answer:First, find the derivative of y=ax^{3}-6x^{2}+12x: y'=3ax^{2}-12x+12 Next, evaluate the derivative at x=1: y'|_{x=1}=3a Now, find the derivative of y=e^{x}: y'=e^{x} Then, evaluate the derivative at x=1: y'|_{x=1}=e Since the tangent lines of curves C_1: y=ax^{3}-6x^{2}+12x and C_2: y=e^{x} at x=1 are parallel, their slopes must be equal: 3a=e Solving for a, we get: a=boxed{frac{e}{3}}

answer:1. **Initial Setup:** - Let ( a ) and ( n ) be two non-zero natural numbers that are coprime (i.e., ( gcd(a, n) = 0 )). - We aim to show that there exists an integer ( d ) such that for every integer ( b in mathbb{N} ), ( n ) divides ( a^b - 1 ) if and only if ( d ) divides ( b ). 2. **Using the Pigeonhole Principle:** - Considering the sequence ( 1, a, a^2, a^3, ldots ) modulo ( n ), it must be periodic because there are only ( n ) possible remainders when dividing by ( n ). - By the pigeonhole principle, there exist integers ( u < v ) such that [ a^u equiv a^v pmod{n} ] - This implies: [ a^v - a^u equiv 0 pmod{n} quad Rightarrow quad n mid (a^v - a^u) ] [ n mid a^u (a^{v-u} - 1) ] - Since ( gcd(a, n) = 1 ), by the theorem of Gauss, if ( n ) divides ( a^u (a^{v-u} - 1) ) and ( gcd(a^u, n) = 1 ), then [ n mid a^{v-u} - 1 ] 3. **Smallest Positive Integer ( d ):** - There exists a smallest positive integer ( d ) such that: [ a^d equiv 1 pmod{n} ] - This ( d ) is known as the order of ( a ) modulo ( n ). 4. **Multiples of the Order:** - For any natural integer ( k ), we have: [ a^{dk} equiv (a^d)^k equiv 1^k equiv 1 pmod{n} ] 5. **General ( b ):** - Let ( b ) be any natural number, and consider the Euclidean division of ( b ) by ( d ): [ b = qd + r quad text{with} quad 0 leq r < d ] - Using this division, we can write: [ a^b = a^{qd+r} = (a^d)^q cdot a^r equiv 1^q cdot a^r equiv a^r pmod{n} ] 6. **Condition for ( b ) :** - If we assume ( n mid (a^b - 1) ), then: [ a^b equiv 1 pmod{n} ] - From our division, ( a^r equiv 1 pmod{n} ) with ( 0 leq r < d ). - Because ( d ) is the smallest positive integer such that ( a^d equiv 1 pmod{n} ), and ( r < d ), ( r ) must be ( 0 ). Thus, ( r = 0 ), and: [ b = qd quad Rightarrow quad d mid b ] 7. **Conclusion:** - We have shown that ( n mid (a^b - 1) ) if and only if ( d mid b ). [ boxed{} ]

answer:1. **Define the function ( f(n) ):** Given ( d(n) ) denotes the number of positive divisors of ( n ), we define ( f(n) ) as: [ f(n) = d(k_1) + d(k_2) + cdots + d(k_m) ] where ( 1 = k_1 < k_2 < cdots < k_m = n ) are all divisors of ( n ). 2. **Establish the multiplicative property of ( f(n) ):** We need to show that ( f(n) ) is multiplicative. Let ( g(n) ) be the number of divisors of ( n ). Therefore, [ f(n) = sum_{d|n} g(d) ] By Dirichlet Convolution, since ( g(n) ) is multiplicative, ( f(n) ) is also multiplicative. 3. **Calculate ( f(n) ) for prime powers:** Consider ( n = p^e ) where ( p ) is a prime and ( e ) is a positive integer. The divisors of ( p^e ) are ( 1, p, p^2, ldots, p^e ). Therefore, [ f(p^e) = d(1) + d(p) + d(p^2) + cdots + d(p^e) ] Since ( d(p^k) = k+1 ) for ( k = 0, 1, 2, ldots, e ), we have: [ f(p^e) = 1 + 2 + 3 + cdots + (e+1) = frac{(e+1)(e+2)}{2} ] 4. **Define a new function ( H_p(e) ):** Let ( n = prod_{i=1}^{k} p_i^{e_i} ). Define: [ H_p(e) coloneqq frac{2p^e}{(e+1)(e+2)} ] By definition, ( H_p(0) = 1 ) for all primes ( p ). 5. **Solve the equation for ( f(n) ):** Plugging in the equation for ( f(n) ) reduces the problem to solving: [ 1 = prod_{i=1}^{k} frac{2p_i^{e_i}}{(e_i+1)(e_i+2)} ] or equivalently, [ 1 = prod_{i=1}^{k} H_{p_i}(e_i) ] 6. **Analyze the behavior of ( H_p(e) ):** Notice that: [ H_p(e) - H_p(e-1) = 2p^{e-1} left( frac{pe - e - 2}{(e)(e+1)(e+2)} right) ] This shows that: - ( H_p(e) ) is strictly increasing for all ( p geq 5 ), - ( H_3(e) ) is strictly increasing for ( e geq 1 ), - ( H_2(e) ) is strictly increasing for ( e geq 2 ). 7. **Casework on ( H_2(e) ) and ( H_3(e) ):** Some lower bounds and casework show that any prime ( geq 5 ) can't divide ( n ). Doing casework on ( H_2(e) ) and ( H_3(e) ) reveals the answers. 8. **Conclusion:** The only almost perfect numbers are ( n in {1, 3, 18, 36} ). The final answer is ( boxed{ n in {1, 3, 18, 36} } )

answer:To solve this, we can consider 2009 as the product of integers in the forms of 1 times 1 times 2009, 1 times 7 times 287, 7 times 7 times 41, and 1 times 49 times 41. There are two scenarios to consider for the three unknowns: all three are positive, or one is positive and two are negative. First, let's consider the case where xyz are all positive. There are three possible sets of solutions for each form, for example, 1 times 1 times 2009, 1 times 2009 times 1, and 2009 times 1 times 1. In the case where one is positive and two are negative, the original x, y, z can have 3 variations; For instance, 1 times 1 times 2009 = 1 times (-1) times (-2009) = (-1) times (-1) times 2009 = (-1) times 1 times (-2009), Thus, in the scenario with one positive and two negatives, the original three possible sets of solutions will generate 9 possible sets of solutions. Therefore, for a grouping like 1 times 1 times 2009, there are in total 12 possible sets of solutions, Similarly, for 7 times 7 times 41, there are also 12 possible sets of solutions. However, for 1 times 7 times 287, since all three numbers are different, unlike the previous two cases, there are 6 possible sets of solutions when all are positive, and 18 possible sets of solutions when two are negative and one is positive. Similarly, 1 times 49 times 41 also has 24 possible sets of solutions. In summary, there are a total of boxed{72} groups of integer solutions for xyz = 2009. By decomposing xyz = 2009 into 1 times 1 times 2009, 1 times 7 times 287, 7 times 7 times 41, 1 times 49 times 41, and considering each case separately, we can find the number of solution sets. This problem tests the method of finding integer solutions to equations, where trial and error is a basic and effective method.