answer:1. **Claim 1:** - We need to show that (1001) is the order of (10) modulo (2003). - The order of an integer (a) modulo (n) is the smallest positive integer (k) such that (a^k equiv 1 pmod{n}). - To prove this, we need to show that (10^{1001} equiv 1 pmod{2003}) and that no smaller positive integer (k) satisfies this condition. - **Proof:** - Consider the set of all positive divisors of (1001). The divisors are (1, 7, 11, 13, 77, 91, 143, 1001). - We need to check if any of these divisors (other than (1001)) is the order of (10) modulo (2003). - After calculations, we find that none of these numbers (except (1001)) is an order modulo (2003). - We verify that (10^{1001} equiv 1 pmod{2003}) through detailed calculations: [ 10^4 equiv -15 pmod{2003} ] [ 15^4 equiv 550 pmod{2003} ] [ 10^{16} equiv 550 pmod{2003} ] [ 550^2 equiv 47 pmod{2003} ] [ 10^{32} equiv 47 pmod{2003} ] [ 10^{64} equiv 206 pmod{2003} ] [ 10^{96} equiv 206 cdot 47 equiv 1670 pmod{2003} ] [ 10^{100} equiv 1670 cdot (-15) equiv 989 pmod{2003} ] [ 10^{200} equiv 989^2 equiv 657 pmod{2003} ] [ 10^{400} equiv 1004 pmod{2003} ] [ 10^{800} equiv 507 pmod{2003} ] [ 10^{1000} equiv 657 cdot 507 equiv 601 pmod{2003} ] [ 10^{1001} equiv 6010 equiv 1 pmod{2003} ] - Therefore, (1001) is indeed the order of (10) modulo (2003). 2. **Claim 2:** - There is no number (n) with (S(n) = 2) that is divisible by (2003). - **Proof:** - Assume the contrary, that there exists (n) of the form (2 cdot 10^k) or (10^k + 1) that is divisible by (2003). - The first case (2 cdot 10^k) is clearly impossible because (2003) is not divisible by (2). - For the second case, if (10^k equiv -1 pmod{2003}), then (2k) must be divisible by (1001). Thus, (k) must be divisible by (1001), but then (10^k equiv 1 pmod{2003}), which is a contradiction. 3. **Claim 3:** - There exist natural numbers (a, b) such that (10^a + 10^b + 1) is divisible by (2003). - **Proof:** - Assume the contrary. Consider the sets of remainders modulo (2003): [ A = {10^a mid a = 1, dots, 1001} ] [ B = {-10^b - 1 mid b = 1, dots, 1001} ] - According to Claim 1, neither of the elements from one set gives the same remainder. - Claim 2 implies that neither of these two sets contains (0) or (2002), so in total, these two sets have at most (2001) elements. - Therefore, (A) and (B) must have some common element, meaning: [ 10^a equiv -10^b - 1 pmod{2003} ] [ 10^a + 10^b + 1 equiv 0 pmod{2003} ] - This number has a sum of digits precisely equal to (3). (blacksquare) The final answer is ( boxed{ 3 } ).

answer:We use the AM-GM inequality to find the minimum of the function (3 sqrt{x} + frac{4}{x}). First, express the function in a form suitable for applying AM-GM: [3 sqrt{x} + frac{4}{x} = sqrt{x} + sqrt{x} + sqrt{x} + frac{4}{x}] Now apply AM-GM inequality: [sqrt{x} + sqrt{x} + sqrt{x} + frac{4}{x} geq 4 sqrt[4]{sqrt{x} cdot sqrt{x} cdot sqrt{x} cdot frac{4}{x}}] Simplifying the right-hand side: [4 sqrt[4]{4} = 4 cdot sqrt[4]{4} = 4 cdot 4^{1/4} = 4 cdot 2^{1/2} = 4 sqrt{2}] Thus, the minimum value of (3 sqrt{x} + frac{4}{x}) is (4 sqrt{2}), which occurs when (sqrt{x} = sqrt[4]{frac{4}{x}}) or (x = 2^{2/3}). Therefore, the minimum value is (boxed{4 sqrt{2}}).

