answer:1. Observe that (10^{2021}) is a 1 followed by 2021 zeros: [ 10^{2021} = 1underbrace{00ldots0}_{2021 text{ zeros}} ] 2. Let (S) be the integer equal to (10^{2021} - 2021). 3. Start by expressing (2021) in base 10: [ 2021 = 2000 + 20 + 1 ] 4. Write (2021) using its digits in a positional notation: [ 2021 = 2 times 10^3 + 0 times 10^2 + 2 times 10^1 + 1 times 10^0 ] 5. We need to compute (10^{2021} - 2021): First, consider simpler calculation: [ 10^4 - 2021 = 10000 - 2021 = 7979 ] 6. Now apply the same concept: [ 10^{2021} - 2021 = 1underbrace{00ldots0}_{2021 text{ zeros}} - 2021 ] 7. To understand the subtraction: [ 10^{2021} = 1underbrace{00ldots0000}_{2021text{ zeros}} ] [ 2021 = 2000 + 20 + 1 ;;text{(positions considered: units, tens, thousands, tens)} ] 8. Conduct the long subtraction in base 10: [ begin{array}{c} 10^{2021}: ; 1underbrace{000 ldots 000}_{2021text{ zeros}} - ;;;; 2021: ;;;; 000ldots 0 2021 (align to match the positions) end{array} ] 9. The value for (S) can be obtained, logically, after borrowing and subtracting step-wise: [ underbrace{999ldots999}_{2019 text{ nines}}7979 ] 10. Now, find the sum of the digits of (999ldots97979): - The 2019 digits (9) provide: [ 2019 times 9 ] - Plus the sum of the digits in 7979: [ = 7 + 9 + 7 + 9 = 32 ] 11. Adding these: [ 2019 times 9 + 7 + 9 + 7 + 9 = 2019 times 9 + 32 ] 12. Hence: [ 18171 + 32 = 18185 ] # Conclusion: [ boxed{E} ]

answer:Expanding each squared norm, we have: begin{align*} &|mathbf{a} - 3 mathbf{b}|^2 + |mathbf{b} - 3 mathbf{c}|^2 + |mathbf{c} - 3 mathbf{a}|^2 &= (mathbf{a} - 3 mathbf{b}) cdot (mathbf{a} - 3 mathbf{b}) + (mathbf{b} - 3 mathbf{c}) cdot (mathbf{b} - 3 mathbf{c}) + (mathbf{c} - 3 mathbf{a}) cdot (mathbf{c} - 3 mathbf{a}) &= |mathbf{a}|^2 - 6 mathbf{a} cdot mathbf{b} + 9 |mathbf{b}|^2 + |mathbf{b}|^2 - 6 mathbf{b} cdot mathbf{c} + 9 |mathbf{c}|^2 + |mathbf{c}|^2 - 6 mathbf{c} cdot mathbf{a} + 9 |mathbf{a}|^2 &= 10 |mathbf{a}|^2 + 10 |mathbf{b}|^2 + 10 |mathbf{c}|^2 - 6 (mathbf{a} cdot mathbf{b} + mathbf{b} cdot mathbf{c} + mathbf{c} cdot mathbf{a}) &= 10 cdot 4 + 10 cdot 9 + 10 cdot 16 - 6 (mathbf{a} cdot mathbf{b} + mathbf{b} cdot mathbf{c} + mathbf{c} cdot mathbf{a}) &= 290 - 6 (mathbf{a} cdot mathbf{b} + mathbf{b} cdot mathbf{c} + mathbf{c} cdot mathbf{a}). end{align*} Using the inequality |mathbf{a} + mathbf{b} + mathbf{c}|^2 ge 0, we expand this as: [ |mathbf{a}|^2 + |mathbf{b}|^2 + |mathbf{c}|^2 + 2 (mathbf{a} cdot mathbf{b} + mathbf{b} cdot mathbf{c} + mathbf{c} cdot mathbf{a}) ge 0. ] Thus, [ 2 (mathbf{a} cdot mathbf{b} + mathbf{b} cdot mathbf{c} + mathbf{c} cdot mathbf{a}) ge -29. ] Therefore, [ |mathbf{a} - 3 mathbf{b}|^2 + |mathbf{b} - 3 mathbf{c}|^2 + |mathbf{c} - 3 mathbf{a}|^2 le 290 + 3 cdot 29 = 377. ] The maximum possible value is boxed{377}, occurring when mathbf{a} + mathbf{b} + mathbf{c} = mathbf{0} and the vectors align to provide maximum dot product values.

