answer：Okay, so I have this problem where there's a 6x6 grid of blocks, and I need to select 4 blocks such that no two are in the same row, column, or diagonal. The answer choices are A) 900, B) 1350, C) 1620, D) 2025, and E) 3600. Hmm, I need to figure out how to approach this.First, I know that selecting blocks so that no two are in the same row or column is similar to placing queens on a chessboard where none attack each other. But here, it's a 6x6 grid, and I need to place 4 blocks with the additional constraint of not being on the same diagonal. That makes it a bit more complicated.Let me break it down. To ensure no two blocks are in the same row or column, I can think of it as choosing one block from each of four different rows and four different columns. So, first, I need to choose 4 rows out of 6 and 4 columns out of 6. The number of ways to choose 4 rows from 6 is given by the combination formula C(6,4), which is 15. Similarly, the number of ways to choose 4 columns from 6 is also C(6,4) = 15. So, up to this point, I have 15 * 15 = 225 ways to choose the rows and columns.But now, within these selected rows and columns, I need to place the blocks such that no two are on the same diagonal. This is where it gets tricky. If I just randomly place one block in each selected row and column, there's a chance they might end up on the same diagonal. So, I need to count the number of ways to arrange the blocks so that no two are on the same diagonal.This seems similar to the problem of derangements, where none of the elements appear in their original position. In this case, though, it's about diagonals. I remember that for permutations, the number of derangements for n elements is denoted by !n. For n=4, the derangement number is 9. So, does that mean that for each selection of rows and columns, there are 9 valid arrangements where no two blocks are on the same diagonal?If that's the case, then the total number of ways would be the number of ways to choose the rows and columns multiplied by the number of derangements. So, 15 * 15 * 9 = 2025. That seems to match one of the answer choices, which is D) 2025.But wait, let me make sure I'm not missing something. Is the derangement number applicable here? In derangements, we're talking about permutations where no element is in its original position, but here, it's about diagonals. Are these two concepts related?Thinking about it, if I arrange the blocks such that no two are on the same diagonal, it's similar to ensuring that in the permutation matrix, no two ones are on the same diagonal. This is indeed a derangement problem because each block must be placed in a unique row and column without overlapping diagonals. So, yes, the derangement number applies here.Therefore, the calculation seems correct: 15 ways to choose rows, 15 ways to choose columns, and 9 derangements for each combination, giving a total of 2025. So, the answer should be D) 2025.

answer：Okay, so I have this polynomial equation:[ 2010 + 2009x + 2008x^2 + 2007x^3 + cdots + 2x^{2008} + x^{2009} = 0 ]And I need to figure out if there's an integer ( x ) that makes this equation true. Hmm, that's a pretty long polynomial. It goes all the way up to ( x^{2009} ). Let me try to break this down step by step.First, I notice that the coefficients are decreasing from 2010 down to 1, and the exponents of ( x ) are increasing from 0 up to 2009. So, it's like each term is ( (2010 - k)x^k ) where ( k ) goes from 0 to 2009. That's an interesting pattern.I wonder if there's a way to simplify this polynomial or maybe factor it somehow. But factoring a 2009-degree polynomial seems really complicated. Maybe I should consider specific integer values for ( x ) and see if any of them satisfy the equation.Let me start by testing some small integer values for ( x ).**Case 1: ( x = 0 )**If I plug in ( x = 0 ), all the terms except the constant term disappear. So, the equation becomes:[ 2010 = 0 ]That's obviously not true. So, ( x = 0 ) is not a solution.**Case 2: ( x = 1 )**Let's try ( x = 1 ). Then, every term becomes just the coefficient because ( 1 ) raised to any power is still ( 1 ). So, the equation becomes:[ 2010 + 2009 + 2008 + cdots + 2 + 1 = 0 ]This is the sum of the first 2010 positive integers. I remember that the sum of the first ( n ) integers is ( frac{n(n+1)}{2} ). So, plugging in ( n = 2010 ):[ frac{2010 times 2011}{2} = 2010 times 1005.5 = 2021055 ]That's definitely not zero. So, ( x = 1 ) is not a solution either.**Case 3: ( x = -1 )**Okay, let's try ( x = -1 ). This might be more interesting because the signs of the terms will alternate. Let's write out the polynomial with ( x = -1 ):[ 2010 + 2009(-1) + 2008(-1)^2 + 2007(-1)^3 + cdots + 2(-1)^{2008} + (-1)^{2009} ]Simplifying each term:- The constant term is ( 2010 ).