answer：Okay, so I have this problem here where I need to find the center of a sphere S. The endpoints of its diameter are given as (2, -3, 4) and (0, -7, -2). Hmm, I remember that the center of a sphere is essentially the midpoint of its diameter. So, if I can find the midpoint between these two points, that should give me the coordinates of the center.Alright, let me recall the midpoint formula. In two dimensions, the midpoint (M) between two points (x₁, y₁) and (x₂, y₂) is calculated as ((x₁ + x₂)/2, (y₁ + y₂)/2). I think it's similar in three dimensions, but we just have an extra z-coordinate to consider. So, the midpoint formula in three dimensions should be ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2). That makes sense because we're just averaging each corresponding coordinate.Let me write down the given points again to make sure I have them right. The first endpoint is (2, -3, 4) and the second one is (0, -7, -2). So, I need to average the x-coordinates, the y-coordinates, and the z-coordinates separately.Starting with the x-coordinates: the first x is 2 and the second x is 0. Adding them together gives 2 + 0 = 2. Then, dividing by 2 gives 2/2 = 1. So, the x-coordinate of the midpoint is 1.Next, the y-coordinates: the first y is -3 and the second y is -7. Adding these together: -3 + (-7) = -10. Dividing by 2 gives -10/2 = -5. So, the y-coordinate of the midpoint is -5.Now, the z-coordinates: the first z is 4 and the second z is -2. Adding them together: 4 + (-2) = 2. Dividing by 2 gives 2/2 = 1. So, the z-coordinate of the midpoint is 1.Putting it all together, the midpoint, which is the center of the sphere, should be (1, -5, 1). Let me double-check my calculations to make sure I didn't make any mistakes.For the x-coordinate: 2 + 0 = 2, divided by 2 is 1. That seems right.For the y-coordinate: -3 + (-7) = -10, divided by 2 is -5. That also looks correct.For the z-coordinate: 4 + (-2) = 2, divided by 2 is 1. Yep, that checks out too.So, I think I've done everything correctly. The center of the sphere S is at the point (1, -5, 1). I don't see any errors in my calculations, so I feel confident about this answer.

answer：First, identify the original number of yellow marbles Sam had, which is 86.Next, determine how many yellow marbles Sam has left after Joan took some, which is 61.To find out how many marbles Joan took, subtract the remaining number of yellow marbles from the original number:86 - 61 = 25.Therefore, Joan took 25 yellow marbles.

answer：First, I need to determine the number of small cubes along one edge of the large cube, denoted as ( n ).When a cube is painted on all faces and then cut into smaller cubes, the small cubes with only one face painted are those located on the surface but not on the edges or corners. Each face of the large cube contributes ( (n-2) times (n-2) ) small cubes with one face painted.Since there are 6 faces on a cube, the total number of small cubes with one face painted is ( 6 times (n-2)^2 ).Given that there are 54 small cubes with one face painted, I can set up the equation:[6 times (n-2)^2 = 54]Solving for ( n ):[(n-2)^2 = 9 n - 2 = 3 n = 5]Therefore, the large cube is divided into ( 5 times 5 times 5 = 125 ) small cubes.

