answer：Alright, so I've got this logic puzzle about Francisco's orchard. There are five types of trees: apple, pear, orange, lemon, and tangerine. They're arranged in five parallel rows, each row having only one variety. The question is asking me to figure out which row the tangerine trees are in, based on the given clues. The options are from the 1st to the 5th row.Let me start by listing out the clues to make sure I don't miss anything:(a) The orange trees are next to the lemon trees.(b) The pear trees are not next to the orange trees or the lemon trees.(c) The apple trees are next to the pear trees, but not next to the orange trees or the lemon trees.So, we have five rows, each with a different tree. Let's denote the rows as Row 1, Row 2, Row 3, Row 4, and Row 5. Each row will have one of the five tree types.I think the best way to approach this is to try to map out the possible positions of each tree based on the clues. Since the clues involve adjacency (next to), it's a matter of figuring out which trees can be next to each other and which cannot.First, let's parse clue (a): Orange trees are next to lemon trees. So, wherever the orange trees are, the lemon trees must be either immediately to their left or right. That means orange and lemon must be in adjacent rows.Clue (b): Pear trees are not next to orange or lemon trees. So, pear trees must be separated by at least one row from both orange and lemon trees.Clue (c): Apple trees are next to pear trees, but not next to orange or lemon trees. So, apple must be adjacent to pear, but apple cannot be next to orange or lemon. That also means that apple and pear must be in adjacent rows, and the rows next to apple cannot be orange or lemon.Let me try to visualize this. Maybe I can represent the rows as five slots:1: ?2: ?3: ?4: ?5: ?Each slot will have one of the five trees: apple (A), pear (P), orange (O), lemon (L), tangerine (T).From clue (a), O and L must be adjacent. So, possible pairs are (O, L) or (L, O). They can be in rows 1-2, 2-3, 3-4, or 4-5.From clue (b), P cannot be next to O or L. So, wherever P is, the rows immediately before and after it cannot be O or L.From clue (c), A must be next to P, but A cannot be next to O or L. So, A and P must be adjacent, and the rows next to A (which is already next to P) cannot be O or L.Let me try to figure out possible positions for O and L first since they have to be adjacent.Possible positions for O and L:1. Rows 1 and 22. Rows 2 and 33. Rows 3 and 44. Rows 4 and 5Let me consider each possibility.Case 1: O and L are in Rows 1 and 2.So, Row 1: O, Row 2: L or Row 1: L, Row 2: O.But then, from clue (b), pear cannot be next to O or L. So, pear cannot be in Row 3 because it's next to Row 2 (which is either O or L). Similarly, pear cannot be in Row 5 because it's only adjacent to Row 4, which is not necessarily O or L yet. Wait, no, if O and L are in Rows 1 and 2, then pear can't be in Row 3 because it's next to Row 2 (L or O). So pear can't be in Row 3. Similarly, pear can't be in Row 5 because it's only adjacent to Row 4, which isn't necessarily O or L yet, but pear can't be next to O or L, so if pear is in Row 5, Row 4 can't be O or L. But O and L are already in Rows 1 and 2, so Row 4 can be anything except O or L. So pear could be in Row 5, but let's see.But from clue (c), A must be next to P. So, if P is in Row 5, A must be in Row 4. But then, from clue (c), A cannot be next to O or L. Since A is in Row 4, it's next to Row 3 and Row 5. Row 5 is P, which is fine, but Row 3 is something else. But O and L are in Rows 1 and 2, so Row 3 is not O or L. So that's okay. So in this case, A would be in Row 4, P in Row 5.But let's check if that works. So:Row 1: ORow 2: LRow 3: ?Row 4: ARow 5: PBut then, Row 3 is the only remaining tree, which is T (tangerine). So Row 3: T.But let's check if this satisfies all clues.Clue (a): O and L are adjacent. Yes, Rows 1 and 2.Clue (b): P is in Row 5, which is not next to O or L (since O and L are in Rows 1 and 2). So that's fine.Clue (c): A is next to P? A is in Row 4, P is in Row 5. So yes, they are adjacent. Also, A is not next to O or L. A is in Row 4, next to Row 3 (T) and Row 5 (P). Neither T nor P are O or L, so that's good.So this arrangement seems to work.But let's see if there are other possibilities.Case 2: O and L are in Rows 2 and 3.So, Row 2: O, Row 3: L or Row 2: L, Row 3: O.From clue (b), pear cannot be next to O or L. So pear cannot be in Rows 1, 4, or 5.Wait, if O and L are in Rows 2 and 3, then pear cannot be in Row 1 (because it's next to Row 2, which is O or L), nor in Row 4 (next to Row 3, which is L or O), nor in Row 5 (next to Row 4, but Row 4 is not necessarily O or L yet). Wait, actually, pear can't be next to O or L, so pear can't be in Row 1 (next to Row 2, which is O or L), nor in Row 4 (next to Row 3, which is O or L). So pear can only be in Row 5.But then, from clue (c), A must be next to P. So if P is in Row 5, A must be in Row 4. But A can't be next to O or L. A is in Row 4, which is next to Row 3 (O