answer:To solve this problem, we need to find the smallest ( n ) such that the sequence of positive integers ( a_1, a_2, ldots, a_n ) each of which is (leq 15), and the sum ( a_1! + a_2! + ldots + a_n! ) has the last four digits 2001. 1. **Identify Relevant Factorials:** First, we examine the factorials of integers from ( 1 ) to ( 15 ) and their last four digits. - ( 1! = 1 ) - ( 2! = 2 ) - ( 3! = 6 ) - ( 4! = 24 ) - ( 5! = 120 ) - ( 6! = 720 ) - ( 7! = 5040 ) (last 4 digits = 5040) - ( 8! = 40320 ) (last 4 digits = 0320) - ( 9! = 362880 ) (last 4 digits = 2880) - ( 10! = 3628800 ) (last 4 digits = 8800) - ( 11! = 39916800 ) (last 4 digits = 6800) - ( 12! = 479001600 ) (last 4 digits = 1600) - ( 13! = 6227020800 ) (last 4 digits = 0800) - ( 14! = 87178291200 ) (last 4 digits = 1200) - ( 15! = 1307674368000 ) (last 4 digits = 8000) 2. **Check for the Possibility to get the Desired Sum:** We want to find ( a_1!, a_2!, ldots, a_n! ) such that their sum modulo 10,000 equals 2001. First, observe: - Only ( 1! ) is odd, yielding the last digit 1. - All other factorials either contribute even summands or end in 4 zeros from ( 10! ) onwards 3. **Derive a Solution:** Let's see possible contributions to make up the last four digits 2001 (need to have three numbers whose sum yields 2001): We observe that ( 1! ) contributes very little but is necessary because 2001's last digit is odd. Considering other factorials: - ( 1! = 1 ) - We need remaining sum ( 2000 = a_1! + a_2! - 1!). Given the noted factorials ending in last four digits: - ( 14! = 1200 ) - ( 13! = 800 ) (1200 + 800 = 2000) 4. **Verify the Correct Sum:** We can construct the sequence: [ 13! + 14! + 1! = 6227020800 + 87178291200 + 1 ] Now, take modulo 10000: [ 800 + 1200 + 1 = 2001 ] 5. **Conclusion:** The sequence ( (13!, 14!, 1!) ) achieves the desired last four digits sum. Thus, the smallest ( n ) such that ( a_1! + a_2! + ldots + a_n! ) yields a sum whose last four digits are 2001 is ( n = 3 ). [ boxed{3} ]

answer:If a^2 = b^2 + c^2, the equation x^2 + 2ax + b^2 = 0 is equivalent to x^2 + 2ax + a^2 - c^2 = 0, which can be factored as [x + (a + c)][x + (a - c)] = 0. The roots of the equation are x = -(a + c) or x = -(a - c). The equation x^2 + 2cx - b^2 = 0 is equivalent to x^2 + 2ax - (a^2 - c^2) = 0, which can be factored as [x + (a + c)][x + (c - a)] = 0. The roots of the equation are x = -(a + c) or x = -(c - a). Clearly, the two equations have a common root -(a + c), so the sufficiency holds. If the equations x^2 + 2ax + b^2 = 0 and x^2 + 2cx - b^2 = 0 have a common root, let the common root be m. Then m^2 + 2am + b^2 = 0 (1) and m^2 + 2cm - b^2 = 0 (2). Adding the two equations, we get 2m^2 + 2(a + c)m = 0. Since m = 0 is not a valid solution, we have m = -(a + c). Substituting m into equation (1), we get (a + c)^2 - 2a(a + c) + b^2 = 0. Simplifying, we obtain a^2 + 2ac + c^2 - 2a^2 - 2ac + b^2 = 0, which reduces to a^2 = b^2 + c^2. Thus, the necessity holds. Therefore, P is a necessary and sufficient condition for Q. Answer: boxed{text{Necessary and sufficient condition}}. This problem primarily tests the ability to judge necessary and sufficient conditions by applying the root-finding method for quadratic equations.

answer:**Analysis** Simplify the original equation to get the result. **Solution** Given the equation, we have begin{cases} x-2=0, y+2=0, end{cases} which leads to begin{cases} x=2, y=-2. end{cases} Therefore, the equation represents the point (2,-2). Hence, the correct choice is boxed{text{B}}.

answer:We start by recognizing the problem structure similar to the difference of squares: [ x^2 - y^2 = (x+y)(x-y) ] Substitute the expressions for ( x ) and ( y ): [ x + y = 125^{63} + 125^{-63} + 125^{63} - 125^{-63} = 2 cdot 125^{63} ] [ x - y = 125^{63} - 125^{-63} - (125^{63} + 125^{-63}) = -2 cdot 125^{-63} ] Now, multiply these results: [ x^2 - y^2 = (2 cdot 125^{63})(-2 cdot 125^{-63}) = -4 cdot (125^{63} cdot 125^{-63}) = -4 cdot 125^{63-63} = -4 cdot 125^0 = -4 cdot 1 ] Thus, the final answer is: [ boxed{-4} ]