answer：Okay, so I have this problem here: I need to find the maximum distance from a point on the circle defined by ρ = 8 sin θ in polar coordinates to the line θ = π/3. The options are A: 4, B: 7, C: 1, D: 6. Hmm, let me try to figure this out step by step.First, I remember that in polar coordinates, the equation ρ = 8 sin θ represents a circle. I think it's a circle that's not centered at the origin. Maybe I should convert this equation to Cartesian coordinates to better visualize it.So, in polar coordinates, ρ = 8 sin θ. I recall that ρ sin θ is equal to y in Cartesian coordinates. So, if I multiply both sides by ρ, I get ρ² = 8ρ sin θ. Since ρ² is x² + y², and ρ sin θ is y, substituting these in, I get x² + y² = 8y.Now, to write this in standard circle form, I need to complete the square for the y terms. So, x² + y² - 8y = 0. Let me rearrange this: x² + (y² - 8y) = 0. To complete the square for y² - 8y, I take half of -8, which is -4, square it to get 16, and add and subtract it inside the equation.So, x² + (y² - 8y + 16 - 16) = 0. This simplifies to x² + (y - 4)² - 16 = 0. Bringing the -16 to the other side, I get x² + (y - 4)² = 16. Okay, so this is a circle with center at (0, 4) and radius 4. Got that.Next, I need to find the maximum distance from any point on this circle to the line θ = π/3. θ = π/3 is a straight line in polar coordinates, which passes through the origin at an angle of 60 degrees from the positive x-axis. I think in Cartesian coordinates, this line can be represented as y = tan(π/3) x, which is y = √3 x.So, now I have a circle centered at (0, 4) with radius 4 and a line y = √3 x. I need to find the maximum distance from any point on the circle to this line.I remember that the distance from a point (x0, y0) to the line ax + by + c = 0 is given by |ax0 + by0 + c| / sqrt(a² + b²). So, first, let me write the line equation in standard form. y = √3 x can be rewritten as √3 x - y = 0.So, a = √3, b = -1, c = 0.Now, the distance from the center of the circle (0, 4) to this line is |√3*0 + (-1)*4 + 0| / sqrt((√3)² + (-1)²) = | -4 | / sqrt(3 + 1) = 4 / 2 = 2.So, the distance from the center to the line is 2 units. Since the radius of the circle is 4, I think the maximum distance from any point on the circle to the line would be the distance from the center to the line plus the radius. That is, 2 + 4 = 6.Wait, but let me make sure. Is it always just adding the radius? I think so because the maximum distance would occur when you go from the center towards the line, and then extend that direction by the radius. So, yes, 2 + 4 = 6.Looking back at the options, D is 6. So, I think the answer is D: 6.But just to double-check, maybe I should visualize this. The circle is centered at (0,4) with radius 4, so it touches the origin. The line θ = π/3 is a 60-degree line from the origin. The distance from the center to the line is 2, so the closest point on the line to the center is 2 units away. Then, moving in the opposite direction from the line, the farthest point on the circle would be 2 + 4 = 6 units away. That makes sense.Alternatively, if I think about parametrizing the circle and computing the distance, but that might be more complicated. I think the method I used is correct.So, yeah, I'm confident the answer is 6, which is option D.**Final Answer**boxed{D}