or L) and Row 5 (P). So A is next to O or L, which violates clue (c). Therefore, this case is impossible.So Case 2 is invalid.Case 3: O and L are in Rows 3 and 4.So, Row 3: O, Row 4: L or Row 3: L, Row 4: O.From clue (b), pear cannot be next to O or L. So pear cannot be in Rows 2, 5, or 4 (if O and L are in 3 and 4). Wait, pear can't be next to O or L, so pear can't be in Row 2 (next to Row 3, which is O or L), Row 4 (next to Row 3, which is O or L), or Row 5 (next to Row 4, which is L or O). So pear can only be in Row 1.From clue (c), A must be next to P. So if P is in Row 1, A must be in Row 2. But A can't be next to O or L. A is in Row 2, which is next to Row 1 (P) and Row 3 (O or L). So A is next to O or L, which violates clue (c). Therefore, this case is impossible.Case 4: O and L are in Rows 4 and 5.So, Row 4: O, Row 5: L or Row 4: L, Row 5: O.From clue (b), pear cannot be next to O or L. So pear cannot be in Rows 3 or 4 (if O and L are in 4 and 5). Wait, pear can't be next to O or L, so pear can't be in Row 3 (next to Row 4, which is O or L) or Row 5 (next to Row 4, which is O or L). So pear can only be in Row 1 or 2.From clue (c), A must be next to P. So if P is in Row 1, A must be in Row 2. If P is in Row 2, A must be in Row 1 or 3. But let's check.If P is in Row 1, then A must be in Row 2. Then, A is in Row 2, which is next to Row 1 (P) and Row 3. Since A can't be next to O or L, Row 3 can't be O or L. But O and L are in Rows 4 and 5, so that's fine. So Row 3 would be T (tangerine). Then, Row 3: T, Row 4: O, Row 5: L.So the arrangement would be:Row 1: PRow 2: ARow 3: TRow 4: ORow 5: LLet's check the clues.Clue (a): O and L are adjacent. Yes, Rows 4 and 5.Clue (b): P is in Row 1, which is not next to O or L (since O and L are in 4 and 5). So that's fine.Clue (c): A is next to P (Row 2 next to Row 1). Also, A is not next to O or L. A is in Row 2, next to Row 1 (P) and Row 3 (T). Neither are O or L, so that's good.Alternatively, if P is in Row 2, then A must be in Row 1 or 3.If A is in Row 1, then Row 1: A, Row 2: P. Then, A is next to P, which is fine. But A can't be next to O or L. A is in Row 1, next to Row 2 (P), which is fine. Then, Row 3 would be T, Row 4: O, Row 5: L.So arrangement:Row 1: ARow 2: PRow 3: TRow 4: ORow 5: LCheck clues:Clue (a): O and L adjacent. Yes.Clue (b): P is in Row 2, not next to O or L. P is next to Row 1 (A) and Row 3 (T). Neither are O or L, so that's fine.Clue (c): A is next to P (Row 1 and 2). A is not next to O or L. A is in Row 1, next to Row 2 (P). Good.Alternatively, if A is in Row 3, then Row 3: A, Row 2: P. But then, A is in Row 3, which is next to Row 2 (P) and Row 4 (O). A can't be next to O, which is in Row 4. So that violates clue (c). Therefore, A can't be in Row 3 if P is in Row 2.So, in Case 4, we have two possible arrangements:1. Row 1: P, Row 2: A, Row 3: T, Row 4: O, Row 5: L2. Row 1: A, Row 2: P, Row 3: T, Row 4: O, Row 5: LBoth seem valid.Wait, but in the first arrangement, Row 1: P, Row 2: A, Row 3: T, Row 4: O, Row 5: L.In this case, tangerine is in Row 3.In the second arrangement, Row 1: A, Row 2: P, Row 3: T, Row 4: O, Row 5: L.Again, tangerine is in Row 3.So in both subcases of Case 4, tangerine is in Row 3.But wait, earlier in Case 1, when O and L were in Rows 1 and 2, tangerine was in Row 3 as well.So in both valid cases, tangerine is in Row 3.Wait, let me confirm.In Case 1:Row 1: O, Row 2: L, Row 3: T, Row 4: A, Row 5: PTangerine is in Row 3.In Case 4:Either Row 1: P, Row 2: A, Row 3: T, Row 4: O, Row 5: LOr Row 1: A, Row 2: P, Row 3: T, Row 4: O, Row 5: LAgain, tangerine in Row 3.So regardless of whether O and L are in Rows 1-2 or 4-5, tangerine ends up in Row 3.Wait, but let me make sure there are no other possibilities.Is there a way for O and L to be in Rows 2-3 or 3-4 without violating the clues?Earlier, I considered Case 2 and Case 3 and found them invalid because they led to contradictions with clue (c). So those cases don't work.Therefore, the only possible arrangements are when O and L are in Rows 1-2 or 4-5, both leading to tangerine being in Row 3.So, the answer should be the 3rd row.But let me just double-check to make sure I didn't miss anything.In Case 1:Rows 1: O, 2: L, 3: T, 4: A, 5: PClues:(a) O and L adjacent: Yes.(b) P is in Row 5, not next to O or L: Correct, since O and L are in 1 and 2.(c) A is in Row 4, next to P in Row 5, and not next to O or L: A is next to T (Row 3) and P (Row 5). Neither are O or L. Good.In Case 4:Rows 1: P, 2: A, 3: T, 4: O, 5: LClues:(a) O and L adjacent: Yes.(b) P is in Row 1, not next to O or L: Correct.(c) A is in Row 2, next to P in Row 1, and not next to O or L: A is next to P (Row 1) and T (Row 3). Neither are O or L. Good.Alternatively, Rows 1: A, 2: P, 3: T, 4: O, 5: LClues:(a) O and L adjacent: Yes.(b) P is in Row 2, not next to O or L: Correct.(c) A is in Row 1, next to P in Row 2, and not next to O or L: A is next to P (Row 2) and nothing else. Good.So in all valid arrangements, tangerine is in Row 3.Therefore, the answer is (c) 3rd row.

