answer:1. **Rearrange the given recursion:** The given recursion is: [ a_{n+1} - a_n = n(a_n - 1) ] Rearrange it to: [ a_{n+1} - 1 = (n+1)(a_n - 1) ] 2. **Telescoping argument:** We can use a telescoping argument to find a general form for (a_n). Start with (a_0 = 3): [ a_0 - 1 = 2 ] For (n = 0): [ a_1 - 1 = 1(a_0 - 1) = 1 cdot 2 = 2 implies a_1 = 3 ] For (n = 1): [ a_2 - 1 = 2(a_1 - 1) = 2 cdot 2 = 4 implies a_2 = 5 ] For (n = 2): [ a_3 - 1 = 3(a_2 - 1) = 3 cdot 4 = 12 implies a_3 = 13 ] We observe that: [ a_n - 1 = 2 cdot n! ] Therefore: [ a_n = 2 cdot n! + 1 ] 3. **Check the gcd condition:** We need to determine all integers (m ge 2) for which (gcd(m, a_n) = 1) for all (n ge 0). Since (a_n = 2 cdot n! + 1), we note that (a_n) is always odd. Therefore, (m) must be odd to ensure (gcd(m, a_n) = 1). 4. **Check for primes (p ge 3):** We need to show that for any prime (p ge 3), there exists an (n) such that (p mid a_n). Consider (p = 3): [ a_0 = 3 implies 3 mid a_0 ] For (p ge 5), by Wilson's theorem, ((p-1)! equiv -1 pmod{p}). Therefore: [ (p-3)! equiv frac{-1}{(p-2)(p-1)} equiv frac{-1}{2 cdot (p-1)} equiv -frac{1}{2} pmod{p} ] Consequently: [ a_{p-3} = 2 cdot (p-3)! + 1 equiv 2 cdot left(-frac{1}{2}right) + 1 equiv -1 + 1 equiv 0 pmod{p} ] Thus, (p mid a_{p-3}). 5. **Conclusion:** Since (a_n) is odd for all (n) and for any prime (p ge 3), there exists an (n) such that (p mid a_n), the only possible values for (m) are powers of 2. The final answer is ( boxed{ m = 2^k } ), where (k ge 1).

answer:From log_{ frac {1}{2}}(3-x)geqslant -2, we have log_{ frac {1}{2}}(3-x)=log_{ frac {1}{2}}4, This implies 0 < 3-xleqslant 4, Solving for x, we get boxed{-1leqslant x < 3}, Hence, boxed{A=[-1,3)}, From 2^{x}-1geqslant 0, we solve for x and get boxed{xgeqslant 0}, Hence, boxed{B=[0,+infty)}, Therefore, boxed{A∩B=[0,3)}.

answer:Given a, b, c are positive real numbers with a+b+c = 12 and ab+bc+ca = 30, assume without loss of generality that a leq b leq c. Thus, 12 = a+b+c leq 3c, which implies c geq frac{12}{3} = 4. By the AM-GM inequality, we have: [ (a+b)^2 geq 4ab. ] Thus, [ (12 - c)^2 geq 4(30 - ac - bc) = 120 - 4(a+b)c = 120 - 4(12 - c)c. ] Rearranging and simplifying yields: [ 3c^2 - 24c + 36 geq 0 Rightarrow (c-4)(3c-9) geq 0. ] The quadratic factors to yield valid c values between 4 and 3 (which does not apply since c geq 4), or c geq 3. Therefore, c leq frac{24}{3} = 8. Now, considering m = min{ab, ac, bc}, we have: [ m = ab = 30 - c(a+b) = 30 - c(12-c) = -c^2 + 12c - 30. ] Setting -c^2 + 12c - 30 = 0, we solve for c: [ c = frac{-12 pm sqrt{144 + 120}}{-2} = frac{-12 pm sqrt{264}}{-2}. ] Solving yields c values, but we need the feasible range frac{12}{3} leq c leq 8 only. The maximum valid value for m maximizing ab is when c = 8, yielding m = -64 + 96 - 30 = 2. Conclusion: The largest possible value of m under these conditions is boxed{2}.

answer:First, identify g^{-1}(4) by finding x such that g(x) = 4. From the table, g(1) = 4, so g^{-1}(4) = 1. Therefore, g^{-1}(g^{-1}(g^{-1}(4))) = g^{-1}(g^{-1}(1)). Next, find g^{-1}(1) by locating x such that g(x) = 1. From the table, g(5) = 1, so g^{-1}(1) = 5. Now, g^{-1}(g^{-1}(1)) = g^{-1}(5). Finally, identify g^{-1}(5) by finding x such that g(x) = 5. From the table, g(2) = 5, so g^{-1}(5) = 2. Thus, g^{-1}(g^{-1}(g^{-1}(4))) = boxed{2}.