answer:1. Determine the prime factorization of (12^{2007}) and (12^{2000}): [ 12 = 2^2 cdot 3 ] Therefore, [ 12^{2007} = (2^2 cdot 3)^{2007} = 2^{4014} cdot 3^{2007} ] and [ 12^{2000} = (2^2 cdot 3)^{2000} = 2^{4000} cdot 3^{2000} ]2. Calculate the number of divisors for (12^{2007}): The number of divisors of a number (n) with prime factorization (p_1^{a_1} cdot p_2^{a_2} cdot ldots cdot p_m^{a_m}) is given by: [ (a_1 + 1)(a_2 + 1) cdots (a_m + 1) ] For (12^{2007} = 2^{4014} cdot 3^{2007}), the number of divisors is: [ (4014 + 1)(2007 + 1) = 4015 cdot 2008 ]3. Calculate the number of divisors for (12^{2000}): Similarly, for (12^{2000} = 2^{4000} cdot 3^{2000}), the number of divisors is: [ (4000 + 1)(2000 + 1) = 4001 cdot 2001 ]4. Determine the probability that a randomly chosen positive divisor of (12^{2007}) is also a divisor of (12^{2000}): A divisor of (12^{2007}) is of the form (2^a cdot 3^b) where (0 leq a leq 4014) and (0 leq b leq 2007). For it to also be a divisor of (12^{2000}), it must satisfy (0 leq a leq 4000) and (0 leq b leq 2000). The number of such divisors is: [ (4000 + 1)(2000 + 1) = 4001 cdot 2001 ] Therefore, the probability is: [ frac{4001 cdot 2001}{4015 cdot 2008} ]5. Simplify the fraction (frac{4001 cdot 2001}{4015 cdot 2008}): We need to find (a) and (b) such that (frac{a}{b} = frac{4001 cdot 2001}{4015 cdot 2008}) and (a) and (b) are relatively prime. Since (4001) and (4015) are relatively prime, and (2001) and (2008) are relatively prime, the fraction is already in its simplest form: [ frac{4001 cdot 2001}{4015 cdot 2008} ]6. Calculate (a + b): [ a + b = 4001 cdot 2001 + 4015 cdot 2008 ]7. Find the remainder when (a + b) is divided by 2007: [ a + b = 4001 cdot 2001 + 4015 cdot 2008 ] We need to compute this modulo 2007. Notice that: [ 4001 equiv 1994 pmod{2007} ] [ 4015 equiv 2008 equiv 1 pmod{2007} ] [ 2001 equiv -6 pmod{2007} ] [ 2008 equiv 1 pmod{2007} ] Therefore, [ 4001 cdot 2001 + 4015 cdot 2008 equiv 1994 cdot (-6) + 1 cdot 1 pmod{2007} ] [ equiv -11964 + 1 pmod{2007} ] [ equiv -11963 pmod{2007} ] [ equiv 2007 - 11963 pmod{2007} ] [ equiv 79 pmod{2007} ]The final answer is (boxed{79})

answer:SolutionBIf Ria's number were 1, she would know, since she and Sylvie have consecutive positive integers written on their pieces of paper, that the number on Sylvie's piece of paper is 2. Therefore, when Ria says she does not know Sylvie's number, Sylvie can deduce that Ria's number is not 1 . Similarly, when Sylvie says she does not know Ria's number, Ria can deduce that Sylvie's number is not 1. Also if Sylvie's number were 2, she would be able to deduce that Ria's number is 3 since she knows it is not 1 . Therefore, since Sylvie says she does not know Ria's number, Ria can deduce that Sylvie's number is not 2 .Therefore, since Ria can now say that she knows Sylvie's number, it follows that either Ria's number is 2 and she can deduce that Slvie's number is 3, or that Ria's number is 3 and she can deduce that Sylvie's number is 4 . If Ria's number is greater than 3 she would not be able to deduce Sylvie's number.Therefore, of the options given, the only possibility for Ria's number is 2 .

answer:First, place n with any other n-1 numbers in the n-th row, the corresponding probability is p_{1}=frac{n}{frac{n(n+1)}{2}}=frac{2}{n+1}; Second, place the largest number among the remaining frac{n(n-1)}{2} numbers with any other n-2 numbers in the (n-1)-th row, the corresponding probability is p_{2}=frac{n-1}{frac{n(n-1)}{2}}=frac{2}{n};Third, place the largest number among the remaining frac{(n-1)(n-2)}{2} numbers with any other n-3 numbers in the (n-2)-th row, the corresponding probability is p_{3}=frac{n-2}{frac{(n-1)(n-2)}{2}}=frac{2}{n-1};And so on, in each step, place the largest number among the remaining numbers with any other k-1 numbers in the k-th row, until the last number is placed in the first row. Therefore, the probability of event A= " M_{1}<M_{2}<cdots<M_{n} " isP(A)=p_{1} p_{2} cdots p_{n}=frac{2}{n+1} cdot frac{2}{n} cdots 1=frac{2^{n}}{(n+1)!} .

answer:SolutionEStarting on the left, the first two boxes add to T so the second box is T-2021. The second and third boxes add to T+1 so the third box is 2022 . Continuing in this way, the numbers obtained are T-2022,2023, T-2023, and 2024. The final box is 2021 and the final two boxes have total T. So T=2024+2021=4045.

answer:Since (z-2)(1-i) = 1+i, we find z by solving the equation:begin{align*}z - 2 &= frac{1+i}{1-i} z &= frac{1+i}{1-i} + 2 end{align*}Next, we simplify the fraction by multiplying the numerator and the denominator by the complex conjugate of the denominator:begin{align*}z &= frac{(1+i)(1+i)}{(1-i)(1+i)} + 2 &= frac{1 + 2i - 1}{2} + 2 &= frac{2i}{2} + 2 &= i + 2.end{align*}Now we compute the modulus of z:begin{align*}|z| &= sqrt{(2)^2 + (1)^2} &= sqrt{4 + 1} &= sqrt{5}.end{align*}Hence the modulus of the complex number z is boxed{sqrt{5}}.

answer:Key observation. AD = 20sqrt{7}.Proof 1. By the [triangle inequality](https://artofproblemsolving.com/wiki/index.php/Triangle_inequality), we can immediately see that AD geq 20sqrt{7}. However, notice that 10sqrt{21} = 20sqrt{7}cdotsinfrac{pi}{3}, so by the law of sines, when AD = 20sqrt{7}, angle ACD is right and the circle centered at A with radius 10sqrt{21}, which we will call omega, is tangent to overline{CD}. Thus, if AD were increased, overline{CD} would have to be moved even farther outwards from A to maintain the angle of frac{pi}{3} and omega could not touch it, a contradiction.Proof 2. Again, use the triangle inequality to obtain AD geq 20sqrt{7}. Let x = AD and y = CD. By the law of cosines on triangle ADC, 2100 = x^2+y^2-xy iff y^2-xy+(x^2-2100) = 0. Viewing this as a quadratic in y, the discriminant Delta must satisfy Delta = x^2-4(x^2-2100) = 8400-3x^2 geq 0 iff x leq 20sqrt{7}. Combining these two inequalities yields the desired conclusion.This observation tells us that E, A, and D are collinear, in that order.Then, triangle ADC and triangle ACF are 30-60-90 triangles. Hence AF = 15sqrt {7}, andEF = EA + AF = 10sqrt {7} + 15sqrt {7} = 25sqrt {7}.Finally, the answer is 25+7=boxed{032}.