answer：Okay, so I have this triangle ABC. The sides opposite angles A, B, and C are a, b, and c respectively. I know that a is 3 and b is 5. Also, it's given that (π/3) - A equals B. I need to find the value of c.First, let me write down what I know:1. a = 3, which is opposite angle A.2. b = 5, which is opposite angle B.3. The relationship between angles A and B: (π/3) - A = B.I think I can use the Law of Sines here because I have two sides and their opposite angles. The Law of Sines states that a/sin(A) = b/sin(B) = c/sin(C). So, maybe I can set up an equation using a and b.Given that a = 3 and b = 5, the ratio would be 3/sin(A) = 5/sin(B). Since I have a relationship between A and B, I can substitute B with (π/3 - A). Let me write that down:3/sin(A) = 5/sin(π/3 - A)Hmm, okay. So I can write this as:3/sin(A) = 5/sin(π/3 - A)I need to solve for angle A. Maybe I can cross-multiply to get rid of the denominators:3 * sin(π/3 - A) = 5 * sin(A)I remember that sin(π/3 - A) can be expanded using the sine subtraction formula. The formula is sin(x - y) = sin(x)cos(y) - cos(x)sin(y). So applying that here:sin(π/3 - A) = sin(π/3)cos(A) - cos(π/3)sin(A)I know that sin(π/3) is √3/2 and cos(π/3) is 1/2. So substituting those values in:sin(π/3 - A) = (√3/2)cos(A) - (1/2)sin(A)So plugging this back into the equation:3 * [(√3/2)cos(A) - (1/2)sin(A)] = 5 * sin(A)Let me distribute the 3:(3√3/2)cos(A) - (3/2)sin(A) = 5 sin(A)Now, I can bring all terms to one side:(3√3/2)cos(A) - (3/2)sin(A) - 5 sin(A) = 0Combine like terms:(3√3/2)cos(A) - (3/2 + 5) sin(A) = 0Wait, 3/2 + 5 is the same as 3/2 + 10/2, which is 13/2. So:(3√3/2)cos(A) - (13/2) sin(A) = 0I can multiply both sides by 2 to eliminate the denominators:3√3 cos(A) - 13 sin(A) = 0Hmm, okay. So:3√3 cos(A) = 13 sin(A)I can divide both sides by cos(A) to get:3√3 = 13 tan(A)So, tan(A) = (3√3)/13Therefore, angle A is arctangent of (3√3)/13. Let me compute that. But maybe I don't need the exact angle, just the sine or cosine of it for further calculations.Alternatively, I can express sin(A) and cos(A) in terms of tan(A). Since tan(A) = opposite/adjacent = (3√3)/13, I can imagine a right triangle where the opposite side is 3√3 and the adjacent side is 13. Then, the hypotenuse would be sqrt[(3√3)^2 + 13^2].Calculating that:(3√3)^2 = 9*3 = 2713^2 = 169So hypotenuse squared is 27 + 169 = 196, so hypotenuse is 14.Therefore, sin(A) = opposite/hypotenuse = (3√3)/14And cos(A) = adjacent/hypotenuse = 13/14Okay, so now I have sin(A) and cos(A). Let me note that down:sin(A) = (3√3)/14cos(A) = 13/14Now, since I know angle A, I can find angle B because B = π/3 - A. So, angle B is π/3 - A.But maybe I can find sin(B) directly since I know angle B is related to angle A. Alternatively, since I have sin(A) and cos(A), maybe I can find sin(B) using the sine of (π/3 - A).Wait, but earlier I used the sine subtraction formula, so maybe I can use that again.Alternatively, since I know angle B is π/3 - A, and I know sin(A) and cos(A), I can compute sin(B) as sin(π/3 - A) which is equal to sin(π/3)cos(A) - cos(π/3)sin(A). I already calculated this earlier when setting up the equation, but maybe now I can compute its exact value.So, sin(B) = sin(π/3 - A) = (√3/2)(13/14) - (1/2)(3√3/14)Let me compute that:First term: (√3/2)(13/14) = (13√3)/28Second term: (1/2)(3√3/14) = (3√3)/28So, sin(B) = (13√3)/28 - (3√3)/28 = (10√3)/28 = (5√3)/14So, sin(B) is 5√3/14.Wait, but earlier, from the Law of Sines, we had 3/sin(A) = 5/sin(B). Let me check if this holds.Compute 3/sin(A) = 3 / (3√3/14) = 3 * (14)/(3√3) = 14/√3 ≈ 8.0829Compute 5/sin(B) = 5 / (5√3/14) = 5 * (14)/(5√3) = 14/√3 ≈ 8.0829Yes, so that checks out. So, that's consistent.Now, I need to find side c. Since I have sides a and b, and I can find angle C, perhaps I can use the Law of Sines again.But first, let me find angle C. Since the sum of angles in a triangle is π, angle C = π - A - B.But since B = π/3 - A, angle C = π - A - (π/3 - A) = π - A - π/3 + A = π - π/3 = 2π/3.Wait, that's interesting. So angle C is 2π/3, which is 120 degrees.So, angle C is 120 degrees, and side c is opposite angle C.So, using the Law of Sines again:c / sin(C) = a / sin(A)So, c = a * sin(C) / sin(A)We know a is 3, sin(C) is sin(2π/3) which is sin(120°) = √3/2, and sin(A) is (3√3)/14.So, plugging in:c = 3 * (√3/2) / (3√3/14)Let me compute that step by step.First, compute the numerator: 3 * (√3/2) = (3√3)/2Then, divide by (3√3)/14:(3√3)/2 ÷ (3√3)/14 = (3√3)/2 * 14/(3√3)Simplify:The 3√3 in numerator and denominator cancel out, so we have 14/2 = 7.So, c = 7.Wait, that seems straightforward. Let me verify.Alternatively, I can compute c using the Law of Cosines since I know angle C is 120°, and sides a and b are 3 and 5.Law of Cosines states that c² = a² + b² - 2ab cos(C)So, plugging in:c² = 3² + 5² - 2*3*5*cos(120°)Compute each term:3² = 95² = 252*3*5 = 30cos(120°) is equal to cos(2π/3) which is -1/2.So, substituting:c² = 9 + 25 - 30*(-1/2) = 34 + 15 = 49Therefore, c = sqrt(49) = 7.Yes, that confirms it. So, c is 7.So, both methods give me c = 7, which is reassuring.Therefore, the value of c is 7.**Final Answer**The value of ( c ) is boxed{7}.

