💡Okay, so I have this problem where I need to find the number of subsets ( B ) of the set ( {1, 2, ldots, 2005} ) such that the sum of the elements of ( B ) is congruent to 2006 modulo 2048. Hmm, that seems a bit tricky, but let me try to break it down step by step.First, I know that a subset can have any number of elements from 0 to 2005, and each element is either included or excluded. So, in total, there are ( 2^{2005} ) possible subsets. But we need only those subsets where the sum of the elements is congruent to 2006 modulo 2048. That means the sum can be 2006, 2006 + 2048, 2006 + 2*2048, and so on, but since we're dealing with modulo 2048, we just need the remainder when the sum is divided by 2048 to be 2006.I remember that when dealing with sums modulo a power of 2, especially 2048 which is ( 2^{11} ), the binary representation of numbers can be useful. Each number can be represented as a sum of powers of 2, and since we're dealing with modulo ( 2^{11} ), the higher bits beyond the 11th bit don't affect the result.Let me consider the set ( {1, 2, ldots, 2005} ). I can separate this set into two parts: one part containing the powers of 2 up to 1024 (since ( 2^{10} = 1024 ) and ( 2^{11} = 2048 ) is larger than 2005), and the other part containing the remaining numbers.So, let me define ( R = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024} ). This set ( R ) has 11 elements, each a power of 2. The remaining elements of ( {1, 2, ldots, 2005} ) form the set ( T = {3, 5, 6, 7, 9, ldots, 2005} ). The number of elements in ( T ) is ( 2005 - 11 = 1994 ).Now, any subset ( B ) can be written as ( B = B_R cup B_T ), where ( B_R subseteq R ) and ( B_T subseteq T ). The sum of the elements of ( B ) is then the sum of the elements of ( B_R ) plus the sum of the elements of ( B_T ).Since we're interested in the sum modulo 2048, the sum of ( B_R ) can be any number from 0 to 2047 because each element in ( R ) is a distinct power of 2, and their sum can represent any number in binary up to ( 2^{11} - 1 = 2047 ). On the other hand, the sum of ( B_T ) can be any integer, but when taken modulo 2048, it can also be any number from 0 to 2047.Here's the key insight: For any subset ( B_T subseteq T ), the sum ( S_T ) modulo 2048 can be adjusted by choosing an appropriate subset ( B_R subseteq R ) such that ( S_T + S_R equiv 2006 pmod{2048} ). Since the elements of ( R ) are powers of 2, each subset ( B_R ) corresponds to a unique number from 0 to 2047. Therefore, for each possible ( S_T mod 2048 ), there is exactly one subset ( B_R ) that will make the total sum congruent to 2006 modulo 2048.This means that the number of valid subsets ( B ) is equal to the number of subsets ( B_T ), which is ( 2^{1994} ). Each of these subsets can be paired with exactly one subset ( B_R ) to achieve the desired sum modulo 2048.Wait, let me make sure I didn't make a mistake here. So, if I fix ( B_T ), then ( S_T ) is determined, and I can find a unique ( B_R ) such that ( S_R equiv 2006 - S_T pmod{2048} ). Since ( R ) has 11 elements, each corresponding to a bit in the binary representation, this should always be possible. Therefore, for each of the ( 2^{1994} ) subsets of ( T ), there is exactly one subset of ( R ) that makes the total sum congruent to 2006 modulo 2048.Yes, that seems correct. So, the total number of subsets ( B ) is ( 2^{1994} ).**Final Answer**The number of such subsets ( B ) is ( boxed{2^{1994}} ).

