answer:(1) Starting from the given equation frac{overline{z}}{1+i} = 2+i, we want to find z. To do this, first multiply both sides by the complex number 1+i and get overline{z} = (1+i)(2+i). Now we evaluate the right side using algebraic expansion: begin{align*} overline{z} &= (1+i)(2+i) &= 1 cdot 2 + 1 cdot i + i cdot 2 + i cdot i &= 2 + i + 2i + i^2 &= 2 + 3i - 1 quad text{(since i^2 = -1)} &= 1 + 3i. end{align*} Since overline{z} = 1 + 3i, z is the conjugate of this number. So, we take the complex conjugate of 1 + 3i to find z: begin{align*} z &= overline{1 + 3i} &= 1 - 3i. end{align*} So the value of z is boxed{1 - 3i}. (2) Since m in mathbb{R} and the complex number z = frac{m(m+2)}{m-1} + (m^2+2m-3)i must be purely imaginary, the real part of z has to be zero. This means that the condition frac{m(m+2)}{m-1} = 0 must be satisfied. However, the denominator m-1 cannot be zero, so we can write this condition as: m(m+2) = 0. Solving this equation gives two possible values for m: begin{align*} m &= 0, m + 2 &= 0 m &= -2. end{align*} However, we must also consider the other constraint in the imaginary part of z, which should not be zero. The imaginary part of z is given by m^2+2m-3, and if we set it to zero, we have: begin{align*} m^2 + 2m - 3 &= 0 (m+3)(m-1) &= 0. end{align*} From this, m = -3 or m = 1. We discard m = 1 since it would make the real part undefined. Combining the values found from the real part and the constraints from the imaginary part, the possible values of m that make z purely imaginary are: m = 0 quad text{and} quad m = -2. However, since m = 0 makes the real part of z zero but also makes the imaginary part zero (which is not allowed), we can exclude m = 0. The only value of m that satisfies the conditions is m = -2. So the solution is boxed{m = -2}.

answer:To find how many times ( 24 ) divides into ( 100! ), we need to determine the highest power of ( 24 ) that divides ( 100! ). Since ( 24 = 2^3 cdot 3 ), we need to count the powers of ( 2^3 ) and ( 3 ) in ( 100! ). 1. **Counting Factors of ( 2 ) in ( 100! ):** The power of a prime ( p ) dividing ( n! ) (denoted ( nu_p(n!) )) can be calculated using the formula: [ nu_p(n!) = leftlfloor frac{n}{p} rightrfloor + leftlfloor frac{n}{p^2} rightrfloor + leftlfloor frac{n}{p^3} rightrfloor + cdots ] For ( p = 2 ) and ( n = 100 ), [ nu_2(100!) = leftlfloor frac{100}{2} rightrfloor + leftlfloor frac{100}{4} rightrfloor + leftlfloor frac{100}{8} rightrfloor + leftlfloor frac{100}{16} rightrfloor + leftlfloor frac{100}{32} rightrfloor + leftlfloor frac{100}{64} rightrfloor ] Calculating each term: [ leftlfloor frac{100}{2} rightrfloor = 50 ] [ leftlfloor frac{100}{4} rightrfloor = 25 ] [ leftlfloor frac{100}{8} rightrfloor = 12 ] [ leftlfloor frac{100}{16} rightrfloor = 6 ] [ leftlfloor frac{100}{32} rightrfloor = 3 ] [ leftlfloor frac{100}{64} rightrfloor = 1 ] Summing these up: [ nu_2(100!) = 50 + 25 + 12 + 6 + 3 + 1 = 97 ] 2. **Counting Factors of ( 3 ) in ( 100! ):** For ( p = 3 ) and ( n = 100 ), [ nu_3(100!) = leftlfloor frac{100}{3} rightrfloor + leftlfloor frac{100}{9} rightrfloor + leftlfloor frac{100}{27} rightrfloor + leftlfloor frac{100}{81} rightrfloor ] Calculating each term: [ leftlfloor frac{100}{3} rightrfloor = 33 ] [ leftlfloor frac{100}{9} rightrfloor = 11 ] [ leftlfloor frac{100}{27} rightrfloor = 3 ] [ leftlfloor frac{100}{81} rightrfloor = 1 ] Summing these up: [ nu_3(100!) = 33 + 11 + 3 + 1 = 48 ] 3. **Counting Factors of ( 24 ) in ( 100! ):** Since ( 24 = 2^3 cdot 3 ), we need to find how many times ( 2^3 ) and ( 3 ) divide ( 100! ): [ leftlfloor frac{nu_2(100!)}{3} rightrfloor = leftlfloor frac{97}{3} rightrfloor = leftlfloor 32.33 rightrfloor = 32 ] [ nu_3(100!) = 48 ] The number of times ( 24 ) divides ( 100! ) is determined by the limiting factor, which in this case is ( leftlfloor frac{nu_2 (100!)}{3} rightrfloor ). Therefore, the number of times ( 24 ) divides into ( 100! ) is ( 32 ). [ boxed{32} ]