answer:1. **Define the integers (a) and (b):** Let (a = 3^m + 2) and (b = 5^m + 2) for any positive integer (m). 2. **Choose (n) appropriately:** We will choose (n) in the form (n = k varphi(a) varphi(b) + m) where (k) is a positive integer and (varphi) is Euler's totient function. The totient function (varphi(n)) counts the number of integers up to (n) that are coprime with (n). 3. **Verify the properties of (a) and (b):** Note that (gcd(3, a) = gcd(3, 3^m + 2) = 1) because 3 and (3^m + 2) are coprime. Similarly, (gcd(5, b) = gcd(5, 5^m + 2) = 1) because 5 and (5^m + 2) are coprime. 4. **Check the congruences:** Since (n = k varphi(a) varphi(b) + m), we have: [ 3^n + 2 = 3^{k varphi(a) varphi(b) + m} + 2 ] By Euler's theorem, (3^{varphi(a)} equiv 1 pmod{a}) and (3^{varphi(b)} equiv 1 pmod{b}). Therefore: [ 3^{k varphi(a) varphi(b) + m} equiv 3^m pmod{a} ] Hence: [ 3^n + 2 equiv 3^m + 2 equiv a pmod{a} ] This implies (3^n + 2) is divisible by (a), making (3^n + 2) composite. 5. **Similarly for (5^n + 2):** [ 5^n + 2 = 5^{k varphi(a) varphi(b) + m} + 2 ] By Euler's theorem, (5^{varphi(a)} equiv 1 pmod{a}) and (5^{varphi(b)} equiv 1 pmod{b}). Therefore: [ 5^{k varphi(a) varphi(b) + m} equiv 5^m pmod{b} ] Hence: [ 5^n + 2 equiv 5^m + 2 equiv b pmod{b} ] This implies (5^n + 2) is divisible by (b), making (5^n + 2) composite. 6. **Conclusion:** Since (n = k varphi(a) varphi(b) + m) can be chosen for any positive integer (k), there are infinitely many such (n) for which both (3^n + 2) and (5^n + 2) are composite. (blacksquare)

answer:Part (a) 1. **Conceptualizing on Flat Surface:** Transform the three-dimensional cube into a two-dimensional representation. Consider unfolding the cube into a 2D structure made of 8 squares, each of side 1.  2. **Labeling Vertices:** The vertices of the resulting cube are labeled in such a way that vertex numbers that appear twice will coincide when the paper is folded back into a cube. 3. **Finding Minimum Path Length:** Notice that any path which passes through 8 different vertices must traverse 7 edges. 4. **Constructing the Path:** Consider the path: [ 1 rightarrow 2 rightarrow 3 rightarrow 4 rightarrow 5 rightarrow 6 rightarrow 7 rightarrow 8 ] which clearly visits all 8 vertices. 5. **Path Length Calculation:** Since each edge has length 1 and the path has 7 edges: [ text{Total Path Length} = 7 times 1 = 7 ] 6. **Conclusion:** Therefore, the minimum possible length for a path that traverses all 8 vertices of the cube is (7). [ boxed{7} ] Part (b) 1. **Extending the Diagram:** Extend the previous 2D layout to include the face centers and the diagonals inside each face square. The cube now contains the original 8 vertices plus 6 additional vertices located at the centers of each face.  2. **Labeling the Face Centers:** Number the original vertices from 1 to 8 and denote the center vertices with letters (a,b,c,d,e,f). 3. **Constructing the Path:** Consider the optimal path: [ 1 rightarrow a rightarrow 2 rightarrow b rightarrow 3 rightarrow c rightarrow 4 rightarrow 5 rightarrow d rightarrow 6 rightarrow e rightarrow 7 rightarrow f rightarrow 8 ] which is highlighted in the diagram below:  4. **Path Length Calculation:** - Edge between a numbered vertex and a lettered vertex spans ( sqrt{2}/2 ) - There are 12 such segments. - One segment (4 to 5) measures 1. Total length: [ 12 times frac{sqrt{2}}{2} + 1 = 6sqrt{2} + 1 ] 5. **Verification of Minimum Path Length:** Since all paths must contain 13 segments and we must include at least one edge of length 1 due to vertex constraints: [ text{Minimum Length} = 6sqrt{2} + 1 ] 6. **Conclusion:** Therefore, the minimum possible length for a path that passes through all 14 vertices is (6 sqrt{2} + 1). [ boxed{6sqrt{2} + 1} ]