💡Okay, so I have this problem here where I need to find the smallest positive perfect cube that has ( n = a^2b^3c^5 ) as a divisor. The options are given as A) ( a^3b^6c^6 ), B) ( a^2b^3c^5 ), C) ( (abc^2)^3 ), D) ( (ab^2c^3)^3 ), and E) ( a^6b^6c^6 ). First, let me understand what a perfect cube is. A perfect cube is a number that can be expressed as some integer raised to the power of three. In terms of prime factorization, each prime's exponent must be a multiple of 3. So, for a number to be a perfect cube, all the exponents in its prime factorization must be divisible by 3.Given that ( n = a^2b^3c^5 ), I need to find the smallest perfect cube that ( n ) divides into. That means the perfect cube must have exponents in its prime factors that are at least as large as those in ( n ), but also multiples of 3.Let me break this down prime by prime:1. **For prime ( a ):** The exponent in ( n ) is 2. To make it a perfect cube, the exponent needs to be a multiple of 3. The smallest multiple of 3 that is greater than or equal to 2 is 3. So, the exponent for ( a ) in the perfect cube should be 3.2. **For prime ( b ):** The exponent in ( n ) is 3. Since 3 is already a multiple of 3, we don't need to increase it. So, the exponent for ( b ) remains 3.3. **For prime ( c ):** The exponent in ( n ) is 5. The smallest multiple of 3 that is greater than or equal to 5 is 6. So, the exponent for ( c ) in the perfect cube should be 6.Putting this all together, the smallest perfect cube that ( n ) divides into should have the prime factorization ( a^3b^3c^6 ). Now, let me look at the options:- **Option A:** ( a^3b^6c^6 ). Here, the exponent for ( b ) is 6, which is more than necessary. Since we only need 3, this might not be the smallest.- **Option B:** ( a^2b^3c^5 ). This is just ( n ) itself, but it's not a perfect cube because the exponents for ( a ) and ( c ) aren't multiples of 3.- **Option C:** ( (abc^2)^3 ). Let me expand this: ( a^3b^3c^6 ). That's exactly what I derived earlier.- **Option D:** ( (ab^2c^3)^3 ). Expanding this gives ( a^3b^6c^9 ). The exponents for ( b ) and ( c ) are higher than needed, so this isn't the smallest.- **Option E:** ( a^6b^6c^6 ). All exponents are 6, which is more than necessary for ( a ) and ( b ), so this isn't the smallest either.So, Option C seems to be the correct answer because it's the smallest perfect cube that includes all the necessary exponents for ( a ), ( b ), and ( c ) while keeping them as low as possible.Wait, let me double-check. If I take ( (abc^2)^3 ), that's ( a^3b^3c^6 ). Comparing this to ( n = a^2b^3c^5 ), yes, each exponent in the cube is greater than or equal to those in ( n ), and all exponents are multiples of 3. So, it's a perfect cube and ( n ) divides into it.I think that's solid. So, the answer should be Option C.

💡Okay, so I have this problem here about replacing a light bulb. The old one is rated at 110 watts, and the new one has 30% higher wattage. I need to find out how many watts the new light has. Hmm, let's see. First, I think I need to understand what it means for the new light to have 30% higher wattage. I guess that means the new light uses 30% more power than the old one. So, if the old one is 110 watts, the new one is 110 plus 30% of 110. Wait, how do I calculate 30% of 110? I remember that percentages can be converted to decimals for calculations. So, 30% is the same as 0.30. To find 30% of 110, I can multiply 110 by 0.30. Let me write that down: 110 * 0.30. Let me do the multiplication. 110 times 0.30. Well, 100 times 0.30 is 30, and 10 times 0.30 is 3, so adding those together, 30 + 3 equals 33. So, 30% of 110 is 33 watts. Now, to find the total wattage of the new light, I need to add this 33 watts to the original 110 watts. So, 110 + 33 equals 143. Therefore, the new light has a wattage of 143 watts. Wait, let me double-check my calculations to make sure I didn't make a mistake. 30% of 110 is indeed 33, and adding that to 110 gives 143. Yeah, that seems right. I think I got it. The new light has 143 watts. That makes sense because increasing the wattage by 30% means it's more powerful than the original 110-watt bulb.

💡First, I need to determine the charge for the first 1/5 of a mile, which I'll denote as ( X ).The total distance of the ride is 8 miles. Since the taxi company charges for every additional 1/5 of a mile after the first 1/5 mile, I need to calculate how many 1/5 mile increments are in 8 miles.There are ( 5 times 8 = 40 ) increments of 1/5 mile in 8 miles.Since the first 1/5 mile is charged at a different rate, I subtract one increment for the first 1/5 mile:( 40 - 1 = 39 ) increments of 1/5 mile are charged at 0.40 each.The total charge for these 39 increments is ( 39 times 0.40 = 15.60 ).The total charge for the ride is 18.40, and I know that 15.60 of this is for the 39 increments after the first 1/5 mile. Therefore, the charge for the first 1/5 mile is the total charge minus the charge for the remaining increments:( X = 18.40 - 15.60 )( X = 2.80 )So, the charge for the first 1/5 of a mile is boxed{2.80}.