answer：Okay, so I have this problem about the function ( f(x) = x - 1 - a ln x ), where ( a > 0 ). There are two parts: part (I) asks me to find the set of values for ( a ) such that ( f(x) geq 0 ) for all ( x ) in ( (0, +infty) ). Part (II) is to prove an inequality involving ( e ) and the sequence ( left(1 + frac{1}{n}right)^n ) and ( left(1 + frac{1}{n}right)^{n+1} ).Starting with part (I). I need to ensure that ( f(x) geq 0 ) for all positive ( x ). Since ( f(x) ) is defined for ( x > 0 ), I can analyze its behavior by looking at its derivative. The function is ( x - 1 - a ln x ). Let me compute the derivative:( f'(x) = 1 - frac{a}{x} ).Setting the derivative equal to zero to find critical points:( 1 - frac{a}{x} = 0 ) ( frac{a}{x} = 1 ) ( x = a ).So, the function has a critical point at ( x = a ). To determine if this is a minimum or maximum, I can look at the second derivative or analyze the sign changes of the first derivative.Let me check the sign of ( f'(x) ) around ( x = a ):- For ( x < a ), ( f'(x) = 1 - frac{a}{x} ). Since ( x < a ), ( frac{a}{x} > 1 ), so ( f'(x) < 0 ).- For ( x > a ), ( f'(x) = 1 - frac{a}{x} ). Since ( x > a ), ( frac{a}{x} < 1 ), so ( f'(x) > 0 ).This means that the function is decreasing on ( (0, a) ) and increasing on ( (a, +infty) ). Therefore, ( x = a ) is a minimum point.Since ( f(x) geq 0 ) for all ( x > 0 ), the minimum value at ( x = a ) must be greater than or equal to zero. Let's compute ( f(a) ):( f(a) = a - 1 - a ln a ).So, we have the inequality:( a - 1 - a ln a geq 0 ).Let me rewrite this:( a(1 - ln a) geq 1 ).Hmm, this seems a bit tricky. Maybe I can rearrange it:( a(1 - ln a) geq 1 ) ( a - a ln a geq 1 ) ( a(1 - ln a) geq 1 ).I need to solve for ( a ) such that this inequality holds. Let me consider the function ( g(a) = a(1 - ln a) ). I need ( g(a) geq 1 ).Let me analyze ( g(a) ):( g(a) = a - a ln a ).Compute its derivative to find its maximum or minimum:( g'(a) = 1 - ( ln a + 1 ) = - ln a ).Setting ( g'(a) = 0 ):( - ln a = 0 ) ( ln a = 0 ) ( a = 1 ).So, ( g(a) ) has a critical point at ( a = 1 ). Let's check the second derivative or the behavior around ( a = 1 ):( g'(a) = - ln a ). For ( a < 1 ), ( ln a < 0 ), so ( g'(a) > 0 ). For ( a > 1 ), ( ln a > 0 ), so ( g'(a) < 0 ). Therefore, ( a = 1 ) is a maximum point.Compute ( g(1) ):( g(1) = 1 - 1 cdot ln 1 = 1 - 0 = 1 ).So, the maximum value of ( g(a) ) is 1 at ( a = 1 ). Since ( g(a) ) increases for ( a < 1 ) and decreases for ( a > 1 ), the maximum is indeed 1. Therefore, ( g(a) leq 1 ) for all ( a > 0 ).But we need ( g(a) geq 1 ). The only point where ( g(a) = 1 ) is at ( a = 1 ). Therefore, the only value of ( a ) that satisfies ( g(a) geq 1 ) is ( a = 1 ).So, for part (I), the set of values for ( a ) is just ( {1} ).Moving on to part (II). I need to prove that:( left(1 + frac{1}{n}right)^n < e < left(1 + frac{1}{n}right)^{n+1} ) for ( n in mathbb{N}^* ).I remember that ( e ) is defined as the limit of ( left(1 + frac{1}{n}right)^n ) as ( n ) approaches infinity. So, the sequence ( a_n = left(1 + frac{1}{n}right)^n ) converges to ( e ) from below, and the sequence ( b_n = left(1 + frac{1}{n}right)^{n+1} ) converges to ( e ) from above.To prove the inequality, I think I need to show that ( a_n ) is increasing and ( b_n ) is decreasing. If both sequences converge to ( e ) and ( a_n < e < b_n ) for all ( n ), then the inequality holds.Let me try to show that ( a_n ) is increasing. That is, ( a_{n+1} > a_n ) for all ( n ).Consider ( a_{n+1} = left(1 + frac{1}{n+1}right)^{n+1} ).I want to compare ( a_{n+1} ) and ( a_n ). Maybe using the AM-GM inequality or some other inequality.Alternatively, I can take the natural logarithm of both sides to simplify the comparison.Let me define ( ln a_n = n ln left(1 + frac{1}{n}right) ).Similarly, ( ln a_{n+1} = (n+1) ln left(1 + frac{1}{n+1}right) ).I need to show that ( ln a_{n+1} > ln a_n ).So, is ( (n+1) ln left(1 + frac{1}{n+1}right) > n