answer：First, I need to solve Paul's equation (|x - 3| = 4). The absolute value equation can be split into two separate linear equations:1. (x - 3 = 4)2. (x - 3 = -4)Solving the first equation:[x - 3 = 4 implies x = 7]Solving the second equation:[x - 3 = -4 implies x = -1]So, the solutions to Paul's equation are (x = 7) and (x = -1).Next, Alice's quadratic equation (x^2 + dx + e = 0) must have these same solutions, 7 and -1. To find the quadratic equation, I can use the factored form:[(x - 7)(x + 1) = 0]Expanding this product:[x^2 + x - 7x - 7 = x^2 - 6x - 7 = 0]From this, I can identify the coefficients (d) and (e):[d = -6 quad text{and} quad e = -7]Therefore, the ordered pair ((d, e)) is ((-6, -7)).

answer：Okay, so I have this trigonometric identity to prove: [(cos alpha - cos beta)^{2} - (sin alpha - sin beta)^{2} = -4 sin^2 frac{alpha - beta}{2} cos (alpha + beta)]Hmm, trigonometric identities can be tricky, but let's take it step by step. I remember that expanding expressions and using known identities can help simplify things. Let me start by expanding both squared terms on the left-hand side.First, let's expand ((cos alpha - cos beta)^2):[(cos alpha - cos beta)^2 = cos^2 alpha - 2 cos alpha cos beta + cos^2 beta]Similarly, expanding ((sin alpha - sin beta)^2):[(sin alpha - sin beta)^2 = sin^2 alpha - 2 sin alpha sin beta + sin^2 beta]Now, subtracting the second expansion from the first:[(cos^2 alpha - 2 cos alpha cos beta + cos^2 beta) - (sin^2 alpha - 2 sin alpha sin beta + sin^2 beta)]Let me distribute the negative sign to the second part:[cos^2 alpha - 2 cos alpha cos beta + cos^2 beta - sin^2 alpha + 2 sin alpha sin beta - sin^2 beta]Now, let's group like terms together:- Terms with (cos^2) and (sin^2): [ cos^2 alpha - sin^2 alpha + cos^2 beta - sin^2 beta ] - Terms with (cos alpha cos beta) and (sin alpha sin beta): [ -2 cos alpha cos beta + 2 sin alpha sin beta ]Looking at the first group, I remember the double-angle identity for cosine:[cos^2 theta - sin^2 theta = cos 2theta]So, applying this identity to both (alpha) and (beta):[cos 2alpha + cos 2beta]Now, the second group of terms:[-2 cos alpha cos beta + 2 sin alpha sin beta]I can factor out a -2:[-2 (cos alpha cos beta - sin alpha sin beta)]Wait, the expression inside the parentheses looks familiar. Isn't that the cosine of a sum formula?Yes! The identity is:[cos(alpha + beta) = cos alpha cos beta - sin alpha sin beta]So, substituting that in:[-2 cos(alpha + beta)]Putting it all together, the entire expression becomes:[cos 2alpha + cos 2beta - 2 cos(alpha + beta)]Hmm, now I need to simplify this further. Maybe I can use another identity for (cos 2alpha + cos 2beta). I recall the sum-to-product identities, which might help here.The sum-to-product identity for cosine is:[cos A + cos B = 2 cosleft(frac{A + B}{2}right) cosleft(frac{A - B}{2}right)]Let me apply this to (cos 2alpha + cos 2beta), where (A = 2alpha) and (B = 2beta):[cos 2alpha + cos 2beta = 2 cosleft(frac{2alpha + 2beta}{2}right) cosleft(frac{2alpha - 2beta}{2}right)]Simplifying the arguments:[2 cos(alpha + beta) cos(alpha - beta)]So, substituting back into our expression:[2 cos(alpha + beta) cos(alpha - beta) - 2 cos(alpha + beta)]I can factor out a (2 cos(alpha + beta)):[2 cos(alpha + beta) [cos(alpha - beta) - 1]]Now, let's look at the term (cos(alpha - beta) - 1). I remember another identity for (cos theta - 1):[cos theta - 1 = -2 sin^2left(frac{theta}{2}right)]Applying this identity with (theta = alpha - beta):[cos(alpha - beta) - 1 = -2 sin^2left(frac{alpha - beta}{2}right)]Substituting back into our expression:[2 cos(alpha + beta) times left(-2 sin^2left(frac{alpha - beta}{2}right)right)]Multiplying the constants:[-4 sin^2left(frac{alpha - beta}{2}right) cos(alpha + beta)]So, putting it all together, we've shown that:[(cos alpha - cos beta)^{2} - (sin alpha - sin beta)^{2} = -4 sin^2 frac{alpha - beta}{2} cos (alpha + beta)]Which is exactly what we needed to prove! I think that covers all the steps. I used several trigonometric identities: the expansion of squared differences, the double-angle identity, the cosine of a sum identity, and the sum-to-product identity. Breaking it down step by step helped me manage each part without getting overwhelmed.**Final Answer**boxed{ -4 sin^2 frac{alpha - beta}{2} cos (alpha + beta) }

answer：First, I need to understand the given equation ( |a - 2| + (2b + 1)^2 = 0 ). Since both terms are non-negative, each must individually equal zero for their sum to be zero.Starting with ( |a - 2| = 0 ), this implies that ( a - 2 = 0 ), so ( a = 2 ).Next, looking at ( (2b + 1)^2 = 0 ), taking the square root of both sides gives ( 2b + 1 = 0 ). Solving for ( b ), I subtract 1 from both sides to get ( 2b = -1 ), and then divide by 2 to find ( b = -frac{1}{2} ).Now, I need to calculate ( (ab)^{2013} ). Substituting the values of ( a ) and ( b ), I have ( ab = 2 times left(-frac{1}{2}right) = -1 ).Finally, raising (-1) to the power of 2013, which is an odd number, results in (-1).

answer：Okay, so I have this problem about a function ( f(x) ) that satisfies the equation ( f(x+y) - f(y) = (x + 2y + 1)x ) for all real numbers ( x ) and ( y ). Also, it's given that ( f(1) = 0 ). There are three parts to this problem: finding ( f(0) ), finding the general form of ( f(x) ), and then determining the range of ( a ) such that the inequality ( f(x) > ax - 5 ) holds for ( 0 < x < 2 ).Starting with part (1): Find ( f(0) ).Hmm, I need to find ( f(0) ). Since the functional equation holds for all ( x ) and ( y ), maybe I can choose specific values for ( x ) and ( y ) to simplify the equation and solve for ( f(0) ).Let me try setting ( x = 1 ) and ( y = 0 ). Plugging these into the equation:( f(1 + 0) - f(0) = (1 + 2*0 + 1)*1 )Simplifying the right side:( (1 + 0 + 1)*1 = 2*1 = 2 )So, the equation becomes:( f(1) - f(0) = 2 )But we know that ( f(1) = 0 ), so substituting that in:( 0 - f(0) = 2 )Which simplifies to:( -f(0) = 2 )Multiplying both sides by -1:( f(0) = -2 )Okay, so that gives me ( f(0) = -2 ). That seems straightforward.Moving on to part (2): Find ( f(x) ).I need to find the general form of the function ( f(x) ). The functional equation is given as:( f(x + y) - f(y) = (x + 2y + 1)x )This looks like a difference equation, and since it holds for all ( x ) and ( y ), perhaps I can fix one variable and vary the other to find a pattern or expression for ( f(x) ).Let me try setting ( y = 0 ). Then the equation becomes:( f(x + 0) - f(0) = (x + 2*0 + 1)x )Simplifying:( f(x) - f(0) = (x + 1)x )We already found ( f(0) = -2 ), so substituting that in:( f(x) - (-2) = (x + 1)x )Which simplifies to:( f(x) + 2 = x^2 + x )Subtracting 2 from both sides:( f(x) = x^2 + x - 2 )Wait, so is that the function? Let me check if this satisfies the original functional equation.Let me compute ( f(x + y) - f(y) ) using ( f(x) = x^2 + x - 2 ):( f(x + y) = (x + y)^2 + (x + y) - 2 )( = x^2 + 2xy + y^2 + x + y - 2 )Subtracting ( f(y) ):( f(x + y) - f(y) = [x^2 + 2xy + y^2 + x + y - 2] - [y^2 + y - 2] )( = x^2 + 2xy + y^2 + x + y - 2 - y^2 - y + 2 )( = x^2 + 2xy + x )Simplify:( x^2 + x + 2xy )Factor out an x:( x(x + 1 + 2y) )Which is equal to:( (x + 2y + 1)x )Yes, that matches the right side of the original equation. So, ( f(x) = x^2 + x - 2 ) is indeed the correct function.Now, moving on to part (3): The inequality ( f(x) > ax - 5 ) holds for ( 0 < x < 2 ). We need to find the range of ( a ).Given that ( f(x) = x^2 + x - 2 ), let's substitute that into the inequality:( x^2 + x - 2 > ax - 5 )Let me rearrange this inequality to collect like terms:( x^2 + x - 2 - ax + 5 > 0 )( x^2 + (1 - a)x + 3 > 0 )So, the inequality simplifies to:( x^2 + (1 - a)x + 3 > 0 ) for all ( x ) in the interval ( 0 < x < 2 ).I need to find the values of ( a ) such that this quadratic expression is positive for all ( x ) in ( (0, 2) ).First, let me denote the quadratic as:( Q(x) = x^2 + (1 - a)x + 3 )We need ( Q(x) > 0 ) for all ( x in (0, 2) ).To ensure that ( Q(x) ) is positive on the interval ( (0, 2) ), we can analyze the quadratic function.Quadratic functions are parabolas. Since the coefficient of ( x^2 ) is positive (1), the parabola opens upwards. Therefore, the minimum value occurs at the vertex.The vertex of a quadratic ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). In our case, ( a = 1 ), ( b = (1 - a) ), so the vertex is at:( x = -frac{1 - a}{2*1} = frac{a - 1}{2} )So, the vertex is at ( x = frac{a - 1}{2} ).Now, we need to consider whether this vertex lies within the interval ( (0, 2) ) or outside of it.Case 1: The vertex is inside the interval ( (0, 2) ).This happens when:( 0 < frac{a - 1}{2} < 2 )Multiplying all parts by 2:( 0 < a - 1 < 4 )Adding 1:( 1 < a < 5 )In this case, the minimum value of ( Q(x) ) occurs at ( x = frac{a - 1}{2} ). To ensure ( Q(x) > 0 ) for all ( x in (0, 2) ), the minimum value must be positive.Compute ( Q ) at the vertex:( Qleft( frac{a - 1}{2} right) = left( frac{a - 1}{2} right)^2 + (1 - a)left( frac{a - 1}{2} right) + 3 )Let me compute this step by step.First, square term:( left( frac{a - 1}{2} right)^2 = frac{(a - 1)^2}{4} )Second term:( (1 - a)left( frac{a - 1}{2} right) = frac{(1 - a)(a - 1)}{2} )Note that ( (1 - a)(a - 1) = -(a - 1)^2 ), so:( frac{-(a - 1)^2}{2} )Third term is 3.Putting it all together:( Qleft( frac{a - 1}{2} right) = frac{(a - 1)^2}{4} - frac{(a - 1)^2}{2} + 3 )Combine the first two terms:( frac{(a - 1)^2}{4} - frac{2(a - 1)^2}{4} = -frac{(a - 1)^2}{4} )So, the expression becomes:( -frac{(a - 1)^2}{4} + 3 )Set this greater than 0:( -frac{(a - 1)^2}{4} + 3 > 0 )Multiply both sides by 4:( -(a - 1)^2 + 12 > 0 )Rearrange:( 12 - (a - 1)^2 > 0 )Which is:( (a - 1)^2 < 12 )Taking square roots:( |a - 1| < sqrt{12} )Simplify ( sqrt{12} ) to ( 2sqrt{3} ):( |a - 1| < 2sqrt{3} )Which implies:( -2sqrt{3} < a - 1 < 2sqrt{3} )Adding 1:( 1 - 2sqrt{3} < a < 1 + 2sqrt{3} )But remember, in Case 1, we have ( 1 < a < 5 ). So, we need to find the intersection of ( 1 < a < 5 ) and ( 1 - 2sqrt{3} < a < 1 + 2sqrt{3} ).Compute ( 1 + 2sqrt{3} approx 1 + 3.464 = 4.464 ). Since ( 4.464 < 5 ), the upper bound is ( 1 + 2sqrt{3} ). The lower bound is ( 1 ), since ( 1 - 2sqrt{3} ) is negative and our Case 1 assumes ( a > 1 ).Therefore, in Case 1, the range of ( a ) is ( 1 < a < 1 + 2sqrt{3} ).Case 2: The vertex is to the left of the interval ( (0, 2) ).This happens when:( frac{a - 1}{2} leq 0 )Multiply both sides by 2:( a - 1 leq 0 )So,( a leq 1 )In this case, the minimum of ( Q(x) ) on ( (0, 2) ) occurs at the left endpoint ( x = 0 ). But since ( x ) is in ( (0, 2) ), we can consider the limit as ( x ) approaches 0 from the right.Compute ( Q(0) ):( Q(0) = 0^2 + (1 - a)*0 + 3 = 3 )Since 3 is positive, ( Q(x) > 0 ) for all ( x ) near 0. However, we need to ensure that ( Q(x) > 0 ) for all ( x ) in ( (0, 2) ). Since the parabola opens upwards and the vertex is to the left of the interval, the function is increasing on ( (0, 2) ). Therefore, the minimum on ( (0, 2) ) is at ( x = 0 ), which is 3, so ( Q(x) > 0 ) for all ( x ) in ( (0, 2) ).Thus, for ( a leq 1 ), the inequality holds.Case 3: The vertex is to the right of the interval ( (0, 2) ).This happens when:( frac{a - 1}{2} geq 2 )Multiply both sides by 2:( a - 1 geq 4 )So,( a geq 5 )In this case, the minimum of ( Q(x) ) on ( (0, 2) ) occurs at the right endpoint ( x = 2 ). Let's compute ( Q(2) ):( Q(2) = 2^2 + (1 - a)*2 + 3 = 4 + 2 - 2a + 3 = 9 - 2a )We need ( Q(2) > 0 ):( 9 - 2a > 0 )Solving for ( a ):( -2a > -9 )Divide both sides by -2 (remembering to reverse the inequality sign):( a < frac{9}{2} )( a < 4.5 )But in Case 3, ( a geq 5 ). However, ( 4.5 < 5 ), so there is no overlap between ( a geq 5 ) and ( a < 4.5 ). Therefore, in this case, there are no solutions because ( Q(2) ) would be negative or zero, which violates the inequality.Putting it all together:From Case 1: ( 1 < a < 1 + 2sqrt{3} ) (approximately 4.464)From Case 2: ( a leq 1 )From Case 3: No solutionsTherefore, combining Case 1 and Case 2, the range of ( a ) is ( a < 1 + 2sqrt{3} ). However, since in Case 2, ( a leq 1 ) is allowed, and in Case 1, ( a ) can go up to approximately 4.464, the overall range is ( a < 1 + 2sqrt{3} ).But wait, let me double-check. If ( a leq 1 ), the inequality holds, and if ( 1 < a < 1 + 2sqrt{3} ), the inequality also holds. So, the combined range is ( a < 1 + 2sqrt{3} ).But let me confirm whether at ( a = 1 + 2sqrt{3} ), the inequality holds.At ( a = 1 + 2sqrt{3} ), the minimum of ( Q(x) ) is zero, because ( Qleft( frac{a - 1}{2} right) = 0 ). So, at that point, ( Q(x) = 0 ) at the vertex, which is inside the interval ( (0, 2) ). Therefore, the inequality ( Q(x) > 0 ) would not hold at that specific point. Hence, ( a ) must be strictly less than ( 1 + 2sqrt{3} ).Therefore, the range of ( a ) is all real numbers less than ( 1 + 2sqrt{3} ).So, summarizing:1. ( f(0) = -2 )2. ( f(x) = x^2 + x - 2 )3. The range of ( a ) is ( a < 1 + 2sqrt{3} )**Final Answer**(1) (boxed{-2})(2) (boxed{x^2 + x - 2})(3) (boxed{(-infty, 1 + 2sqrt{3})})

answer：Alright, so I'm trying to understand and prove these properties of nim-sum. Nim-sum is essentially the bitwise XOR operation, right? It's used a lot in game theory, especially in the game of Nim, which is where it gets its name. But I need to make sure I really grasp what's going on here.Starting with property (a): m XOR m equals 0. Hmm, that makes sense because when you XOR a number with itself, every bit cancels out. For example, if I take the number 5, which is 101 in binary, and XOR it with itself, I get 000, which is 0. So, yeah, that seems straightforward. I think I can formalize this by looking at each bit individually and showing that m_i XOR m_i is 0 for all bits i.Moving on to property (b): m XOR k equals k XOR m. This is about the commutative property of XOR. I remember that XOR is commutative, meaning the order doesn't matter. So, whether I XOR m with k or k with m, the result should be the same. To prove this, I can look at each corresponding bit of m and k. Since XOR is commutative on individual bits, the entire operation should