💡Okay, so I have this quadratic equation: (x^2 + (k^2 + ak)x + 1999 + k^2 + ak = 0). The problem says that both roots are prime numbers, and I need to find the value of (a). Hmm, let me think about how to approach this.First, I remember that for a quadratic equation (x^2 + bx + c = 0), the sum of the roots is (-b) and the product is (c). So in this case, if the roots are prime numbers (p) and (q), then:1. (p + q = -(k^2 + ak))2. (pq = 1999 + k^2 + ak)Hmm, interesting. So if I add these two equations together, I get:(p + q + pq = 1999)Wait, that looks like it can be factored. Let me see:(p + q + pq + 1 = 2000)Which is ((p + 1)(q + 1) = 2000). Oh, that's a useful factorization!So now, I need to find two primes (p) and (q) such that when I add 1 to each, their product is 2000. Let me factorize 2000 to find possible pairs.2000 factors into (2^4 times 5^3). So the factors of 2000 are combinations of these primes. Let me list the factors:1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000, 2000.Now, since (p) and (q) are primes, (p + 1) and (q + 1) must be factors of 2000. Also, (p + 1) and (q + 1) must be greater than 2 because the smallest prime is 2, so (p + 1) would be at least 3.Let me look for pairs of factors of 2000 where both factors minus 1 are primes.Starting with the smaller factors:- 2 and 1000: (2 - 1 = 1) (not prime), so discard.- 4 and 500: (4 - 1 = 3) (prime), (500 - 1 = 499) (prime). Okay, that's a valid pair.- 5 and 400: (5 - 1 = 4) (not prime), discard.- 8 and 250: (8 - 1 = 7) (prime), (250 - 1 = 249) (249 is divisible by 3, not prime). Discard.- 10 and 200: (10 - 1 = 9) (not prime), discard.- 20 and 100: (20 - 1 = 19) (prime), (100 - 1 = 99) (not prime). Discard.- 25 and 80: (25 - 1 = 24) (not prime), discard.- 40 and 50: (40 - 1 = 39) (not prime), discard.So the only valid pair is 4 and 500, which gives primes 3 and 499. Let me check if 499 is a prime. Yeah, 499 is a prime number because it's not divisible by any number except 1 and itself.So, the roots are 3 and 499. Now, let's go back to Vieta's formulas.Sum of roots: (3 + 499 = 502 = -(k^2 + ak))Product of roots: (3 times 499 = 1497 = 1999 + k^2 + ak)Wait, so from the sum, we have:(502 = -(k^2 + ak)) => (k^2 + ak = -502)From the product:(1497 = 1999 + k^2 + ak)Substitute (k^2 + ak) from the sum into the product equation:(1497 = 1999 + (-502))Calculate that: (1999 - 502 = 1497). Yep, that checks out.So, we have (k^2 + ak = -502). Let me write that as:(k^2 + ak + 502 = 0)This is a quadratic equation in terms of (k). The problem states that there's a unique real number (k), so the discriminant of this quadratic must be zero.Discriminant (D = a^2 - 4 times 1 times 502 = 0)So:(a^2 - 2008 = 0)Therefore, (a^2 = 2008)Since (a) is a positive real number, (a = sqrt{2008})Simplify (sqrt{2008}):2008 divided by 4 is 502, so (sqrt{2008} = sqrt{4 times 502} = 2sqrt{502})So, (a = 2sqrt{502})Let me double-check everything to make sure I didn't make a mistake.1. Factored 2000 correctly into ((p + 1)(q + 1)).2. Found the only valid prime pair: 3 and 499.3. Calculated sum and product correctly.4. Set up the quadratic in (k) correctly.5. Calculated discriminant correctly, leading to (a = 2sqrt{502}).Everything seems to check out. I think that's the correct answer.**Final Answer**boxed{2sqrt{502}}