answer：Alright, so I have this problem: I need to find the smallest positive multiple of 23 that is 4 more than a multiple of 89. Hmm, okay. Let me try to break this down.First, let me translate the problem into an equation. If I'm looking for a multiple of 23, let's say that multiple is 23k, where k is a positive integer. The problem states that this multiple is 4 more than a multiple of 89. So, that means 23k = 89m + 4, where m is also a positive integer. So, in equation form:23k ≡ 4 mod 89I think this is a modular equation. So, I need to solve for k in this congruence. Once I find k, I can just multiply by 23 to get the multiple.So, to solve 23k ≡ 4 mod 89, I need to find the multiplicative inverse of 23 modulo 89. Because if I can find a number, let's call it x, such that 23x ≡ 1 mod 89, then I can multiply both sides of the equation by x to get k ≡ 4x mod 89. That should give me the smallest positive k.Okay, so how do I find the inverse of 23 modulo 89? I remember that the inverse exists because 23 and 89 are coprime. Since 89 is a prime number, and 23 is less than 89 and doesn't divide 89, their greatest common divisor is 1. So, an inverse must exist.To find the inverse, I can use the Extended Euclidean Algorithm. Let me recall how that works. The algorithm helps find integers x and y such that:23x + 89y = 1The x here will be the inverse of 23 modulo 89.So, let's perform the Euclidean Algorithm on 89 and 23.First, divide 89 by 23:89 ÷ 23 = 3 with a remainder of 89 - 23*3 = 89 - 69 = 20.So, 89 = 23*3 + 20.Now, take 23 and divide by the remainder 20:23 ÷ 20 = 1 with a remainder of 23 - 20*1 = 3.So, 23 = 20*1 + 3.Next, take 20 and divide by the remainder 3:20 ÷ 3 = 6 with a remainder of 20 - 3*6 = 20 - 18 = 2.So, 20 = 3*6 + 2.Now, take 3 and divide by the remainder 2:3 ÷ 2 = 1 with a remainder of 3 - 2*1 = 1.So, 3 = 2*1 + 1.Next, take 2 and divide by the remainder 1:2 ÷ 1 = 2 with a remainder of 0.So, 2 = 1*2 + 0.Since the remainder is now 0, the last non-zero remainder is 1, which is the GCD, as expected.Now, to find the coefficients x and y, we'll work backwards using the Extended Euclidean Algorithm.Starting from the last non-zero remainder:1 = 3 - 2*1But 2 is from the previous step: 2 = 20 - 3*6So, substitute that into the equation:1 = 3 - (20 - 3*6)*1Simplify:1 = 3 - 20*1 + 3*6Combine like terms:1 = 3*(1 + 6) - 20*11 = 3*7 - 20*1But 3 is from an earlier step: 3 = 23 - 20*1Substitute that in:1 = (23 - 20*1)*7 - 20*1Expand:1 = 23*7 - 20*7 - 20*1Combine like terms:1 = 23*7 - 20*(7 + 1)1 = 23*7 - 20*8But 20 is from the first step: 20 = 89 - 23*3Substitute that in:1 = 23*7 - (89 - 23*3)*8Expand:1 = 23*7 - 89*8 + 23*24Combine like terms:1 = 23*(7 + 24) - 89*81 = 23*31 - 89*8So, this gives us the equation:23*31 - 89*8 = 1Which means:23*31 ≡ 1 mod 89Therefore, the multiplicative inverse of 23 modulo 89 is 31.So, going back to our original congruence:23k ≡ 4 mod 89Multiply both sides by 31:k ≡ 4*31 mod 89Calculate 4*31:4*31 = 124Now, 124 mod 89: 89*1 = 89, so 124 - 89 = 35So, 124 ≡ 35 mod 89Therefore, k ≡ 35 mod 89Since we're looking for the smallest positive k, k = 35.Therefore, the smallest positive multiple of 23 is 23*35.Let me compute that:23*35: 20*35 = 700, 3*35=105, so 700 + 105 = 805So, 23*35 = 805Let me verify that 805 is indeed 4 more than a multiple of 89.Compute 805 - 4 = 801Check if 801 is a multiple of 89.Divide 801 by 89:89*9 = 801, because 89*10=890, which is too big, so 89*9=801Yes, that's correct.So, 805 is 4 more than 801, which is 89*9.Therefore, 805 is the smallest positive multiple of 23 that is 4 more than a multiple of 89.Wait, let me just make sure I didn't make any calculation errors.First, when I found the inverse, I had:1 = 23*31 - 89*8So, 23*31 ≡ 1 mod 89, that's correct.Then, multiplying both sides by 4:k ≡ 4*31 mod 894*31 is 124, which is 35 mod 89, since 124 - 89 = 35.So, k=35.23*35: Let's compute 23*35 again.23*30=690, 23*5=115, so 690+115=805. Correct.805 -4=801, and 801 ÷89=9, since 89*9=801.Yes, that's correct.So, all steps seem to check out.Therefore, the answer is 805.**Final Answer**The smallest positive multiple of 23 that is 4 more than a multiple of 89 is boxed{805}.

answer：Alright, so I have this problem here: For how many three-digit numbers can you subtract 297 and obtain a second three-digit number which is the original three-digit number reversed? The options are A) 5, B) 10, C) 20, D) 40, E) 60.Hmm, okay. Let me try to break this down step by step. I need to find all three-digit numbers such that when I subtract 297 from them, I get their reverse. So, if the original number is ABC (where A, B, C are digits), then subtracting 297 should give me CBA. First, let me represent the original number as a three-digit number. Let's denote the original number as N = 100*A + 10*B + C, where A, B, C are digits from 0 to 9, and A ≠ 0 because it's a three-digit number. Similarly, the reversed number would be M = 100*C + 10*B + A.According to the problem, N - 297 = M. So, substituting the expressions for N and M, we get:100*A + 10*B + C - 297 = 100*C + 10*B + ALet me simplify this equation. Let's subtract 100*C + 10*B + A from both sides to bring all terms to one side:100*A + 10*B + C - 297 - 100*C - 10*B - A = 0Simplify the terms:(100*A - A) + (10*B - 10*B) + (C - 100*C) - 297 = 0Which simplifies to:99*A - 99*C - 297 = 0Hmm, okay. So, 99*(A - C) - 297 = 0. Let me factor out 99:99*(A - C - 3) = 0Wait, no, 99*(A - C) = 297So, 99*(A - C) = 297Divide both sides by 99:A - C = 3So, A - C = 3. That means the hundreds digit minus the units digit is 3. So, A = C + 3.Alright, so now we know that the hundreds digit is 3 more than the units digit. So, let's think about the possible values for C and A.Since A and C are digits, A must be between 1 and 9 (since it's the hundreds digit and can't be zero), and C must be between 0 and 9. But since A = C + 3, C can be at most 6 because if C were 7, then A would be 10, which isn't a digit. So, C can be from 0 to 6, which would make A from 3 to 9.So, C can be 0,1,2,3,4,5,6, and correspondingly, A would be 3,4,5,6,7,8,9.Now, what about the middle digit B? In the equation above, the B terms canceled out. So, B can be any digit from 0 to 9, right? Because in the equation, 10*B - 10*B = 0, so B doesn't affect the equation. So, B can be 0 through 9.Therefore, for each possible value of C (from 0 to 6), A is determined as C + 3, and B can be any digit from 0 to 9. So, the number of such three-digit numbers would be the number of possible Cs multiplied by the number of possible Bs.Number of Cs: 7 (0 through 6)Number of Bs: 10 (0 through 9)So, total numbers = 7 * 10 = 70.Wait, but hold on. The options given are 5,10,20,40,60. 