be commutative as well.Property (c) is about associativity: (m XOR t) XOR k equals m XOR (t XOR k). Associativity means that the way we group the operations doesn't affect the outcome. I know that XOR is associative, so this should hold true. To prove it, I can consider the bits of m, t, and k. Since XOR is associative on individual bits, the overall operation should maintain associativity.Now, property (d) is a bit trickier. It states that if n is the nim-sum of several numbers m1, m2, ..., ml and n is not zero, then there exists at least one number mj such that mj XOR n is less than mj. This seems to relate to the idea that if you have a non-zero nim-sum, you can always find a number in the set whose XOR with the nim-sum reduces its value.Let me think about this. If n is not zero, it means that there's at least one bit where the nim-sum has a 1. Let's say the highest bit where n has a 1 is at position s. Now, for the nim-sum to have a 1 at position s, there must be an odd number of mj's that have a 1 at that position. So, there must be at least one mj with a 1 at position s.If I take such an mj and XOR it with n, the bit at position s will flip from 1 to 0 because n has a 1 there. This means that the resulting number, mj XOR n, will have a 0 at position s, making it smaller than mj, which had a 1 there. Therefore, mj XOR n must be less than mj.Wait, does this always hold? Let me test it with an example. Suppose I have numbers 4 (100), 7 (111), and 3 (011). The nim-sum of these is 4 XOR 7 XOR 3. Let's compute that:4 XOR 7 is 3 (as given), and 3 XOR 3 is 0. So, n is 0 in this case. But the problem states that n is not zero, so this example doesn't apply. Let's try another set.Take numbers 5 (101), 6 (110), and 3 (011). The nim-sum is 5 XOR 6 XOR 3. Let's compute:5 XOR 6 is 3 (011), and 3 XOR 3 is 0. Again, n is 0. Hmm, maybe I need a different set where n is not zero.Let's try 5 (101), 6 (110), and 2 (010). The nim-sum is 5 XOR 6 XOR 2.First, 5 XOR 6 is 3 (011), then 3 XOR 2 is 1 (001). So, n is 1. Now, according to property (d), there should be at least one number mj such that mj XOR 1 is less than mj.Looking at the numbers:5 XOR 1 = 4 (100), which is less than 5.6 XOR 1 = 7 (111), which is greater than 6.2 XOR 1 = 3 (011), which is greater than 2.So, in this case, 5 XOR n (which is 1) is less than 5. Therefore, the property holds.Another example: numbers 7 (111), 3 (011), and 4 (100). The nim-sum is 7 XOR 3 XOR 4.7 XOR 3 is 4 (100), then 4 XOR 4 is 0. So, n is 0 again. Not helpful.Let me pick numbers where n is not zero. How about 8 (1000), 5 (0101), and 12 (1100). The nim-sum is 8 XOR 5 XOR 12.8 XOR 5 is 13 (1101), then 13 XOR 12 is 1 (0001). So, n is 1.Now, check each number:8 XOR 1 = 9 (1001), which is greater than 8.5 XOR 1 = 4 (0100), which is less than 5.12 XOR 1 = 13 (1101), which is greater than 12.So, 5 XOR n is less than 5. Therefore, the property holds again.It seems consistent. So, the key idea is that if n is not zero, it has a highest bit set to 1. Since n is the nim-sum, there must be an odd number of mj's with that bit set. Therefore, at least one mj has that bit set, and XORing that mj with n will unset that bit, making the result smaller.I think I've got the gist of it. Now, to formalize this into a proof.For property (a), I can argue that for each bit, m_i XOR m_i is 0, so the entire number becomes 0.For property (b), since XOR is commutative on each bit, the overall operation is commutative.For property (c), associativity holds because XOR is associative on each bit.For property (d), since n is not zero, it has a highest bit set. There must be an mj with that bit set, and XORing with n will unset that bit, making mj XOR n less than mj.I think that covers all the properties.