ln left(1 + frac{1}{n}right) )?Let me denote ( x = frac{1}{n} ), so ( n = frac{1}{x} ), and ( n+1 = frac{1}{x} + 1 = frac{1 + x}{x} ).Wait, maybe that's complicating things. Alternatively, let's consider the function ( h(x) = x ln left(1 + frac{1}{x}right) ).We can analyze whether ( h(x) ) is increasing or decreasing.Compute the derivative ( h'(x) ):( h(x) = x ln left(1 + frac{1}{x}right) ).Let me rewrite ( h(x) ) as ( x ln left(frac{x + 1}{x}right) = x left( ln(x + 1) - ln x right) ).Then,( h'(x) = ln(x + 1) - ln x + x left( frac{1}{x + 1} - frac{1}{x} right) ).Simplify:( h'(x) = lnleft(1 + frac{1}{x}right) + x left( frac{1}{x + 1} - frac{1}{x} right) ).Compute the second term:( x left( frac{1}{x + 1} - frac{1}{x} right) = x left( frac{x - (x + 1)}{x(x + 1)} right) = x left( frac{-1}{x(x + 1)} right) = - frac{1}{x + 1} ).So,( h'(x) = lnleft(1 + frac{1}{x}right) - frac{1}{x + 1} ).Now, I need to determine the sign of ( h'(x) ). If ( h'(x) > 0 ), then ( h(x) ) is increasing, which would imply that ( a_n ) is increasing.So, is ( lnleft(1 + frac{1}{x}right) > frac{1}{x + 1} )?Let me consider the function ( k(t) = ln(1 + t) - frac{t}{1 + t} ) for ( t > 0 ).Compute ( k(t) ):( k(t) = ln(1 + t) - frac{t}{1 + t} ).Compute its derivative:( k'(t) = frac{1}{1 + t} - left( frac{(1 + t) - t}{(1 + t)^2} right) = frac{1}{1 + t} - frac{1}{(1 + t)^2} = frac{(1 + t) - 1}{(1 + t)^2} = frac{t}{(1 + t)^2} > 0 ).Since ( k'(t) > 0 ) for all ( t > 0 ), ( k(t) ) is increasing. Also, ( k(0) = ln(1) - 0 = 0 ). Therefore, for ( t > 0 ), ( k(t) > 0 ).Thus, ( ln(1 + t) > frac{t}{1 + t} ) for ( t > 0 ). Letting ( t = frac{1}{x} ), we have:( lnleft(1 + frac{1}{x}right) > frac{frac{1}{x}}{1 + frac{1}{x}} = frac{1}{x + 1} ).Therefore, ( h'(x) > 0 ), which means ( h(x) ) is increasing. Hence, ( a_n ) is increasing because ( h(n) = ln a_n ) is increasing.Similarly, I need to show that ( b_n = left(1 + frac{1}{n}right)^{n+1} ) is decreasing. Let's analyze ( b_n ).Take the natural logarithm:( ln b_n = (n + 1) ln left(1 + frac{1}{n}right) ).Let me define ( m(n) = (n + 1) ln left(1 + frac{1}{n}right) ).I need to show that ( m(n) ) is decreasing, which would imply ( b_n ) is decreasing.Compute ( m(n) - m(n + 1) ):( m(n) - m(n + 1) = (n + 1) ln left(1 + frac{1}{n}right) - (n + 2) ln left(1 + frac{1}{n + 1}right) ).I need to show this is positive, which would mean ( m(n) > m(n + 1) ).Alternatively, let's consider the function ( p(x) = (x + 1) ln left(1 + frac{1}{x}right) ).Compute its derivative:( p'(x) = lnleft(1 + frac{1}{x}right) + (x + 1) cdot frac{-1}{x(x + 1)} ).Simplify:( p'(x) = lnleft(1 + frac{1}{x}right) - frac{1}{x} ).We need to determine if ( p'(x) ) is positive or negative.Consider the function ( q(t) = ln(1 + t) - t ) for ( t > 0 ).Compute ( q(t) ):( q(t) = ln(1 + t) - t ).Compute its derivative:( q'(t) = frac{1}{1 + t} - 1 = frac{1 - (1 + t)}{1 + t} = frac{-t}{1 + t} < 0 ).Since ( q'(t) < 0 ), ( q(t) ) is decreasing. Also, ( q(0) = 0 ). Therefore, for ( t > 0 ), ( q(t) < 0 ).Thus, ( ln(1 + t) - t < 0 ) for ( t > 0 ). Letting ( t = frac{1}{x} ), we have:( lnleft(1 + frac{1}{x}right) - frac{1}{x} < 0 ).Therefore, ( p'(x) < 0 ), which means ( p(x) ) is decreasing. Hence, ( m(n) = p(n) ) is decreasing, so ( b_n ) is decreasing.Since ( a_n ) is increasing and ( b_n ) is decreasing, and both converge to ( e ), it follows that ( a_n < e < b_n ) for all ( n in mathbb{N}^* ).Therefore, the inequality ( left(1 + frac{1}{n}right)^n < e < left(1 + frac{1}{n}right)^{n+1} ) holds for all positive integers ( n ).**Final Answer**(I) The set of values for ( a ) is (boxed{1}).(II) The inequality is proven as required.

