ) such that:[2 leq x leq 2017]and[x = 15k + 1]for some integer ( k ).Let me write that inequality out:[2 leq 15k + 1 leq 2017]Subtracting 1 from all parts:[1 leq 15k leq 2016]Then, dividing all parts by 15:[frac{1}{15} leq k leq frac{2016}{15}]Calculating ( frac{2016}{15} ):[2016 ÷ 15 = 134.4]So, ( k ) must satisfy:[0.0667 leq k leq 134.4]But ( k ) has to be an integer, so the smallest integer ( k ) can be is 1 (since 0.0667 is less than 1, but ( k=0 ) would give ( x=1 ), which is less than 2, so we start at ( k=1 )), and the largest integer ( k ) can be is 134 (since 134.4 is just a bit more than 134, so 134 is the maximum integer value).Wait, hold on. Let me check that. If ( k=0 ), then ( x=1 ), which is less than 2, so we can't include that. So the first valid ( k ) is 1, giving ( x=16 ). Then, the last term is when ( k=134 ), giving ( x=15*134 +1 ). Let me compute that:[15*134 = 2010]So, ( x=2010 +1 = 2011 ). Is 2011 less than or equal to 2017? Yes, it is. So, 2011 is included.Wait, but if ( k=134.4 ), which is approximately 134.4, so 134 is the integer part. So, is 134 the maximum ( k )?But let me check ( k=134 ):[x=15*134 +1=2010 +1=2011]Which is within the range. What about ( k=135 )?[x=15*135 +1=2025 +1=2026]But 2026 is greater than 2017, so that's outside our range. So, ( k ) can only go up to 134.So, the values of ( k ) start at 1 and go up to 134. Therefore, the number of terms in the sequence is 134.Wait, hold on again. Let me verify. If ( k=1 ) gives 16, which is within 2 to 2017, and ( k=134 ) gives 2011, which is also within the range. So, the number of terms is 134.But, hold on, let me think again. The formula is ( x=15k +1 ). So, when ( k=0 ), ( x=1 ), which is less than 2, so we don't include that. So, starting from ( k=1 ), which gives 16, up to ( k=134 ), which gives 2011.So, the number of terms is 134.But wait, let me think about how sequences work. If ( k ) starts at 1 and goes to 134, that's 134 terms. But sometimes, when counting inclusive ranges, we have to be careful. For example, if you have numbers from 1 to 10, that's 10 numbers. Similarly, from 1 to n is n numbers.So, in this case, if ( k ) starts at 1 and ends at 134, that's 134 terms. So, the number of terms is 134.But wait, let me double-check by calculating how many numbers satisfy ( x equiv 1 pmod{15} ) between 2 and 2017.The first number in this range is 16, as when ( k=1 ). The last number is 2011, when ( k=134 ). So, the sequence is 16, 31, 46, ..., 2011.To find the number of terms, we can use the formula for the nth term of an arithmetic sequence:[a_n = a_1 + (n-1)d]Where:- ( a_n ) is the nth term,- ( a_1 ) is the first term,- ( d ) is the common difference,- ( n ) is the number of terms.In our case:- ( a_1 = 16 ),- ( d = 15 ),- ( a_n = 2011 ).Plugging into the formula:[2011 = 16 + (n-1)*15]Subtract 16 from both sides:[1995 = (n-1)*15]Divide both sides by 15:[133 = n - 1]Add 1 to both sides:[n = 134]So, that confirms it. There are 134 terms in the sequence.Wait, but hold on. Let me think about whether 1 is included or not. The problem says numbers from 2 to 2017. So, 1 is excluded. So, the first term is 16, which is correct. So, starting at 16, each subsequent term is 15 more than the previous, up to 2011. So, 134 terms.Therefore, the number of terms in the sequence is 134.But just to make sure, let me compute the number of terms another way. The total number of numbers congruent to 1 mod 15 between 1 and 2017 is equal to the floor of (2017 -1)/15, which is floor(2016/15)=134.4, so 134. But since we're starting from 2, we need to check if 1 is included or not. Since 1 is less than 2, we don't include it, so the count remains 134.Alternatively, the formula for the number of terms in an arithmetic progression is:[n = frac{(a_n - a_1)}{d} + 1]Where:- ( a_n ) is the last term,- ( a_1 ) is the first term,- ( d ) is the common difference.Plugging in the values:[n = frac{(2011 - 16)}{15} + 1 = frac{1995}{15} + 1 = 133 + 1 = 134]So, again, we get 134.Therefore, I'm confident that the number of terms in the sequence is 134.**Final Answer**The number of terms in the sequence is boxed{134}.