70 isn't an option. Hmm, so I must have made a mistake somewhere.Let me double-check my reasoning.Starting from N - 297 = M, where N is ABC and M is CBA.So, N = 100A + 10B + CM = 100C + 10B + ASo, N - M = 297Which gives:100A + 10B + C - (100C + 10B + A) = 297Simplify:100A + 10B + C - 100C - 10B - A = 297Which becomes:99A - 99C = 297Divide both sides by 99:A - C = 3So, that's correct. So, A = C + 3So, C can be 0-6, A can be 3-9. So, 7 possibilities for C and A.But wait, the problem says that subtracting 297 gives a three-digit number. So, N - 297 must also be a three-digit number. So, N must be greater than or equal to 297 + 100 = 397, because the smallest three-digit number is 100.Wait, no. Wait, N is a three-digit number, so N is between 100 and 999. When we subtract 297, the result must also be a three-digit number. So, N - 297 >= 100.Therefore, N >= 100 + 297 = 397.So, N must be at least 397. So, N is between 397 and 999.But N is a three-digit number where A = C + 3. So, let's see. Since N >= 397, A must be at least 4 because 397 has A=3, but 397 - 297 = 100, which is still a three-digit number. Wait, 397 - 297 = 100, which is a three-digit number, but 397 is a three-digit number. So, actually, N can be as low as 397, but N must be such that N - 297 is also a three-digit number, which is CBA.But wait, CBA is the reverse of ABC. So, if N is 397, then M would be 793. But 397 - 297 = 100, which is not 793. So, that doesn't hold. So, my mistake earlier was assuming that N >= 397, but actually, N - 297 must equal the reverse of N, which is a three-digit number. So, N - 297 must be a three-digit number, so N must be at least 297 + 100 = 397, as I thought.But also, since N is a three-digit number, N <= 999, so N - 297 <= 999 - 297 = 702. So, M = N - 297 must be a three-digit number between 100 and 702.But M is the reverse of N, so M = reverse(N). So, M is also a three-digit number, so M must be between 100 and 999. But since N - 297 = M, and N <= 999, M <= 702. So, M is between 100 and 702.But M is the reverse of N, so let's think about the constraints on M.If M is between 100 and 702, then the hundreds digit of M (which is the units digit of N) must be between 1 and 7, because M is a three-digit number. So, the units digit of N, which is C, must be between 1 and 7? Wait, no.Wait, M is reverse(N). So, if N is ABC, then M is CBA. So, M is a three-digit number, so C cannot be zero because M would start with C, which is the units digit of N. So, C cannot be zero because then M would start with zero, which is not a three-digit number. Therefore, C must be at least 1.So, C is from 1 to 6, because A = C + 3, and A must be at most 9. So, C can be 1,2,3,4,5,6. So, 6 possibilities for C, not 7.Wait, earlier I thought C can be 0-6, but since M must be a three-digit number, C cannot be zero. Therefore, C is from 1 to 6, which gives A from 4 to 9.Therefore, number of Cs is 6, and Bs are 10, so total numbers would be 6*10=60.But wait, 60 is an option, option E. But let me check again.Wait, if C is from 1 to 6, then A is from 4 to 9, which is correct because A = C + 3. So, C=1, A=4; C=2, A=5; up to C=6, A=9.So, 6 choices for C, 10 choices for B, so 60 numbers. So, is the answer 60? But wait, let me check if all these numbers satisfy N - 297 = reverse(N).Wait, let's take an example. Let me pick C=1, B=0, A=4. So, N=401. Reverse(N)=104. 401 - 297 = 104. Yes, that works.Another example: C=6, B=9, A=9. N=996. Reverse(N)=699. 996 - 297=700-1=699. Yes, that works.Wait, but what about C=1, B=9, A=4. N=491. Reverse(N)=194. 491 - 297=194. Correct.Wait, so all these numbers seem to satisfy the condition.But hold on, earlier I thought N must be at least 397, but with C=1, A=4, N=401, which is above 397. So, all these numbers are above 397, so N - 297 is at least 104, which is a three-digit number.Wait, but what about C=1, B=0, A=4: N=401, which is 401 - 297=104, which is okay.But wait, if C=0, then A=3, but then M would be 0B3, which is not a three-digit number. So, C cannot be zero, so we have 6 choices for C.Therefore, total numbers are 6*10=60. So, the answer is E) 60.But wait, hold on, the options are A)5, B)10, C)20, D)40, E)60.So, 60 is an option, E. But let me think again.Wait, is there any other constraint that I might have missed? For example, does the subtraction N - 297 necessarily result in a three-digit number? Well, we already considered that N must be at least 397, so N - 297 is at least 100, which is a three-digit number. So, that's covered.But let me think about the digits. Since N is a three-digit number, A is from 1-9, C is from 1-6, and B is from 0-9.Wait, but in the equation, we have A - C = 3. So, A is determined by C. So, for each C from 1-6, A is fixed as C + 3, and B is free to vary from 0-9. So, 6*10=60 numbers.Therefore, the answer should be 60, which is option E.Wait, but the initial options didn't have 70, which I thought earlier, but then I realized that C can't be zero, so it's 60.But let me check again with another example. Let's take C=2, A=5, B=5. So, N=552. Reverse(N)=255. 552 - 297=255. Correct.Another one: C=3, A=6, B=7. N=673. Reverse(N)=376. 