answer：Okay, so I have this problem: I need to prove that there exists a prime number ( p ) such that the sum of its digits is a composite odd integer. Then, I have to find the smallest such ( p ). Hmm, let's break this down step by step.First, let's understand what a composite odd integer is. A composite number is a positive integer that has at least one positive divisor other than 1 and itself. So, composite odd integers would be numbers like 9, 15, 21, 25, etc. These are odd numbers that aren't prime.Now, I need to find a prime number ( p ) where the sum of its digits is one of these composite odd numbers. Also, I need the smallest such prime. So, my goal is to find the smallest prime number where the sum of its digits is, say, 9, 15, 21, 25, etc.Let me start by checking the smallest composite odd integers and see if there's a prime number with that digit sum.Starting with 9: The smallest composite odd integer is 9. So, I need to find a prime number whose digits add up to 9. Let me think of some primes with digit sums of 9.For example, 18 is not prime, but 18 is not prime. Wait, 18 is not prime, but 19 is prime. The sum of digits of 19 is 1 + 9 = 10, which is even, so that's not it. Next, 27: 2 + 7 = 9, but 27 is not prime. 36: 3 + 6 = 9, but 36 is not prime. 45: 4 + 5 = 9, but 45 is not prime. 54: 5 + 4 = 9, not prime. 63: 6 + 3 = 9, not prime. 72: 7 + 2 = 9, not prime. 81: 8 + 1 = 9, not prime. 90: 9 + 0 = 9, not prime. 99: 9 + 9 = 18, which is even, so that's not it.Wait, maybe I need to look at three-digit numbers. Let's see: 108: 1 + 0 + 8 = 9, but 108 is not prime. 117: 1 + 1 + 7 = 9, not prime. 126: 1 + 2 + 6 = 9, not prime. 135: 1 + 3 + 5 = 9, not prime. 144: 1 + 4 + 4 = 9, not prime. 153: 1 + 5 + 3 = 9, not prime. 162: 1 + 6 + 2 = 9, not prime. 171: 1 + 7 + 1 = 9, not prime. 180: 1 + 8 + 0 = 9, not prime. 189: 1 + 8 + 9 = 18, which is even. Hmm, this isn't working.Maybe I need to look at primes where the sum of digits is 9. Let's think about primes like 199: 1 + 9 + 9 = 19, which is prime, not composite. 199 is prime, but the digit sum is 19, which is prime, so that's not it. 289: 2 + 8 + 9 = 19, same issue. 379: 3 + 7 + 9 = 19, same. 469: 4 + 6 + 9 = 19, same. 559: 5 + 5 + 9 = 19, same. 649: 6 + 4 + 9 = 19, same. 739: 7 + 3 + 9 = 19, same. 829: 8 + 2 + 9 = 19, same. 919: 9 + 1 + 9 = 19, same. 997: 9 + 9 + 7 = 25, which is composite and odd. Okay, so 997 is a prime number, and its digit sum is 25, which is composite and odd. So, that seems to fit.But wait, is 997 the smallest prime with a composite odd digit sum? Let me check smaller primes. Let's see, primes less than 997. Let's check primes in the 100s. For example, 101: 1 + 0 + 1 = 2, which is prime. 