answer：First, I need to find the measure of angle ( A ) in triangle ( ABC ). Since the sum of the angles in any triangle is ( 180^circ ), I can calculate angle ( A ) by subtracting the known angles ( B ) and ( C ) from ( 180^circ ).Next, I determine which side is the shortest. In a triangle, the side opposite the smallest angle is the shortest side. Here, angle ( B ) is ( 45^circ ), which is smaller than angles ( A ) and ( C ), so side ( b ) is the shortest side.To find the length of side ( b ), I use the Law of Sines, which relates the sides of a triangle to the sines of their opposite angles. The formula is:[frac{b}{sin B} = frac{c}{sin C}]Plugging in the known values:[frac{b}{sin 45^circ} = frac{1}{sin 60^circ}]Solving for ( b ):[b = frac{sin 45^circ}{sin 60^circ} = frac{sqrt{2}/2}{sqrt{3}/2} = frac{sqrt{2}}{sqrt{3}} = frac{sqrt{6}}{3}]Therefore, the length of the shortest side is ( frac{sqrt{6}}{3} ).

answer：Okay, so I have this quadratic function ( g(x) = ax^2 + bx + c ), and I know that ( a ), ( b ), and ( c ) are integers. The problem gives me a few conditions:1. ( g(2) = 0 )2. ( 20 < g(10) < 30 )3. ( 30 < g(11) < 40 )4. ( 7000k < g(200) < 7000(k+1) ) for some integer ( k )I need to find the value of ( k ).Alright, let's start with the first condition: ( g(2) = 0 ). That means when I plug in ( x = 2 ) into the quadratic, it equals zero. So,( g(2) = a(2)^2 + b(2) + c = 4a + 2b + c = 0 )From this equation, I can solve for ( c ):( c = -4a - 2b )Okay, so now I can express ( c ) in terms of ( a ) and ( b ). That might be helpful for the other conditions.Next, let's look at ( g(10) ). The condition is ( 20 < g(10) < 30 ). Let's compute ( g(10) ):( g(10) = a(10)^2 + b(10) + c = 100a + 10b + c )But I already know ( c = -4a - 2b ), so substitute that in:( g(10) = 100a + 10b + (-4a - 2b) = 100a - 4a + 10b - 2b = 96a + 8b )Hmm, that simplifies to ( 96a + 8b ). I can factor out an 8:( g(10) = 8(12a + b) )So, ( 20 < 8(12a + b) < 30 ). Let's divide all parts by 8 to make it easier:( frac{20}{8} < 12a + b < frac{30}{8} )Which simplifies to:( 2.5 < 12a + b < 3.75 )Since ( a ) and ( b ) are integers, ( 12a + b ) must also be an integer. The only integer between 2.5 and 3.75 is 3. So,( 12a + b = 3 )Alright, that's a key equation. Let me write that down:1. ( 12a + b = 3 )Now, moving on to the next condition: ( 30 < g(11) < 40 ). Let's compute ( g(11) ):( g(11) = a(11)^2 + b(11) + c = 121a + 11b + c )Again, substitute ( c = -4a - 2b ):( g(11) = 121a + 11b + (-4a - 2b) = 121a - 4a + 11b - 2b = 117a + 9b )Factor out a 9:( g(11) = 9(13a + b) )So, ( 30 < 9(13a + b) < 40 ). Dividing all parts by 9:( frac{30}{9} < 13a + b < frac{40}{9} )Simplify:( 3.overline{3} < 13a + b < 4.overline{4} )Again, since ( a ) and ( b ) are integers, ( 13a + b ) must be an integer. The only integer between approximately 3.333 and 4.444 is 4. So,( 13a + b = 4 )Now I have two equations:1. ( 12a + b = 3 )2. ( 13a + b = 4 )I can subtract the first equation from the second to eliminate ( b ):( (13a + b) - (12a + b) = 4 - 3 )Simplify:( a = 1 )Okay, so ( a = 1 ). Now plug this back into the first equation to find ( b ):( 12(1) + b = 3 )( 12 + b = 3 )( b = 3 - 12 )( b = -9 )Now that I have ( a = 1 ) and ( b = -9 ), I can find ( c ) using the equation ( c = -4a - 2b ):( c = -4(1) - 2(-9) = -4 + 18 = 14 )Wait, hold on, that can't be right. Let me double-check:( c = -4a - 2b = -4(1) - 2(-9) = -4 + 18 = 14 )Hmm, that seems correct. So ( c = 14 ).Wait, but let me verify if this satisfies the original conditions.First, ( g(2) = 4a + 2b + c = 4(1) + 2(-9) + 14 = 4 - 18 + 14 = 0 ). That's correct.Next, ( g(10) = 100a + 10b + c = 100(1) + 10(-9) + 14 = 100 - 90 + 14 = 24 ). Which is between 20 and 30. Good.Then, ( g(11) = 121a + 11b + c = 121(1) + 11(-9) + 14 = 121 - 99 + 14 = 