💡Okay, so I have this problem here where I need to find the algebraic cofactor of the element 7 in a third-order determinant. The determinant is given as:begin{vmatrix} 3 & 7 & 4 x+a & 5 & 2 1 & 0 & x end{vmatrix}And they denote this cofactor as ( f(x) ). Then, it says that the solution set of the inequality ( f(x) > 0 ) is ( (-1, b) ), and I need to find the value of ( a + b ).Alright, let's break this down step by step. First, I need to recall what an algebraic cofactor is. From what I remember, the algebraic cofactor of an element in a determinant is calculated by taking the determinant of the minor matrix (which is the matrix obtained by deleting the row and column of that element) and then multiplying it by ( (-1)^{i+j} ), where ( i ) and ( j ) are the row and column indices of the element.So, in this case, the element is 7, which is in the first row and second column. So, ( i = 1 ) and ( j = 2 ). Therefore, the cofactor should be ( (-1)^{1+2} ) times the determinant of the minor matrix.Let me write that out. The minor matrix for element 7 is obtained by removing the first row and second column. So, the minor matrix would be:begin{vmatrix} x + a & 2 1 & x end{vmatrix}So, the determinant of this minor matrix is:( (x + a)(x) - (2)(1) = x(x + a) - 2 = x^2 + a x - 2 )Then, the algebraic cofactor is ( (-1)^{1+2} ) times this determinant. Since ( 1 + 2 = 3 ), ( (-1)^3 = -1 ). So, the cofactor is:( - (x^2 + a x - 2) = -x^2 - a x + 2 )So, that's ( f(x) ). So, ( f(x) = -x^2 - a x + 2 ).Now, the problem states that the solution set of the inequality ( f(x) > 0 ) is ( (-1, b) ). So, I need to find the values of ( a ) and ( b ) such that when I solve ( -x^2 - a x + 2 > 0 ), the solution is ( (-1, b) ).Let me write the inequality:( -x^2 - a x + 2 > 0 )I can multiply both sides by -1 to make it easier, but I have to remember to reverse the inequality sign. So, that gives:( x^2 + a x - 2 < 0 )So, now, the inequality is ( x^2 + a x - 2 < 0 ). The solution set of this inequality is ( (-1, b) ), which means that the quadratic equation ( x^2 + a x - 2 = 0 ) has roots at ( x = -1 ) and ( x = b ).So, if the roots are ( x = -1 ) and ( x = b ), then the quadratic can be written as ( (x + 1)(x - b) = 0 ). Let me expand this:( (x + 1)(x - b) = x^2 - b x + x - b = x^2 + (1 - b)x - b )But we also have the quadratic ( x^2 + a x - 2 ). So, these two quadratics must be equal. Therefore, their coefficients must be equal.So, equating the coefficients:1. Coefficient of ( x^2 ): 1 = 1. That's fine.2. Coefficient of ( x ): ( 1 - b = a )3. Constant term: ( -b = -2 )From the constant term, ( -b = -2 ) implies ( b = 2 ).Then, from the coefficient of ( x ), ( 1 - b = a ). Since ( b = 2 ), this becomes ( 1 - 2 = a ), so ( a = -1 ).Therefore, ( a = -1 ) and ( b = 2 ). So, ( a + b = -1 + 2 = 1 ).Wait, let me double-check my steps to make sure I didn't make a mistake.1. Calculated the minor matrix correctly by removing the first row and second column.2. Calculated the determinant of the minor matrix as ( x^2 + a x - 2 ).3. Applied the cofactor sign correctly with ( (-1)^{1+2} = -1 ), so cofactor is ( -x^2 - a x + 2 ).4. Set up the inequality ( -x^2 - a x + 2 > 0 ), which becomes ( x^2 + a x - 2 < 0 ).5. Since the solution is ( (-1, b) ), the quadratic must factor as ( (x + 1)(x - b) ).6. Expanded and equated coefficients, found ( b = 2 ) and ( a = -1 ).7. Sum ( a + b = 1 ).Everything seems to check out. I think that's the correct answer.**Final Answer**The value of ( a + b ) is boxed{1}.