103: 1 + 0 + 3 = 4, composite but even. 107: 1 + 0 + 7 = 8, composite but even. 109: 1 + 0 + 9 = 10, composite but even. 113: 1 + 1 + 3 = 5, prime. 127: 1 + 2 + 7 = 10, composite but even. 131: 1 + 3 + 1 = 5, prime. 137: 1 + 3 + 7 = 11, prime. 139: 1 + 3 + 9 = 13, prime. 149: 1 + 4 + 9 = 14, composite but even. 151: 1 + 5 + 1 = 7, prime. 157: 1 + 5 + 7 = 13, prime. 163: 1 + 6 + 3 = 10, composite but even. 167: 1 + 6 + 7 = 14, composite but even. 173: 1 + 7 + 3 = 11, prime. 179: 1 + 7 + 9 = 17, prime. 181: 1 + 8 + 1 = 10, composite but even. 191: 1 + 9 + 1 = 11, prime. 193: 1 + 9 + 3 = 13, prime. 197: 1 + 9 + 7 = 17, prime. 199: 1 + 9 + 9 = 19, prime.Okay, moving on to the 200s. 211: 2 + 1 + 1 = 4, composite but even. 223: 2 + 2 + 3 = 7, prime. 227: 2 + 2 + 7 = 11, prime. 229: 2 + 2 + 9 = 13, prime. 233: 2 + 3 + 3 = 8, composite but even. 239: 2 + 3 + 9 = 14, composite but even. 241: 2 + 4 + 1 = 7, prime. 251: 2 + 5 + 1 = 8, composite but even. 257: 2 + 5 + 7 = 14, composite but even. 263: 2 + 6 + 3 = 11, prime. 269: 2 + 6 + 9 = 17, prime. 271: 2 + 7 + 1 = 10, composite but even. 277: 2 + 7 + 7 = 16, composite but even. 281: 2 + 8 + 1 = 11, prime. 283: 2 + 8 + 3 = 13, prime. 293: 2 + 9 + 3 = 14, composite but even.Moving to the 300s. 307: 3 + 0 + 7 = 10, composite but even. 311: 3 + 1 + 1 = 5, prime. 313: 3 + 1 + 3 = 7, prime. 317: 3 + 1 + 7 = 11, prime. 331: 3 + 3 + 1 = 7, prime. 337: 3 + 3 + 7 = 13, prime. 347: 3 + 4 + 7 = 14, composite but even. 349: 3 + 4 + 9 = 16, composite but even. 353: 3 + 5 + 3 = 11, prime. 359: 3 + 5 + 9 = 17, prime. 367: 3 + 6 + 7 = 16, composite but even. 373: 3 + 7 + 3 = 13, prime. 379: 3 + 7 + 9 = 19, prime. 383: 3 + 8 + 3 = 14, composite but even. 389: 3 + 8 + 9 = 20, composite but even. 397: 3 + 9 + 7 = 19, prime.400s: 401: 4 + 0 + 1 = 5, prime. 409: 4 + 0 + 9 = 13, prime. 419: 4 + 1 + 9 = 14, composite but even. 421: 4 + 2 + 1 = 7, prime. 431: 4 + 3 + 1 = 8, composite but even. 433: 4 + 3 + 3 = 10, composite but even. 439: 4 + 3 + 9 = 16, composite but even. 443: 4 + 4 + 3 = 11, prime. 449: 4 + 4 + 9 = 17, prime. 457: 4 + 5 + 7 = 16, composite but even. 461: 4 + 6 + 1 = 11, prime. 463: 4 + 6 + 3 = 13, prime. 467: 4 + 6 + 7 = 17, prime. 479: 4 + 7 + 9 = 20, composite but even. 487: 4 + 8 + 7 = 19, prime. 491: 4 + 9 + 1 = 14, composite but even. 499: 4 + 9 + 9 = 22, composite but even.500s: 503: 5 + 0 + 3 = 8, composite but even. 509: 5 + 0 + 9 = 14, composite but even. 521: 5 + 2 + 1 = 8, composite but even. 523: 5 + 2 + 3 = 10, composite but even. 541: 5 + 4 + 1 = 10, composite but even. 547: 5 + 4 + 7 = 16, composite but even. 557: 5 + 5 + 7 = 17, prime. 