673 - 297=376. Correct.Wait, so all these seem to work. So, 60 numbers is correct.But hold on, the problem is from a competition, and the options are A)5, B)10, C)20, D)40, E)60.So, 60 is an option, so I think that's the answer.Wait, but wait a second. Let me think about the range of N.N is a three-digit number, so N >= 100. But N - 297 must also be a three-digit number, so N >= 397.But N is also equal to 100*A + 10*B + C, and A = C + 3.So, A is at least 4 (since C is at least 1), so N is at least 400 + 0 + 1 = 401, which is above 397, so all such N satisfy N >= 397.Therefore, all 60 numbers satisfy the condition.But wait, let me think about another angle. Maybe I'm overcounting because when you subtract 297, the reversed number might have some constraints on the digits.Wait, for example, when you subtract 297, you have to make sure that the subtraction doesn't cause any borrowing issues or something.Wait, but since N is a three-digit number, and M is also a three-digit number, and N - M = 297, which is fixed, so as long as A - C = 3, the subtraction works out.But let me think about the subtraction digit by digit.N is ABC, M is CBA.So, subtracting M from N: A B C- C B A= 2 9 7So, let's do this subtraction column by column, starting from the units place.Units place: C - A = 7. But wait, C is less than A because A = C + 3. So, C - A would be negative, so we need to borrow from the tens place.So, units place: (C + 10) - A = 7.But A = C + 3, so substituting:(C + 10) - (C + 3) = 7Simplify:C + 10 - C - 3 = 77 = 7So, that works out. So, the units digit subtraction is okay, and we have to borrow 1 from the tens place.Now, moving to the tens place. Since we borrowed 1, the tens digit becomes B - 1. So, we have:(B - 1) - B = 9Wait, that can't be right. Wait, let's do it step by step.Wait, in the tens place, after borrowing, the digit is (B - 1). Then, subtracting the tens digit of M, which is B.So, (B - 1) - B = -1. But in subtraction, we can't have negative digits, so we need to borrow from the hundreds place.So, we borrow 1 from the hundreds place, making the tens digit (B - 1 + 10) - B = 9.Which is exactly what we have: the tens digit in the result is 9.So, that works out.Now, moving to the hundreds place. We had borrowed 1 earlier, so the hundreds digit becomes (A - 1). Then, subtracting the hundreds digit of M, which is C.So, (A - 1) - C = 2.But A = C + 3, so substituting:(C + 3 - 1) - C = 2Simplify:(C + 2) - C = 22 = 2Which is correct.So, all the digit subtractions check out. Therefore, as long as A = C + 3, the subtraction works out, and the result is 297.Therefore, my initial conclusion that there are 60 such numbers is correct.But wait, the options don't have 60 as the first thought, but it's option E. So, I think that's the answer.Wait, but let me think again. Maybe I'm missing something. For example, if C=6, A=9, then N=9B6. Reverse(N)=6B9. 9B6 - 297=6B9.Let me compute 9B6 - 297.9B6 - 297 = (900 + 10B + 6) - 297 = 900 + 10B + 6 - 297 = (900 - 297) + 10B + 6 = 603 + 10B + 6 = 609 + 10B.But reverse(N)=6B9, which is 600 + 10B + 9.So, 609 + 10B = 600 + 10B + 9.Yes, that's equal. So, that works.Similarly, take C=1, A=4, N=4B1. Reverse(N)=1B4.Compute N - 297: 400 + 10B + 1 - 297 = 401 + 10B - 297 = 104 + 10B.Reverse(N)=100 + 10B + 4 = 104 + 10B.So, that's equal. So, that works.Therefore, all such numbers satisfy the condition.So, with C from 1 to 6, and B from 0 to 9, that's 6*10=60 numbers.Therefore, the answer is E) 60.Wait, but the initial options didn't have 60 as the first thought, but it's there as option E. So, I think that's correct.But wait, let me think if there's any case where even though A = C + 3, the subtraction might not result in the reverse number.Wait, for example, take N=401, reverse is 104. 401 - 297=104, which is correct.Another example: N=412, reverse=214. 412 - 297=115, which is not 214. Wait, that's a problem.Wait, hold on. Wait, N=412, reverse is 214. But 412 - 297=115, which is not equal to 214.Wait, that contradicts my earlier conclusion. So, what's wrong here?Wait, because in this case, A=4, C=2, so A - C=2, which is not equal to 3. So, this number shouldn't be counted because A - C must be 3.Wait, but in my earlier reasoning, I considered that A - C=3, so in this case, N=412, A=4, C=2, which is A - C=2, which is not 3, so this number shouldn't be considered.Wait, so maybe my earlier reasoning is correct, but I have to make sure that A - C=3.So, in the case of N=412, A=4, C=2, which is A - C=2≠3, so it's not part of the 60 numbers. So, in my count, only numbers where A - C=3 are considered, so N=412 is not in the list.Wait, but let me take a number where A - C=3. For example, N=431, reverse=134. 431 - 297=134. Correct.Another example: N=521, reverse=125. 521 - 297=224≠125. Wait, that's a problem.Wait, N=521, A=5, C=1, so A - C=4≠3. So, this number shouldn't be considered.Wait, so N=521 is not in the list because A - C=4≠3.Wait, so maybe my earlier reasoning is correct, but I have to ensure that A - C=3.Wait, let me take N=431, which is A=4, C=1, so A - C=3. Reverse is 134. 431 - 297=134. Correct.Another example: N=542, reverse=245. 542 - 297=245. Correct.Wait, N=542, A=5, C=2, so A - C=3. Correct.Wait, N=653, reverse=356. 653 - 297=356. Correct.Wait, N=653, A=6, C=3, A - C=3. Correct.Wait, N=764, reverse=467. 764 - 297=467. Correct.N=764, A=7, C=4, A - C=3.N=875, reverse=578. 875 - 297=578. Correct.N=875, A=8, C=5, A - C=3.N=986, reverse=689. 986 - 297=689. Correct.N=986, A=9, C=6, A - C=3.So, all these examples work.Wait, but earlier, when I took N=412, which is A=4, C=2, A - C=2≠3, so it's not in the list, and indeed, 412 - 297=115≠214.So, my reasoning is correct: only numbers where A - C=3 are considered, and for each such number, the subtraction works.Therefore, the total number is 6*10=60.Wait, but in my initial count, I considered C from 1 to 6, which gives A from 4 to 9, and B from 0 to 9, so 6*10=60 numbers.But wait, let me think about the case when C=6, A=9. N=9B6. Reverse(N)=6B9. 9B6 - 297=6B9.Compute 9B6 - 297:900 + 10B + 6 - 297 = (900 - 297) + 10B + 6 = 603 + 10B + 6 = 609 + 10B.Reverse(N)=600 + 10B + 9 = 609 + 10B.So, equal. So, that works.Similarly, for C=1, A=4, N=4B1. Reverse(N)=1B4.Compute N - 297: 400 + 10B + 1 - 297 = 401 + 10B - 297 = 104 + 10B.Reverse(N)=100 + 10B + 4 = 104 + 10B.Equal. So, that works.So, all these cases work.Therefore, the answer is 60, which is option E.But wait, the problem is from a competition, and the options are A)5, B)10, C)20, D)40, E)60.So, 60 is an option, so I think that's correct.Wait, but I'm a bit confused because in my initial thought, I thought maybe the answer is 10, but now I'm convinced it's 60.Wait, let me think again. Maybe the problem is that when you reverse the number, the hundreds digit becomes the units digit, and the units digit becomes the hundreds digit, but in the subtraction, we have to consider the borrowing.Wait, but I already did the digit-wise subtraction earlier, and it worked out as long as A - C=3.So, I think my reasoning is correct.Therefore, the answer is E) 60.**Final Answer**boxed{E}