- The next term is ( -2009 ).- The next term is ( +2008 ).- The next term is ( -2007 ).- And so on, alternating signs.So, the polynomial becomes:[ 2010 - 2009 + 2008 - 2007 + cdots + 2 - 1 ]Let me group the terms in pairs:[ (2010 - 2009) + (2008 - 2007) + cdots + (2 - 1) ]Each pair simplifies to ( 1 ). How many such pairs are there? Well, starting from 2010 down to 1, that's 2010 terms. Since we're pairing them, the number of pairs is ( frac{2010}{2} = 1005 ).So, the sum becomes:[ 1005 times 1 = 1005 ]Which is definitely not zero. So, ( x = -1 ) is also not a solution.**Case 4: ( x = 2 )**Let me try ( x = 2 ). Plugging in ( x = 2 ):[ 2010 + 2009(2) + 2008(2)^2 + 2007(2)^3 + cdots + 2(2)^{2008} + (2)^{2009} ]This is going to be a huge number. Each term is positive and increasing rapidly because of the powers of 2. So, the sum is definitely positive and way larger than zero. Therefore, ( x = 2 ) is not a solution.**Case 5: ( x = -2 )**Now, let's try ( x = -2 ). This might be more interesting because the signs will alternate, and the terms might cancel each other out to some extent.The polynomial becomes:[ 2010 + 2009(-2) + 2008(-2)^2 + 2007(-2)^3 + cdots + 2(-2)^{2008} + (-2)^{2009} ]Simplifying each term:- The constant term is ( 2010 ).- The next term is ( -4018 ).- The next term is ( +2008 times 4 = +8032 ).- The next term is ( -2007 times 8 = -16056 ).- And so on, with alternating signs and increasing magnitudes.This seems complicated, but maybe I can find a pattern or see if the terms can cancel out. However, given the alternating signs and the rapidly increasing magnitudes, it's hard to see how they would sum to zero. The higher-degree terms will dominate, and since the highest degree term is ( (-2)^{2009} ), which is a very large negative number, the overall sum is likely negative. But let's check a few more terms to see.But wait, calculating all these terms manually is impractical. Maybe there's a smarter way to approach this.**Looking for a General Approach**Instead of testing specific values, maybe I can analyze the polynomial's behavior for different ranges of ( x ).1. **For ( x > 1 ):** - All terms are positive because ( x ) is positive, and the coefficients are positive. - So, the sum is definitely positive, meaning no solution exists for ( x > 1 ).2. **For ( x = 1 ):** - As calculated earlier, the sum is positive.3. **For ( x = 0 ):** - The sum is 2010, which is positive.4. **For ( x = -1 ):** - The sum is 1005, which is positive.5. **For ( x < -1 ):** - Here, ( x ) is negative, so the terms will alternate in sign. - The highest degree term is ( x^{2009} ), which is negative because 2009 is odd. - The term before that is ( 2x^{2008} ), which is positive because 2008 is even. - So, the polynomial will have alternating signs, but the highest degree term will dominate because it's a very large negative number. Let me see if I can bound the polynomial for ( x leq -2 ). Consider the polynomial: [ P(x) = 2010 + 2009x + 2008x^2 + cdots + 2x^{2008} + x^{2009} ] For ( x leq -2 ), let's try to group the terms: [ P(x) = (2010 + 2009x) + x^2(2008 + 2007x) + x^4(2006 + 2005x) + cdots + x^{2008}(2 + x) ] Each group inside the parentheses is of the form ( (a + 1) + a x ) where ( a ) decreases from 2009 to 1. Let's evaluate ( (a + 1) + a x ) for ( x leq -2 ): [ (a + 1) + a x = a(1 + x) + 1 ] Since ( x leq -2 ), ( 1 + x leq -1 ). Therefore: [ a(1 + x) + 1 leq a(-1) + 1 = -a + 1 ] Since ( a geq 1 ), ( -a + 1 leq 0 ). So, each group inside the parentheses is non-positive. Moreover, each term ( x^{2k} ) is positive because it's an even power. So, each group multiplied by ( x^{2k} ) is non-positive. Therefore, the entire polynomial ( P(x) ) is a sum of non-positive terms. The only way the sum could be zero is if each term is zero, but that's not possible because the coefficients are positive and ( x ) is negative. Hence, for ( x leq -2 ), ( P(x) ) is negative. Wait, but earlier when I tried ( x = -1 ), the sum was positive. So, between ( x = -1 ) and ( x = -2 ), the polynomial changes from positive to negative. That suggests there might be a root between ( -2 ) and ( -1 ). But we're only interested in integer solutions, so ( x = -2 ) is the next integer to check. But as I saw earlier, for ( x = -2 ), the polynomial is negative. So, between ( x = -2 ) and ( x = -1 ), the polynomial goes from negative to positive, implying a root in that interval. However, since we're only looking for integer roots, and there are no integers between ( -2 ) and ( -1 ), that doesn't help us. Therefore, there's no integer ( x ) for which ( P(x) = 0 ).**Conclusion**After testing specific integer values and analyzing the behavior of the polynomial for different ranges of ( x ), I conclude that there is no integer ( x ) that satisfies the given equation.