36 ). Which is between 30 and 40. Perfect.So, the quadratic is ( g(x) = x^2 - 9x + 14 ).Now, the last condition is ( 7000k < g(200) < 7000(k+1) ). Let's compute ( g(200) ):( g(200) = (200)^2 - 9(200) + 14 = 40000 - 1800 + 14 = 40000 - 1800 is 38200, plus 14 is 38214.Wait, hold on, 40000 - 1800 is 38200, plus 14 is 38214. So, ( g(200) = 38214 ).Now, we need to find integer ( k ) such that ( 7000k < 38214 < 7000(k+1) ).Let's divide 38214 by 7000 to find ( k ):( 38214 ÷ 7000 ≈ 5.459 )So, ( k ) must be 5 because 5.459 is between 5 and 6.Let me verify:( 7000 * 5 = 35000 )( 7000 * 6 = 42000 )So, 35000 < 38214 < 42000. Yes, that's correct. Therefore, ( k = 5 ).Wait, but in the initial steps, I found ( c = 14 ), but in the original problem, the user had ( c = 48 ). Did I make a mistake?Wait, let me check my calculations again.From ( c = -4a - 2b ), with ( a = 1 ) and ( b = -9 ):( c = -4(1) - 2(-9) = -4 + 18 = 14 ). That's correct.But in the initial problem statement, the user had ( c = 48 ). Hmm, that might have been a different problem or a miscalculation. But in my case, with ( a = 1 ), ( b = -9 ), ( c = 14 ), all conditions are satisfied.Wait, let me compute ( g(200) ) again:( g(200) = (200)^2 - 9(200) + 14 = 40000 - 1800 + 14 = 40000 - 1800 is 38200, plus 14 is 38214.Yes, that's correct.So, 38214 divided by 7000 is approximately 5.459, so ( k = 5 ).Wait, but in the initial problem, the user had ( c = 48 ), which would lead to a different ( g(200) ). Let me check that.If ( c = 48 ), then ( g(200) = 200^2 + (-9)(200) + 48 = 40000 - 1800 + 48 = 38248 ).Then, 38248 divided by 7000 is approximately 5.464, which is still between 5 and 6, so ( k = 5 ).Wait, so regardless of whether ( c = 14 ) or ( c = 48 ), ( k ) is still 5. But in my calculation, ( c = 14 ) satisfies all conditions, so why did the user have ( c = 48 )?Wait, perhaps I made a mistake in solving for ( c ). Let me go back.From ( c = -4a - 2b ), with ( a = 1 ) and ( b = -9 ):( c = -4(1) - 2(-9) = -4 + 18 = 14 ). That's correct.But in the initial problem, the user had ( c = 48 ). Maybe they had a different ( a ) and ( b ). Wait, let me check their equations.Wait, in the initial problem, the user had:From ( g(2) = 0 ), ( 4a + 2b + c = 0 ), so ( c = -4a - 2b ).Then, for ( g(10) ), they got ( 96a + 8b = 8(12a + b) ), and since ( 20 < g(10) < 30 ), they concluded ( 12a + b = 3 ).Similarly, for ( g(11) ), they got ( 117a + 9b = 9(13a + b) ), and since ( 30 < g(11) < 40 ), they concluded ( 13a + b = 4 ).Then, solving ( 12a + b = 3 ) and ( 13a + b = 4 ), they found ( a = 1 ), ( b = -9 ), and then ( c = -4a - 2b = -4(1) - 2(-9) = -4 + 18 = 14 ).Wait, but in the initial problem, the user had ( c = 48 ). That must have been a mistake. Because with ( a = 1 ), ( b = -9 ), ( c = 14 ), all conditions are satisfied.But in the initial problem, the user had ( c = 48 ), which would imply different ( a ) and ( b ). Let me check that.If ( c = 48 ), then from ( c = -4a - 2b ), we have:( 48 = -4a - 2b )Which can be rewritten as:( -4a - 2b = 48 )Divide both sides by -2:( 2a + b = -24 )But earlier, we had ( 12a + b = 3 ) and ( 13a + b = 4 ). If ( 2a + b = -24 ), then subtracting from ( 12a + b = 3 ):( (12a + b) - (2a + b) = 3 - (-24) )( 10a = 27 )Which would give ( a = 2.7 ), which is not an integer. So, that's a contradiction. Therefore, ( c = 48 ) is not possible with integer ( a ) and ( b ).Therefore, the correct value of ( c ) is 14, not 48. So, the initial problem had a mistake in the value of ( c ).But regardless, in my calculation, ( g(200) = 38214 ), which is between ( 7000*5 = 35000 ) and ( 7000*6 = 42000 ), so ( k = 5 ).Wait, but in the initial problem, the user had ( g(200) = 38248 ), which is also between 35000 and 42000, so ( k = 5 ) in both cases.Therefore, the answer is ( k = 5 ).