answer：Okay, so I have this problem here: I need to find the coefficient of ( x^6 ) in the expansion of ( (x + 1)^6 cdot sum_{i=0}^{6} x^i ). Hmm, let me try to figure this out step by step.First, let me understand what each part of this expression is. The first part is ( (x + 1)^6 ), which I know can be expanded using the binomial theorem. The second part is a finite geometric series: ( sum_{i=0}^{6} x^i ). I remember that the sum of a geometric series can be expressed in a closed form, which might make things easier.Let me write down both parts separately.Starting with ( (x + 1)^6 ). Using the binomial theorem, this expands to:[sum_{k=0}^{6} binom{6}{k} x^k]So, each term is ( binom{6}{k} x^k ) where ( k ) ranges from 0 to 6.Next, the sum ( sum_{i=0}^{6} x^i ) is a geometric series with the first term 1 and common ratio ( x ). The formula for the sum of a geometric series is ( frac{1 - r^{n+1}}{1 - r} ) when ( r neq 1 ). Applying that here, we get:[sum_{i=0}^{6} x^i = frac{1 - x^{7}}{1 - x}]So, the original expression becomes:[(x + 1)^6 cdot frac{1 - x^{7}}{1 - x}]Hmm, that might be useful. Alternatively, I could consider multiplying the two series together directly. Let me think about which approach is better.If I multiply ( (x + 1)^6 ) with ( sum_{i=0}^{6} x^i ), it's equivalent to convolving the coefficients of the two series. So, the coefficient of ( x^6 ) in the product will be the sum of products of coefficients from ( (x + 1)^6 ) and the geometric series whose exponents add up to 6.Let me denote ( A(x) = (x + 1)^6 ) and ( B(x) = sum_{i=0}^{6} x^i ). Then, the product ( A(x) cdot B(x) ) will have coefficients ( c_n = sum_{k=0}^{n} a_k b_{n - k} ), where ( a_k ) is the coefficient of ( x^k ) in ( A(x) ) and ( b_{n - k} ) is the coefficient of ( x^{n - k} ) in ( B(x) ).Since we're interested in ( c_6 ), the coefficient of ( x^6 ), we can write:[c_6 = sum_{k=0}^{6} a_k b_{6 - k}]Where ( a_k = binom{6}{k} ) and ( b_{6 - k} ) is 1 if ( 6 - k ) is between 0 and 6, inclusive, otherwise 0. But since ( k ) ranges from 0 to 6, ( 6 - k ) will also range from 0 to 6, so ( b_{6 - k} = 1 ) for all ( k ) in this case.Therefore, ( c_6 = sum_{k=0}^{6} binom{6}{k} cdot 1 = sum_{k=0}^{6} binom{6}{k} ).Wait, but the sum of the binomial coefficients ( sum_{k=0}^{n} binom{n}{k} ) is equal to ( 2^n ). So, in this case, ( sum_{k=0}^{6} binom{6}{k} = 2^6 = 64 ). So, does that mean the coefficient of ( x^6 ) is 64?Hold on, let me make sure I didn't miss anything. Because ( B(x) ) is ( sum_{i=0}^{6} x^i ), which is a finite series, so when we multiply it with ( (x + 1)^6 ), the product is a polynomial of degree ( 6 + 6 = 12 ). But we're only looking for the coefficient of ( x^6 ). So, in the convolution, each term ( a_k cdot b_{6 - k} ) is non-zero only when ( 6 - k ) is between 0 and 6, which it is for all ( k ) from 0 to 6.Therefore, the coefficient ( c_6 ) is indeed the sum of all the binomial coefficients from ( k = 0 ) to ( k = 6 ), which is 64.But wait, let me think again. If I consider ( A(x) cdot B(x) = (x + 1)^6 cdot frac{1 - x^7}{1 - x} ), maybe I can compute this product in another way.Let me write ( A(x) cdot B(x) = (x + 1)^6 cdot frac{1 - x^7}{1 - x} ). Then, ( frac{1}{1 - x} ) is the generating function for the sequence of all 1's, so ( frac{1}{1 - x} = sum_{n=0}^{infty} x^n ). Therefore, multiplying ( (x + 1)^6 ) by ( frac{1}{1 - x} ) gives the generating function ( (x + 1)^6 cdot sum_{n=0}^{infty} x^n ), which is the same as ( sum_{n=0}^{infty} left( sum_{k=0}^{6} binom{6}{k} right) x^n ). Wait, no, that's not quite right.Actually, when you multiply two generating functions, the coefficients convolve. So, ( (x + 1)^6 cdot frac{1}{1 - x} ) is equal to ( sum_{n=0}^{infty} left( sum_{k=0}^{6} binom{6}{k} right) x^n ) only if the coefficients of ( (x + 1)^6 ) are zero beyond ( x^6 ). But actually, ( (x + 1)^6 ) is a polynomial of degree 6, so when multiplied by ( frac{1}{1 - x} ), it's like convolving a finite sequence with an infinite sequence of 1's. So, the coefficient of ( x^n ) in the product is the sum of the coefficients of ( (x + 1)^6 ) from ( x^0 ) to ( x^n ), but since ( (x + 1)^6 ) only goes up to ( x^6 ), for ( n leq 6 ), the coefficient is just the sum of the first ( n + 1 ) coefficients of ( (x + 1)^6 ). For ( n > 6 ), it's the sum of all 7 coefficients.But in our case, we have ( (x + 1)^6 cdot frac{1 - x^7}{1 - x} ), which is equal to ( (x + 1)^6 cdot sum_{i=0}^{6} x^i ). So, this is equivalent to ( (x + 1)^6 cdot frac{1 - x^7}{1 - x} ). Therefore, the product is ( (x + 1)^6 cdot frac{1 - x^7}{1 - x} ).But perhaps another approach is to note that ( (x + 1)^6 cdot sum_{i=0}^{6} x^i ) is equal to ( frac{(x + 1)^6 (1 - x^7)}{1 - x} ). So, if I can compute this, maybe I can find the coefficient of ( x^6 ) by expanding or simplifying.Alternatively, maybe I can compute the product directly. Let me try that.First, ( (x + 1)^6 ) is ( x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1 ). The sum ( sum_{i=0}^{6} x^i ) is ( 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 ).So, multiplying these two polynomials together, the coefficient of ( x^6 ) will be the sum of the products of coefficients whose exponents add up to 6.So, let's denote ( A(x) = a_0 + a_1 x + a_2 x^2 + dots + a_6 x^6 ), where ( a_k = binom{6}{k} ). And ( B(x) = b_0 + b_1 x + dots + b_6 x^6 ), where each ( b_i = 1 ).Then, the coefficient ( c_6 ) in ( A(x) cdot B(x) ) is ( sum_{k=0}^{6} a_k b_{6 - k} ). Since ( b_{6 - k} = 1 ) for ( k = 0 ) to ( 6 ), this is just ( sum_{k=0}^{6} a_k ), which is the sum of the coefficients of ( A(x) ).As I thought earlier, the sum of the coefficients of ( (x + 1)^6 ) is ( 2^6 = 64 ). Therefore, the coefficient of ( x^6 ) is 64.Wait, but let me verify this by actually performing the multiplication.Let me write out the multiplication step by step.