answer：Okay, so I have this problem where I need to arrange the odd positive integers in five columns. The first column is column 1, and they go from left to right. I need to figure out in which column the number 1985 appears. The options are columns 1 through 5.First, let me try to understand how these numbers are arranged. The problem mentions that the first column is column 1, and the columns are numbered sequentially from left to right. So, column 1 is the first one on the left, column 2 is next to it, and so on up to column 5 on the right.Now, the numbers given are the odd positive integers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, and so on. These numbers are arranged in five columns. I need to figure out the pattern of how these numbers are placed in the columns.I think the arrangement is such that each row contains five numbers, one in each column. So, the first row would have the first five odd numbers: 1, 3, 5, 7, 9. The second row would have the next five: 11, 13, 15, 17, 19, and so on.But wait, the problem mentions that the first column is column 1, and columns are numbered from left to right. So, in the first row, column 1 has 1, column 2 has 3, column 3 has 5, column 4 has 7, and column 5 has 9. Then, the second row would start with 11 in column 1, 13 in column 2, and so on.But I also need to consider whether the direction of numbering changes for each row. Sometimes, in such problems, the numbering alternates direction for each row, like left to right for the first row, right to left for the second row, and so on. This is similar to how a snake moves, hence sometimes called a "snake-like" arrangement.If that's the case, then the first row would be left to right: 1, 3, 5, 7, 9. The second row would be right to left: 19, 17, 15, 13, 11. The third row would again be left to right: 21, 23, 25, 27, 29, and so on.This alternating pattern is important because it affects the position of the numbers in the columns. If the direction alternates, the same column number would have different positions in different rows.So, to find out where 1985 is, I need to determine two things:1. Which row does 1985 fall into?2. In that row, is the direction left to right or right to left, and thus, which column does 1985 occupy?Let me start by figuring out how many numbers are there before 1985. Since we're dealing with odd numbers, the nth odd number is given by the formula:[ a_n = 2n - 1 ]So, if I set ( a_n = 1985 ), I can solve for n:[ 1985 = 2n - 1 ][ 2n = 1986 ][ n = 993 ]So, 1985 is the 993rd odd number.Since each row contains 5 numbers, the number of rows needed to reach the 993rd number is:[ text{Number of rows} = leftlceil frac{993}{5} rightrceil ]Calculating that:[ frac{993}{5} = 198.6 ]So, we need 199 rows to reach the 993rd number.Wait, let me double-check that. If each row has 5 numbers, then 198 rows would contain:[ 198 times 5 = 990 ]numbers. So, the 991st number would be in row 199, and so would the 993rd number. Therefore, 1985 is in row 199.Now, I need to determine the direction of the 199th row. Since the direction alternates starting from left to right for the first row, the odd-numbered rows (1st, 3rd, 5th, etc.) are left to right, and the even-numbered rows (2nd, 4th, 6th, etc.) are right to left.Since 199 is an