answer:Step 1: Establish the maximum possible value The sum of the digits of any four-digit number is at most 36 (since we are considering numbers less than (1000)). So, the maximum number to be considered as (7) times the sum of its digits is (7 times 36 = 252). Step 2: Simplify the problem using divisibility Any number (n) that equals (7 times text{sum of its digits}) must be divisible by 7, and its sum of digits divisible by 7. This simplifies our check to sums like (7, 14, 21, 28, 35). Step 3: Calculate each candidate We calculate each possible sum and do a digit sum check: - (7 times 7 = 49) (Sum of digits is (4 + 9 = 13)) - (14 times 7 = 98) (Sum of digits is (9 + 8 = 17)) - (21 times 7 = 147) (Sum of digits is (1 + 4 + 7 = 12)) - (28 times 7 = 196) (Sum of digits is (1 + 9 + 6 = 16)) - (35 times 7 = 245) (Sum of digits is (2 + 4 + 5 = 11)) Step 4: Identifying valid numbers From the calculations above, none of the sums matches, hence there are no numbers that are seven times the sum of their digits. Conclusion: No integers less than or equal to (1000) meet the condition that they are (7) times the sum of their digits. Thus, the correct answer is 0. The final answer is boxed{textbf{(A)} 0}

answer:To find the imaginary part of z=frac{1}{{(2+i)}^{2}}, we first simplify the expression step by step: 1. Calculate the square of (2+i): (2+i)^2 = 2^2 + 2cdot2cdot i + i^2 = 4 + 4i - 1 = 3 + 4i 2. Invert the result to find z: z = frac{1}{(2+i)^2} = frac{1}{3+4i} 3. To simplify the fraction, multiply the numerator and the denominator by the conjugate of the denominator: z = frac{1}{3+4i} cdot frac{3-4i}{3-4i} = frac{3-4i}{(3+4i)(3-4i)} 4. Calculate the denominator using the difference of squares formula: (3+4i)(3-4i) = 3^2 - (4i)^2 = 9 - 16(-1) = 9 + 16 = 25 5. Substitute the denominator back into the expression: z = frac{3-4i}{25} = frac{3}{25} - frac{4}{25}i Thus, the imaginary part of z is -frac{4}{25}. Encapsulating the final answer: boxed{-frac{4}{25}} Therefore, the correct choice is boxed{C}.

answer:1. **Compute the Prime Factorization of the Product of Selected Factorials:** The product is 1! cdot 3! cdot 5! cdot 7! cdot 9! cdot 11!. We will calculate the exponent of each prime in this product: - **Prime 2:** (Only even numbers contribute a factor of 2) - 3! contributes 1 factor of 2 - 5! contributes 3 factors of 2 - 7! contributes 4 factors of 2 - 9! contributes 7 factors of 2 - 11! contributes 8 factors of 2 - Summing these, we get 1+3+4+7+8 = 23 factors of 2. - **Prime 3:** - 3! contributes 1 factor of 3 - 5! contributes 1 factor of 3 - 7! contributes 2 factors of 3 - 9! contributes 4 factors of 3 - 11! contributes 3 factors of 3 - Summing these, we get 1+1+2+4+3 = 11 factors of 3. - **Prime 5:** - 5! contributes 1 factor of 5 - 7! contributes 1 factor of 5 - 9! contributes 1 factor of 5 - 11! contributes 2 factors of 5 - Summing these, we get 1+1+1+2 = 5 factors of 5. - **Prime 7:** - 7! contributes 1 factor of 7 - 9! contributes 1 factor of 7 - 11! contributes 1 factor of 7 - Summing these, we get 1+1+1 = 3 factors of 7. 2. **Calculate the Number of Perfect Square Divisors:** - For 2^{23}, the exponents can be 0, 2, 4, ldots, 22, giving 12 choices. - For 3^{11}, the exponents can be 0, 2, 4, ldots, 10, giving 6 choices. - For 5^5, the exponents can be 0, 2, 4, giving 3 choices. - For 7^3, the exponents can be 0, 2, giving 2 choices. Multiplying these choices gives the total number of perfect square divisors: [ 12 times 6 times 3 times 2 = 432 ] 3. **Conclusion:** The number of perfect square divisors of the product 1! cdot 3! cdot 5! cdot ldots cdot 11! is 432. The final answer is boxed{textbf{(A)} 432}

answer:1. **Setup Basic Geometry**: Given that ABCD is a trapezoid with AB parallel DC, drop perpendiculars from points A and B to line DC, meeting at points F and E respectively. 2. **Analysis of Triangles and Rectangle**: Dropping perpendiculars gives us two right triangles, triangle BFE and triangle AFD, and a rectangle ABEF, with AB = EF = 7 due to parallel sides. 3. **30-60-90 Triangle at BCD**: With angle BCD = 30^circ, triangle BCE is a 30-60-90 triangle. The shortest leg EC across from the 30^circ angle is half the hypotenuse BC: [ EC = frac{1}{2} times BC = frac{1}{2} times 4sqrt{3} = 2sqrt{3}. ] 4. **60-90-30 Triangle at CDA**: angle CDA = 60^circ makes triangle AFD a 30-60-90 triangle. Here, AF (height of rectangle ABEF) equals EB = 7. Hence, the longer leg DF opposite the 60^circ angle: [ DF = sqrt{3} times AF = sqrt{3} times 7 = 7sqrt{3}. ] 5. **Calculate DC**: Summing up the lengths EC, EF, and FD gives: [ DC = EC + EF + FD = 2sqrt{3} + 7 + 7sqrt{3} = 9sqrt{3} + 7. ] 6. **Conclusion**: The length of DC is 7 + 9sqrt{3}. The final answer is boxed{C}.