answer：First, I need to simplify the expression ( x^{-1} - x + 2 ).1. **Understand Negative Exponents:** - The term ( x^{-1} ) can be rewritten as ( frac{1}{x} ).2. **Rewrite the Expression:** - Substitute ( x^{-1} ) with ( frac{1}{x} ): [ frac{1}{x} - x + 2 ]3. **Combine Terms Over a Common Denominator:** - To combine the terms, I'll express them with a common denominator of ( x ): [ frac{1}{x} - frac{x^2}{x} + frac{2x}{x} = frac{1 - x^2 + 2x}{x} ]4. **Factor the Numerator:** - The numerator ( 1 - x^2 + 2x ) can be rearranged as ( 1 - (x^2 - 2x) ). - Recognize that ( x^2 - 2x + 1 = (x - 1)^2 ), so: [ 1 - (x^2 - 2x) = 1 - (x - 1)^2 ]5. **Final Simplified Expression:** - Substitute back into the expression: [ frac{1 - (x - 1)^2}{x} ]6. **Conclusion:** - The simplified form of the expression matches option B.

answer：Okay, so I have this geometry problem here. Let me try to understand it step by step. We have an acute isosceles triangle ABC inscribed in a circle. That means triangle ABC is inside a circle, and all its vertices lie on the circumference. Since it's isosceles, two of its sides are equal. The problem says that through points B and C, tangents to the circle are drawn, and these tangents meet at point D. So, point D is outside the circle, and lines DB and DC are tangent to the circle.The problem also states that angles ABC and ACB are each three times angle D. So, angle ABC = angle ACB = 3 * angle D. We need to find the measure of angle BAC in terms of π, specifically as kπ, and determine the value of k.Let me try to visualize this. Triangle ABC is inscribed in a circle, so it's a circumscribed circle or circumcircle. Since ABC is isosceles, either AB = AC, AB = BC, or AC = BC. But since angles ABC and ACB are equal, that tells me that sides AB and AC are equal because in a triangle, equal angles correspond to equal sides. So, AB = AC, making triangle ABC an isosceles triangle with AB = AC.Now, tangents are drawn from points B and C to meet at point D. So, DB and DC are tangent to the circle at points B and C respectively. Point D is the external point from which two tangents are drawn to the circle.I remember that the angle between a tangent and a chord is equal to the angle in the alternate segment. That might be useful here. Also, the angle between two tangents from an external point can be related to the central angle subtended by the arc between the points of contact.Let me denote angle D as θ. So, angle D is the angle at point D between the two tangents DB and DC. According to the problem, angle ABC and angle ACB are each equal to 3θ. Since triangle ABC is isosceles with AB = AC, angles at B and C are equal, which is given as 3θ each.Since the sum of angles in a triangle is π radians, angle BAC can be found as:angle BAC = π - angle ABC - angle ACB= π - 3θ - 3θ= π - 6θSo, angle BAC is π - 6θ.Now, I need to relate angle D (which is θ) with the angles in triangle ABC. Since DB and DC are tangents from point D to the circle, the lengths DB and DC are equal. So, triangle DBC is isosceles with DB = DC.The angle at D is θ, so the base angles at B and C of triangle DBC are equal. Let me denote each of these angles as φ. So, angle DBC = angle DCB = φ.In triangle DBC, the sum of angles is π:θ + φ + φ = πθ + 2φ = π2φ = π - θφ = (π - θ)/2Now, I need to relate φ with the angles in triangle ABC. Let me consider the tangent at B and the chord BC. The angle between tangent DB and chord BC is equal to the angle in the alternate segment. The alternate segment theorem states that the angle between a tangent and a chord is equal to the angle in the opposite segment.So, angle between tangent DB and chord BC is equal to angle BAC. Similarly, angle between tangent DC and chord CB is also equal to angle BAC.Wait, let me clarify. The angle between tangent DB and chord BC is equal to the angle in the alternate segment, which would be angle