answer：Okay, so I have this problem where I need to assign the numbers 1 through 9 to the variables p1, p2, p3, q1, q2, q3, r1, r2, r3. Each number must be used exactly once. Then, I have to calculate P, which is the sum of three products: p1*p2*p3, q1*q2*q3, and r1*r2*r3. My goal is to find the smallest possible value of P.Hmm, let me think about how to approach this. Since I need to minimize the sum of these three products, I should probably try to make each of these products as small as possible. But wait, if I make one product really small, the others might end up being larger because I'm using the remaining numbers. So, it's a balance between keeping all three products relatively small.Maybe I can use some mathematical principle here. I remember that for a given sum, the product is maximized when the numbers are equal, and for a given product, the sum is minimized when the numbers are equal. But in this case, it's a bit different because I have three separate products to sum up. Maybe I can think about distributing the numbers in such a way that the products are as balanced as possible.Let me calculate the total product of all numbers from 1 to 9, which is 9 factorial. 9! = 362880. If I take the cube root of this, it's approximately 71. So, if I could somehow make each of the three products equal to 71, their sum would be around 213. But since 71 isn't an integer and the products have to be integers, I can't actually achieve that. However, this gives me a lower bound. So, the minimal P should be at least 214, right?Wait, let me double-check that. The cube root of 362880 is indeed approximately 71. So, 3 times 71 is 213. Since P has to be an integer, the minimal possible value is 214. But is this achievable? I need to find an arrangement where the sum of the three products is exactly 214.Let me try to construct such an arrangement. I need to split the numbers 1 through 9 into three groups of three numbers each, such that the sum of their products is minimized. To do this, I should try to make the products as close to each other as possible because that tends to give a lower sum.Let me list the numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9.I need to group them into three triplets. Let's start by considering the smallest numbers. If I put 1 in a group, I should pair it with larger numbers to balance the product. For example, 1, 8, 9. Their product is 72. That's a decently large product, but maybe necessary to balance the other groups.Then, for the next group, I can take 2, 6, 7. Their product is 84. Hmm, that's a bit larger than 72. Maybe I can rearrange to make the products closer.Wait, let me try another combination. How about 1, 2, 8? That product is 16. Then, 3, 6, 7 is 126. And 4, 5, 9 is 180. Adding those up: 16 + 126 + 180 = 322. That's way too high. So, that's not a good arrangement.What if I try to make the products more balanced? Let's see. Maybe 2, 5, 7. Their product is 70. Then, 1, 8, 9 is 72. And 3, 4, 6 is 72. Adding those up: 70 + 72 + 72 = 214. Oh, that works! So, this arrangement gives me exactly 214, which matches the lower bound I calculated earlier.Let me verify that I've used all numbers from 1 to 9 without repetition. In the first group: 2, 5, 7. Second group: 1, 8, 9. Third group: 3, 4, 6. Yep, all numbers from 1 to 9 are used exactly once. So, this arrangement is valid.Is there a way to get a sum lower than 214? Let me see. If I try another combination, say 1, 7, 8. Their product is 56. Then, 2, 5, 9 is 90. And 3, 4, 6 is 72. Adding those: 56 + 90 + 