odd number, the 199th row is left to right. That means the numbers in this row start from the first column and go to the fifth column.Now, within the 199th row, 1985 is the 993rd number. Since each row has 5 numbers, the position of 1985 within the row is:[ text{Position in row} = 993 - (198 times 5) = 993 - 990 = 3 ]So, 1985 is the 3rd number in the 199th row.Since the 199th row is left to right, the first number is in column 1, the second in column 2, and the third in column 3. Therefore, 1985 should be in column 3.Wait, hold on. Let me make sure I didn't make a mistake here. If each row has 5 numbers, and 1985 is the 993rd number, then the position within the row is 3. So, in a left-to-right row, that would be column 3.But let me think again about the arrangement. If the first row is left to right, the second row is right to left, the third row is left to right, and so on. So, in the 199th row, which is odd, it's left to right.But let's confirm the starting number of the 199th row. The first number of the first row is 1. The first number of the second row is 11. The first number of the third row is 21. So, each new row starts 10 more than the previous row's start.Wait, that might not be accurate. Let me think. The first row has 1, 3, 5, 7, 9. The second row has 11, 13, 15, 17, 19. The third row has 21, 23, 25, 27, 29, and so on. So, the first number of each row increases by 10 each time.So, the first number of the nth row is:[ a_n = 1 + 10(n - 1) ]So, for the 199th row:[ a_{199} = 1 + 10(199 - 1) = 1 + 10 times 198 = 1 + 1980 = 1981 ]So, the first number in the 199th row is 1981.Since the 199th row is left to right, the numbers are:1981 (column 1), 1983 (column 2), 1985 (column 3), 1987 (column 4), 1989 (column 5).Therefore, 1985 is indeed in column 3.Wait, but let me cross-verify this. If the first number of the 199th row is 1981, then adding 2 each time gives:1981, 1983, 1985, 1987, 1989.So, 1985 is the third number in the row, which is column 3.Therefore, the answer should be column 3, which is option (C).But just to make sure, let me think about another approach.Another way to approach this is to realize that the columns cycle every 10 numbers because each column has numbers that are 10 apart. For example, column 1 has 1, 11, 21, 31, etc., which are 10 apart. Similarly, column 2 has 3, 13, 23, 33, etc.So, if I can find the position of 1985 modulo 10, that might help. Let's see:1985 divided by 10 is 198 with a remainder of 5. So, 1985 ≡ 5 mod 10.Looking at the columns:- Column 1: numbers ≡1 mod 10- Column 2: numbers ≡3 mod 10- Column 3: numbers ≡5 mod 10- Column 4: numbers ≡7 mod 10- Column 5: numbers ≡9 mod 10So, 1985 ≡5 mod 10, which corresponds to column 3. Therefore, this method also leads me to column 3.Hmm, that seems consistent with my earlier conclusion.Wait, but I should also consider the direction of the rows. Since the direction alternates, does that affect the modulo result? Let me think.If the direction alternates, then in even-numbered rows, the columns are filled from right to left. So, in even rows, the first number would be in column 5, the next in column 4, and so on.But in this case, since 199 is odd, the row is filled left to right, so the modulo approach directly applies.If 1985 were in an even row, the column would be different because the direction is reversed. But since it's in an odd row, the column is determined directly by the modulo.Therefore, both methods confirm that 1985 is in column 3.I think I'm confident with this answer now.