563: 5 + 6 + 3 = 14, composite but even. 569: 5 + 6 + 9 = 20, composite but even. 571: 5 + 7 + 1 = 13, prime. 577: 5 + 7 + 7 = 19, prime. 587: 5 + 8 + 7 = 20, composite but even. 593: 5 + 9 + 3 = 17, prime. 599: 5 + 9 + 9 = 23, prime.600s: 601: 6 + 0 + 1 = 7, prime. 607: 6 + 0 + 7 = 13, prime. 613: 6 + 1 + 3 = 10, composite but even. 617: 6 + 1 + 7 = 14, composite but even. 619: 6 + 1 + 9 = 16, composite but even. 631: 6 + 3 + 1 = 10, composite but even. 641: 6 + 4 + 1 = 11, prime. 643: 6 + 4 + 3 = 13, prime. 647: 6 + 4 + 7 = 17, prime. 653: 6 + 5 + 3 = 14, composite but even. 659: 6 + 5 + 9 = 20, composite but even. 661: 6 + 6 + 1 = 13, prime. 673: 6 + 7 + 3 = 16, composite but even. 677: 6 + 7 + 7 = 20, composite but even. 683: 6 + 8 + 3 = 17, prime. 691: 6 + 9 + 1 = 16, composite but even. 701: 7 + 0 + 1 = 8, composite but even. 709: 7 + 0 + 9 = 16, composite but even. 719: 7 + 1 + 9 = 17, prime. 727: 7 + 2 + 7 = 16, composite but even. 733: 7 + 3 + 3 = 13, prime. 739: 7 + 3 + 9 = 19, prime. 743: 7 + 4 + 3 = 14, composite but even. 751: 7 + 5 + 1 = 13, prime. 757: 7 + 5 + 7 = 19, prime. 761: 7 + 6 + 1 = 14, composite but even. 769: 7 + 6 + 9 = 22, composite but even. 773: 7 + 7 + 3 = 17, prime. 787: 7 + 8 + 7 = 22, composite but even. 797: 7 + 9 + 7 = 23, prime.800s: 809: 8 + 0 + 9 = 17, prime. 811: 8 + 1 + 1 = 10, composite but even. 821: 8 + 2 + 1 = 11, prime. 823: 8 + 2 + 3 = 13, prime. 827: 8 + 2 + 7 = 17, prime. 829: 8 + 2 + 9 = 19, prime. 839: 8 + 3 + 9 = 20, composite but even. 853: 8 + 5 + 3 = 16, composite but even. 857: 8 + 5 + 7 = 20, composite but even. 859: 8 + 5 + 9 = 22, composite but even. 863: 8 + 6 + 3 = 17, prime. 877: 8 + 7 + 7 = 22, composite but even. 881: 8 + 8 + 1 = 17, prime. 883: 8 + 8 + 3 = 19, prime. 887: 8 + 8 + 7 = 23, prime. 907: 9 + 0 + 7 = 16, composite but even. 911: 9 + 1 + 1 = 11, prime. 919: 9 + 1 + 9 = 19, prime. 929: 9 + 2 + 9 = 20, composite but even. 937: 9 + 3 + 7 = 19, prime. 941: 9 + 4 + 1 = 14, composite but even. 947: 9 + 4 + 7 = 20, composite but even. 953: 9 + 5 + 3 = 17, prime. 967: 9 + 6 + 7 = 22, composite but even. 971: 9 + 7 + 1 = 17, prime. 977: 9 + 7 + 7 = 23, prime. 983: 9 + 8 + 3 = 20, composite but even. 991: 9 + 9 + 1 = 19, prime. 997: 9 + 9 + 7 = 25, which is composite and odd.Okay, so after checking all primes less than 997, I didn't find any prime where the sum of digits is a composite odd integer. The first such prime I found is 997, where the sum of digits is 25, which is composite and odd. Therefore, 997 is the smallest prime number satisfying the given condition.