answer:Given the problem: [ sum_{i=0}^{p} C_{p}^{i} C_{p}^{i} a^{p-i} b^{i} = sum_{i=0}^{p} C_{p}^{i} C_{q+i}^{i}(a-b)^{p-i} b^{i} ] 1. Start by considering the expansion of ((ax + b)^p (1 + x)^q): [ (ax + b)^p (1 + x)^q = left[ sum_{i=0}^{p} C_p^i (ax)^{p-i} b^i right] left[ sum_{i=0}^{q} C_q^j x^j right] ] 2. Simplify the left-hand side expression to observe the coefficient of (x^p): [ (ax + b)^p (1 + x)^q = left[ sum_{i=0}^{p} C_p^i a^{p-i} x^{p-i} b^i right] left[ sum_{j=0}^{q} C_q^j x^j right] ] 3. The coefficient of (x^p) in this expansion can be found by matching the powers of (x): [ text{Coefficient of } x^p = sum_{i=0}^{p} C_p^i a^{p-i} b^i C_q^{p-i} = sum_{i=0}^{p} C_p^i C_q^{p-i} a^{p-i} b^i ] 4. Now let's consider another way of expanding the original expression: [ (ax + b)^p (1 + x)^q = left[ x(a-b) + b(1 + x) right]^p (1 + x)^q ] 5. We can expand this new expression using the binomial theorem: [ left[ x(a-b) + b(1 + x) right]^p (1 + x)^q = sum_{i=0}^{p} C_p^i left[ x(a - b) right]^{p-i} left[ b(1 + x) right]^i (1 + x)^q ] 6. Simplify the terms inside the binomial expansion: [ = sum_{i=0}^{p} C_p^i (a - b)^{p-i} x^{p-i} b^i (1 + x)^{q+i} ] 7. Further, observe that we need the coefficient of (x^p): [ sum_{i=0}^{p} C_p^i (a - b)^{p-i} b^i (1 + x)^{q+i} x^{p-i} ] 8. Find the coefficient of (x^p) in this final expansion: [ text{Coefficient of } x^p = sum_{i=0}^{p} C_p^i (a - b)^{p-i} b^i C_{q+i}^{p-i} ] 9. Comparing the coefficients of (x^p) from both expansions: [ sum_{i=0}^{p} C_p^i C_{q}^{p-i} a^{p-i} b^i = sum_{i=0}^{p} C_p^i C_{q+i}^{p-i} (a - b)^{p-i} b^i ] Thus: [ boxed{sum_{i=0}^{p} C_p^i C_{q+i}^{p-i} a^{p-i} b^i = sum_{i=0}^{p} C_p^i C_{q+i}^{p-i} (a - b)^{p-i} b^i} ]

answer:Let's denote the man's speed in still water (without the current of the stream) as ( v ) kmph. When the man goes downstream, the speed of the stream adds to his speed in still water, so his downstream speed is ( v + 3 ) kmph. When the man goes upstream, the speed of the stream subtracts from his speed in still water, so his upstream speed is ( v - 3 ) kmph. We are given that the man's upstream speed is 8 kmph. Therefore, we can set up the following equation: [ v - 3 = 8 ] Now, we solve for ( v ): [ v = 8 + 3 ] [ v = 11 ] kmph So the man's speed in still water is 11 kmph. Now we can find his downstream speed: [ text{Downstream speed} = v + 3 ] [ text{Downstream speed} = 11 + 3 ] [ text{Downstream speed} = 14 ] kmph The man's speed downstream is boxed{14} kmph.