answer:1. Let n be a natural number, and let a_d a_{d-1} ldots a_1 a_0 represent its decimal representation, where a_0 is the units digit, a_1 is the tens digit, etc. 2. First, consider the case where n has only one digit: n = a_0 Then: s(n) = a_0 Therefore: n = a_0 leq 2a_0 implies n text{ is a solution} So, all single-digit numbers 0, 1, 2, ldots, 9 are solutions. 3. Now consider the case where n has at least two digits: n = a_d cdot 10^d + a_{d-1} cdot 10^{d-1} + ldots + a_1 cdot 10^1 + a_0 cdot 10^0 and: s(n) = a_d + a_{d-1} + ldots + a_1 + a_0 4. If d geq 2: For each 1 leq i leq d-1: a_i cdot 10^i geq a_i cdot 2 Moreover, since a_d is non-zero: a_d cdot 10^d = 2a_d + (10^d - 2) cdot a_d geq 2a_d + 10^d - 2 5. Since a_0 < 10: begin{align*} a_d cdot 10^d & > 2a_d + a_0 end{align*} 6. Therefore: begin{align*} n = a_d cdot 10^d + a_{d-1} cdot 10^{d-1} + ldots + a_1 cdot 10^1 + a_0 cdot 10^0 & > 2a_d + a_0 + 2a_{d-1} + ldots + 2a_1 + a_0 & = 2s(n) end{align*} which implies that if n has three or more digits, then n is not a solution. 7. Next, consider the case where n has exactly two digits: begin{align*} n & = a_1 cdot 10 + a_0 end{align*} 8. If a_1 geq 2: begin{align*} n = a_1 cdot 10 + a_0 & = 2a_1 + a_0 + 8 cdot a_1 & geq 2a_1 + a_0 + 16 & > 2a_1 + a_0 + a_0 & = 2 s(n) end{align*} Thus, n is not a solution when a_1 geq 2. 9. Consequently, we need a_1 = 1. Let us test numbers from 10 to 19: begin{align*} n & = 10 + a_0 s(n) & = 1 + a_0 end{align*} Thus: begin{align*} n leq 2s(n) & implies 10 + a_0 leq 2(1 + a_0) & implies 10 + a_0 leq 2 + 2a_0 & implies 8 leq a_0 end{align*} Hence, a_0 geq 8. 10. The only valid two-digit solutions are 18 and 19: begin{align*} 18 & leq 2(1 + 8) quad text{and} quad 19 leq 2(1 + 9) end{align*} # Conclusion: The solutions to the problem are: boxed{{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 18, 19}}

answer:1. **Analyzing Equation I: y = x + 3** - A linear equation with slope 1 and a y-intercept of +3. - Graph: straight line. 2. **Analyzing Equation II: y = frac{x^2 - 9}{x - 3}** - Factor the numerator: x^2 - 9 = (x - 3)(x + 3). - Simplify: y = frac{(x - 3)(x + 3)}{x - 3} becomes y = x + 3 when x neq 3. - At x = 3, the function is undefined due to division by zero. - Graph: same as y = x + 3, but with a hole at x = 3. 3. **Analyzing Equation III: (x - 3)y = x^2 - 9** - Rearrange: y = frac{x^2 - 9}{x - 3}. - Same simplification as Equation II, so y = x + 3 when x neq 3. - At x = 3, the equation leads to 0 = 0, valid for all y values. Thus, x = 3 is a vertical line including all y-values at x = 3 in the graph. **Comparison:** - **Equation I** is a continuous straight line. - **Equation II** follows Equation I but includes a hole at x = 3. - **Equation III** follows Equation I but includes a vertical line at x = 3. **Conclusion:** Since Equations I, II, and III do not share an entirely identical graph because of different characteristics at x = 3, the correct response is text{(E) None. All of the equations have different graphs}. The final answer is boxed{text{(E) None. All of the equations have different graphs}}

answer:Given that a, b, and c are all integers, we know that a-b and a-c are also integers. The equation left(a-bright)^{10}+left(a-cright)^{10}=1 implies that both terms on the left side must be non-negative integers (since any integer raised to the 10th power is non-negative) and that at least one of them must be 1 while the other must be 0 (since their sum is 1 and these are the only non-negative integers whose sum is 1). Thus, we have two cases to consider: 1. left{begin{array}{l} |a-b|=1 |a-c|=0 end{array}right. 2. left{begin{array}{l} |a-b|=0 |a-c|=1 end{array}right. For the first case, |a-c|=0 implies a=c. Substituting c for a in |a-b|+|b-c|+|c-a|, we get |a-b|+|b-a|+|a-a|. Since |a-a|=0 and |a-b|=|b-a|, this simplifies to 2|a-b|. Given |a-b|=1, we find 2|a-b|=2cdot1=2. For the second case, |a-b|=0 implies a=b. Substituting b for a in |a-b|+|b-c|+|c-a|, we get |a-a|+|a-c|+|c-a|. Since |a-a|=0 and |a-c|=|c-a|, this simplifies to 2|a-c|. Given |a-c|=1, we find 2|a-c|=2cdot1=2. In both cases, we have |a-b|+|b-c|+|c-a|=2. Therefore, the correct answer is boxed{text{B: }2}.

answer:1. **Introduction and Setup**: - Let sigma_1 and sigma_2 be circles intersecting at points A and B. - Tangents l_1 and l_2 are drawn at point A to circles sigma_1 and sigma_2, respectively. - Points T_1 and T_2 are chosen on sigma_1 and sigma_2 such that the arc angles T_1A and AT_2 are equal (measured clockwise). - The tangent t_1 at T_1 to sigma_1 intersects l_2 at M_1, and the tangent t_2 at T_2 to sigma_2 intersects l_1 at M_2. 2. **Application of Homothety**: - We perform a homothety with center at A and a positive coefficient k that maps circle sigma_1 to circle sigma_1', which equals sigma_2. 3. **Notation**: - Denote the images of points and lines under this homothety with a prime (e.g., T_1' and t_1'). 4. **Points of Intersection**: - Let B_1 be the second point of intersection (other than A) of sigma_1' and sigma_2. 5. **Symmetry Property**: - Circles sigma_1' and sigma_2 are symmetric with respect to the line AB_1. - Under this symmetry, T_1' maps to T_2, t_1' maps to t_2, and M_1' maps to M_2. 6. **Homothety Relations**: - Consequently, AM_2 = AM_1' and AM_1' = k AM_1. - The homothety ensures that the ratio AM_2: AM_1 is constant regardless of the positions of points T_1 and T_2. 7. **Conclusion on Midpoints**: - Thus, all triangles AM_1M_2 are homothetic with center A. - Since the triangles are homothetic, the midpoints of segments M_1M_2 must lie on a fixed line passing through point A. Hence, the conclusion follows: [ boxed{text{The midpoints of segments }M_1M_2 text{ lie on a fixed line passing through point }A.} ]