answer:First, let's calculate the total expenses Ramu incurred on the car. 1. Purchase price of the car: Rs. 42000 2. Repair costs: 35% of Rs. 42000 Repair costs = 0.35 * 42000 = Rs. 14700 3. Sales taxes: 8% of Rs. 42000 Sales taxes = 0.08 * 42000 = Rs. 3360 4. Registration fee: 6% of Rs. 42000 Registration fee = 0.06 * 42000 = Rs. 2520 Now, let's add up all the expenses to find the total cost: Total cost = Purchase price + Repair costs + Sales taxes + Registration fee Total cost = 42000 + 14700 + 3360 + 2520 Total cost = 42000 + 14700 + 3360 + 2520 Total cost = 42000 + 14700 + 3360 + 2520 Total cost = 42000 + 14700 + 3360 + 2520 Total cost = Rs. 62580 Next, let's calculate the net profit Ramu made from selling the car: Selling price of the car: Rs. 64900 Net profit = Selling price - Total cost Net profit = 64900 - 62580 Net profit = Rs. 2320 To find the net profit percentage, we use the formula: Net profit percentage = (Net profit / Total cost) * 100 Net profit percentage = (2320 / 62580) * 100 Net profit percentage = 0.03706 * 100 Net profit percentage = 3.706% Ramu's net profit percentage considering all the expenses is approximately boxed{3.71%} .

answer:(1) Solution: Given that the minimum distance from any point on the ellipse to point F(-1,0) is sqrt{2}-1, we have the following system of equations: begin{cases} c=1 a-c=sqrt{2}-1 end{cases} Solving this system, we get: begin{cases} a=sqrt{2} c=1 end{cases} Since b^{2}=a^{2}-c^{2}=1, the equation of the ellipse is frac{x^{2}}{2}+y^{2}=1. (2) Proof: text{(i)} When line l is vertical, its equation is x=-1. We can find that A(-1, frac{sqrt{2}}{2}) and B(-1,-frac{sqrt{2}}{2}). In this case, overline{MA}cdot overline{MB}=(-1+frac{5}{4}, frac{sqrt{2}}{2})cdot (-1+frac{5}{4},-frac{sqrt{2}}{2})=-frac{7}{16}. text{(ii)} When line l is not vertical, let the equation of line l be y=k(x+1). Solving the system of equations: begin{cases} y=k(x+1) frac{x^{2}}{2}+y^{2}=1 end{cases} We get the quadratic equation (1+2k^{2})x^{2}+4k^{2}x+2k^{2}-2=0. Let A(x_{1},y_{1}) and B(x_{2},y_{2}), then x_{1}+x_{2}=-frac{4k^{2}}{1+2k^{2}} and x_{1}x_{2}=frac{2k^{2}-2}{1+2k^{2}}. Thus, overline{MA}cdot overline{MB}=(x_{1}+frac{5}{4},y_{1})cdot (x_{2}+frac{5}{4},y_{2})=(x_{1}+frac{5}{4})(x_{2}+frac{5}{4})+y_{1}y_{2} =x_{1}x_{2}+frac{5}{4}(x_{1}+x_{2})+frac{25}{16}+k(x_{1}+1)(kx_{2}+1) =(1+k^{2})x_{1}x_{2}+(k^{2}+frac{5}{4})(x_{1}+x_{2})+k^{2}+frac{25}{16} =(1+k^{2})frac{2k^{2}-2}{1+2k^{2}}+(k^{2}+frac{5}{4})(-frac{4k^{2}}{1+2k^{2}})+k^{2}+frac{25}{16} =frac{-4k^{2}-2}{1+2k^{2}}+frac{25}{16} =-2+frac{25}{16} =-frac{7}{16}. Therefore, overline{MA}cdot overline{MB} is constant, and its value is boxed{-frac{7}{16}}.