72 = 218. That's higher than 214.Another try: 1, 5, 9. Product is 45. Then, 2, 7, 8 is 112. And 3, 4, 6 is 72. Total: 45 + 112 + 72 = 229. Still higher.How about 1, 4, 9. Product is 36. Then, 2, 6, 8 is 96. And 3, 5, 7 is 105. Total: 36 + 96 + 105 = 237. Nope, that's worse.Wait, maybe I can try another combination where the products are even closer. Let's see: 3, 5, 7 is 105. Then, 1, 6, 8 is 48. And 2, 4, 9 is 72. Adding up: 105 + 48 + 72 = 225. Still higher than 214.Alternatively, 4, 5, 6 is 120. Then, 1, 7, 8 is 56. And 2, 3, 9 is 54. Total: 120 + 56 + 54 = 230. No improvement.Hmm, seems like 214 is the minimal I can get. Let me check if there's another arrangement that gives 214. Maybe 1, 7, 8 is 56; 2, 5, 9 is 90; and 3, 4, 6 is 72. Wait, that adds up to 56 + 90 + 72 = 218, which is higher.Alternatively, 1, 3, 8 is 24; 2, 5, 9 is 90; and 4, 6, 7 is 168. That's way too high.Wait, maybe 1, 2, 9 is 18; 3, 5, 8 is 120; and 4, 6, 7 is 168. Total is 18 + 120 + 168 = 306. That's even worse.I think the initial arrangement I found is the best. Let me just recap: 2, 5, 7 gives 70; 1, 8, 9 gives 72; and 3, 4, 6 gives 72. Total is 214. That seems to be the minimal possible value.I wonder if there's a different way to arrange the numbers to get the same total. Let me try another combination. How about 1, 5, 8 is 40; 2, 6, 7 is 84; and 3, 4, 9 is 108. Adding those: 40 + 84 + 108 = 232. Still higher.Another try: 1, 4, 8 is 32; 2, 5, 9 is 90; and 3, 6, 7 is 126. Total: 32 + 90 + 126 = 248. Nope.Wait, maybe 1, 3, 9 is 27; 2, 5, 8 is 80; and 4, 6, 7 is 168. That's 27 + 80 + 168 = 275. Still higher.I think I've tried several combinations, and the only one that gives me 214 is the initial arrangement. So, I'm pretty confident that 214 is indeed the minimal possible value of P.Just to be thorough, let me check another arrangement: 1, 2, 7 is 14; 3, 5, 8 is 120; and 4, 6, 9 is 216. That's way too high.Alternatively, 1, 2, 6 is 12; 3, 5, 9 is 135; and 4, 7, 8 is 224. Total is 12 + 135 + 224 = 371. That's way over.Wait, maybe I can try to make two products small and one medium. For example, 1, 2, 3 is 6; 4, 5, 6 is 120; and 7, 8, 9 is 504. That's 6 + 120 + 504 = 630. That's way too high.Alternatively, 1, 2, 4 is 8; 3, 5, 7 is 105; and 6, 8, 9 is 432. Total: 8 + 105 + 432 = 545. Still way too high.So, it seems that making one product too small forces the others to be much larger, which isn't helpful. Therefore, the balanced approach is better.Let me try another balanced arrangement: 3, 5, 7 is 105; 1, 6, 8 is 48; and 2, 4, 9 is 72. Total: 105 + 48 + 72 = 225. Still higher than 214.Wait, how about 2, 4, 9 is 72; 1, 5, 8 is 40; and 3, 6, 7 is 126. Total: 72 + 40 + 126 = 238. Still higher.Hmm, maybe 2, 3, 9 is 54; 1, 5, 8 is 40; and 4, 6, 7 is 168. Total: 54 + 40 + 168 = 262. Nope.I think I've exhausted most of the possibilities, and the only arrangement that gives me 214 is the one with products 70, 72, and 72. Therefore, I'm confident that 214 is the minimal possible value of P.Just to make sure, let me check if there's a way to rearrange the numbers to get a lower sum. Suppose I try 1, 3, 8 is 24; 2, 5, 9 is 90; and 4, 6, 7 is 168. That's 24 + 90 + 168 = 282. Still higher.Alternatively, 1, 4, 7 is 28; 2, 5, 9 is 90; and 3, 6, 8 is 144. Total: 28 + 90 + 144 = 262. Still higher.Wait, maybe 1, 5, 7 is 35; 2, 4, 9 is 72; and 3, 6, 8 is 144. Total: 35 + 72 + 144 = 251. Still higher than 214.I think I've tried enough combinations. It seems that 214 is indeed the smallest possible value. So, the answer should be 214.