BAC. Similarly, the angle between tangent DC and chord CB is also equal to angle BAC.But in triangle DBC, angle DBC is φ, which is equal to the angle between tangent DB and chord BC. So, φ = angle BAC.Wait, is that correct? Let me think again.The angle between tangent DB and chord BC is equal to the angle in the alternate segment. The alternate segment for tangent at B is the segment opposite to angle DBC, which would be angle BAC.So, angle between tangent DB and chord BC is equal to angle BAC. Therefore, angle DBC = angle BAC.But angle DBC is φ, and angle BAC is π - 6θ. So, φ = π - 6θ.But earlier, we found that φ = (π - θ)/2.So, setting these equal:(π - θ)/2 = π - 6θLet me solve this equation for θ.Multiply both sides by 2:π - θ = 2π - 12θBring all terms to one side:π - θ - 2π + 12θ = 0-π + 11θ = 011θ = πθ = π/11So, angle D is π/11 radians.Now, recall that angle BAC = π - 6θ.Substituting θ = π/11:angle BAC = π - 6*(π/11)= π - (6π/11)= (11π/11 - 6π/11)= 5π/11Therefore, angle BAC is 5π/11 radians, so k = 5/11.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Triangle ABC is isosceles with AB = AC, so angles at B and C are equal, each 3θ.2. Sum of angles in triangle ABC: angle BAC + 3θ + 3θ = π => angle BAC = π - 6θ.3. Tangents from D to B and C, so DB = DC, triangle DBC is isosceles.4. Angle at D is θ, so angles at B and C in triangle DBC are (π - θ)/2 each.5. By alternate segment theorem, angle between tangent DB and chord BC is equal to angle BAC.6. Therefore, angle DBC = angle BAC => (π - θ)/2 = π - 6θ.7. Solving gives θ = π/11.8. Substituting back, angle BAC = 5π/11, so k = 5/11.Yes, that seems correct. I think I didn't make any mistakes in the reasoning.

answer：To determine the probability of drawing a red ball on the second draw given that the first ball drawn was red, we can use conditional probability.1. **Total number of balls:** 10 (6 red, 4 white)2. **First draw:** A red ball is drawn. This leaves 5 red balls and 4 white balls, totaling 9 balls remaining.3. **Second draw:** The probability of drawing a red ball is the number of remaining red balls divided by the total remaining balls.Therefore, the probability is:P(text{Second red} | text{First red}) = frac{5}{9}

answer：Alright, so I've got this problem here about transferring coins between pockets and figuring out which pocket originally had more crowns. Let me try to wrap my head around it step by step.First, the problem says: "From the right pocket of my pants, I transferred 4 five-crown coins to the left pocket, and from the left pocket, I transferred 16 two-crown coins to the right pocket. Now I have 13 crowns less in the left pocket than in the right pocket. In which pocket did I originally have more crowns and by how much?"Okay, so there are two pockets: left and right. I'm transferring coins between them, and after these transfers, there's a difference of 13 crowns between the two pockets, with the left pocket having less.Let me break this down.First, I transferred 4 five-crown coins from the right pocket to the left pocket. Each five-crown coin is worth 5 crowns, so 4 of them would be 4 * 5 = 20 crowns. So, the right pocket loses 20 crowns, and the left pocket gains 20 crowns.Then, I transferred 16 two-crown coins from the left pocket to the right pocket. Each two-crown coin is worth 2 crowns, so 16 of them would be 16 * 2 = 32 crowns. So, the left pocket loses 32 crowns, and the right pocket gains 32 crowns.After these transfers, the left pocket has 13 crowns less than the right pocket.I need to find out which pocket originally had more crowns and by how much.Let me assign some variables to make this clearer.Let me denote:- L as the original amount of crowns in the left pocket.