( A(x) = (x + 1)^6 = 1 + 6x + 15x^2 + 20x^3 + 15x^4 + 6x^5 + x^6 )( B(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 )Multiplying these together, the coefficient of ( x^6 ) is obtained by:- ( 1 times x^6 ) (from ( A(x) ) and ( B(x) )): coefficient is 1- ( 6x times x^5 ): coefficient is 6- ( 15x^2 times x^4 ): coefficient is 15- ( 20x^3 times x^3 ): coefficient is 20- ( 15x^4 times x^2 ): coefficient is 15- ( 6x^5 times x ): coefficient is 6- ( x^6 times 1 ): coefficient is 1Adding all these up: 1 + 6 + 15 + 20 + 15 + 6 + 1.Calculating this:1 + 6 = 77 + 15 = 2222 + 20 = 4242 + 15 = 5757 + 6 = 6363 + 1 = 64Yes, so that gives 64. So, the coefficient of ( x^6 ) is indeed 64.Alternatively, thinking about the generating functions, since ( (x + 1)^6 ) is a polynomial and ( sum_{i=0}^{6} x^i ) is another polynomial, their product is a polynomial where each coefficient is the convolution of the coefficients of the two polynomials. So, for ( x^6 ), we're summing over all pairs ( (k, 6 - k) ) where ( k ) ranges from 0 to 6, and each term is ( binom{6}{k} times 1 ), which again sums to 64.Another way to think about it is that ( (x + 1)^6 ) is the generating function for the number of ways to choose subsets of a 6-element set, and ( sum_{i=0}^{6} x^i ) is the generating function for the number of ways to choose a number of items from 0 to 6. When you multiply them, the coefficient of ( x^6 ) represents the number of ways to choose a subset of the 6-element set and also choose a number of items such that the total is 6. But since the subset size can be from 0 to 6, and the number of items is from 0 to 6, the total number of ways is the sum over all possible subset sizes, which is ( 2^6 = 64 ). So, that's another combinatorial interpretation.Wait, actually, hold on. Is that accurate? Let me think again.If I interpret ( (x + 1)^6 ) as the generating function for subsets, where the exponent represents the size of the subset, and ( sum_{i=0}^{6} x^i ) as the generating function for choosing a number of items from 0 to 6, then the product would represent the number of ways to choose a subset and a number such that their sum is 6. But actually, the coefficient of ( x^6 ) would be the number of pairs (subset, number) where the size of the subset plus the number equals 6.But since the number can be from 0 to 6, and the subset size can be from 0 to 6, the total number of such pairs is indeed the sum over all subset sizes ( k ) from 0 to 6 of the number of subsets of size ( k ) times the number of ways to choose the number ( 6 - k ). Since for each ( k ), the number of subsets is ( binom{6}{k} ) and the number of ways to choose the number ( 6 - k ) is 1 (since it's fixed), the total is ( sum_{k=0}^{6} binom{6}{k} = 64 ).So, that's another way to see it combinatorially. Therefore, the coefficient is 64.Alternatively, if I think about the generating function ( (x + 1)^6 cdot frac{1 - x^7}{1 - x} ), and I want the coefficient of ( x^6 ), I can write this as ( (x + 1)^6 cdot left( sum_{n=0}^{infty} x^n - sum_{n=7}^{infty} x^n right) ). So, that's ( (x + 1)^6 cdot sum_{n=0}^{6} x^n ). But since ( (x + 1)^6 ) is a polynomial of degree 6, when multiplied by ( sum_{n=0}^{6} x^n ), the coefficient of ( x^6 ) is the same as the sum of the coefficients of ( (x + 1)^6 ), which is 64.Wait, but actually, if I consider ( (x + 1)^6 cdot frac{1}{1 - x} ), that would be the generating function for the convolution of the coefficients of ( (x + 1)^6 ) with the sequence of all 1's. So, the coefficient of ( x^6 ) in that product would be the sum of the coefficients of ( (x + 1)^6 ) from ( x^0 ) to ( x^6 ), which is 64. However, in our case, we have ( (x + 1)^6 cdot frac{1 - x^7}{1 - x} ), which is ( (x + 1)^6 cdot sum_{n=0}^{6} x^n ). So, the coefficient of ( x^6 ) is the same as in ( (x + 1)^6 cdot frac{1}{1 - x} ), because the ( x^7 ) term and higher would contribute to ( x^6 ) only if we have negative exponents, which we don't. So, effectively, the coefficient of ( x^6 ) is still 64.Therefore, all these different approaches lead me to the conclusion that the coefficient of ( x^6 ) is 64.But just to make sure, let me compute it directly by multiplying the two polynomials.Given ( A(x) = 1 + 6x + 15x^2 + 20x^3 + 15x^4 + 6x^5 + x^6 )and ( B(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 )Multiplying these, the coefficient of ( x^6 ) is:- 1 (from 1 * x^6)- 6x * x^5 = 6x^6- 15x^2 * x^4 = 15x^6- 20x^3 * x^3 = 20x^6- 15x^4 * x^2 = 15x^6- 6x^5 * x = 6x^6- x^6 * 1 = x^6Adding all these coefficients: 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64.Yep, that's consistent. So, I think I can confidently say that the coefficient is 64.**Final Answer**The coefficient of ( x^6 ) is boxed{64}.

answer：First, I need to understand the function of the special key. It replaces the current number ( x ) with ( frac{1}{1 - x} ).Starting with ( x = 5 ), I'll apply the special key repeatedly to observe any patterns or cycles.1. After the first press: ( frac{1}{1 - 5} = -0.25 )2. After the second press: ( frac{1}{1 - (-0.25)} = 0.8 )3. After the third press: ( frac{1}{1 - 0.8} = 5 )I notice that after three presses, the value returns to 5. This indicates a cycle of length 3.To find the result after 100 presses, I'll determine where 100 falls within the cycle. Dividing 100 by 3 gives a remainder of 1 (since ( 100 = 3 times 33 + 1 )).This means after 99 presses, the value will be back to 5, and the 100th press will be the first step of the cycle, resulting in ( -0.25 ).Therefore, after pressing the special key 100 times, the calculator will display (-0.25).