- R as the original amount of crowns in the right pocket.Now, after transferring 20 crowns from the right to the left, the amounts become:- Left pocket: L + 20- Right pocket: R - 20Then, transferring 32 crowns from the left to the right, the amounts become:- Left pocket: (L + 20) - 32 = L - 12- Right pocket: (R - 20) + 32 = R + 12Now, according to the problem, after these transfers, the left pocket has 13 crowns less than the right pocket. So, we can set up the equation:Left pocket = Right pocket - 13Substituting the expressions we have:L - 12 = (R + 12) - 13Simplify the right side:L - 12 = R + 12 - 13L - 12 = R - 1Now, let's solve for L in terms of R:L = R - 1 + 12L = R + 11So, L = R + 11This means that the original amount in the left pocket was 11 crowns more than the original amount in the right pocket.Wait, let me double-check that.Starting with L and R.After transferring 20 crowns from right to left:Left: L + 20Right: R - 20Then transferring 32 crowns from left to right:Left: L + 20 - 32 = L - 12Right: R - 20 + 32 = R + 12Given that Left = Right - 13:L - 12 = (R + 12) - 13L - 12 = R - 1L = R - 1 + 12L = R + 11Yes, that seems correct. So, the left pocket originally had 11 more crowns than the right pocket.But wait, let me think about this again. If I transferred 20 crowns from right to left, and then 32 crowns from left to right, the net transfer from left to right is 32 - 20 = 12 crowns. So, effectively, the right pocket gained 12 crowns, and the left pocket lost 12 crowns.But the problem states that after these transfers, the left pocket has 13 crowns less than the right pocket. So, the difference is 13 crowns, but the net transfer was 12 crowns. That seems a bit confusing.Let me try to visualize this.Let me denote the original amounts as L and R.After transferring 20 crowns from right to left:Left: L + 20Right: R - 20Then transferring 32 crowns from left to right:Left: L + 20 - 32 = L - 12Right: R - 20 + 32 = R + 12Now, the difference between right and left is 13 crowns:(R + 12) - (L - 12) = 13Simplify:R + 12 - L + 12 = 13R - L + 24 = 13R - L = 13 - 24R - L = -11So, R = L - 11Which means L = R + 11So, the left pocket originally had 11 more crowns than the right pocket.Wait, that's consistent with what I had before.But let me think about the net transfer again. The net transfer from left to right is 12 crowns, but the difference is 13 crowns. So, the net transfer is 12, but the difference is 13. That seems like an extra crown somewhere.Is there a mistake in my calculations?Let me check:After transferring 20 from right to left:Left: L + 20Right: R - 20Then transferring 32 from left to right:Left: L + 20 - 32 = L - 12Right: R - 20 + 32 = R + 12Difference: Right - Left = (R + 12) - (L - 12) = R + 12 - L + 12 = R - L + 24Given that this difference is 13:R - L + 24 = 13R - L = 13 - 24R - L = -11So, L = R + 11Yes, that's correct. So, the left pocket had 11 more crowns originally.But why is the net transfer 12 crowns, but the difference is 13? Because the net transfer is 12, but the original difference was 11, so the total difference becomes 13.Wait, maybe I'm overcomplicating it.Let me think of it in terms of equations.Let me denote:After all transfers:Left = L - 12Right = R + 12Given that Left = Right - 13So,L - 12 = (R + 12) - 13Simplify:L - 12 = R - 1So,L = R - 1 + 12L = R + 11Yes, that's correct.So, the left pocket originally had 11 more crowns than the right pocket.I think that's the answer.But just to make sure, let me plug in some numbers.Suppose the right pocket originally had R crowns.Then the left pocket had R + 11 crowns.After transferring 20 crowns from right to left:Right: R - 20Left: R + 11 + 20 = R + 31Then transferring 32 crowns from left to right:Right: R - 20 + 32 = R + 12Left: R + 31 - 32 = R - 1Now, the difference between right and left is:(R + 12) - (R - 1) = R + 12 - R + 1 = 13Which matches the problem statement.So, yes, the left pocket originally had 11 more crowns than the right pocket.Therefore, the answer is that the left pocket originally had 11 more crowns than the right pocket.