answer:1. **Determine Total Income:** [ text{Total Income} = text{Planned Revenue} + text{Advances Received} = 120{,}000{,}000 + 30{,}000{,}000 = 150{,}000{,}000 , text{rub.} ] 2. **Calculate Monthly Planned Expenses:** [ text{Monthly Expenses} = begin{aligned} &, 770{,}000 , text{(Rent)} &+ 1{,}450{,}000 , text{(Various Oils)} &+ 4{,}600{,}000 , text{(Salaries)} &+ 1{,}380{,}000 , text{(Insurance Contributions)} &+ 340{,}000 , text{(Accounting Services)} &+ 1{,}800{,}000 , text{(Advertising)} &+ 800{,}000 , text{(Re-Training)} &+ 650{,}000 , text{(Miscellaneous)} &quad = 11{,}790{,}000 , text{rub.} quad text{(monthly)} end{aligned} ] 3. **Calculate Annual Planned Expenses:** [ text{Annual Planned Expenses} = 11{,}790{,}000 times 12 = 141{,}480{,}000 , text{rub.} ] 4. **Determine Tax Base for Different Taxation Regimes:** **Option (a) - Income Tax Regime:** [ text{Income Tax Base} = 150{,}000{,}000 , text{rub.} ] **Option (b) - Income Minus Expenses Tax Regime:** [ text{Expenditure Tax Base} = 150{,}000{,}000 - 141{,}480{,}000 = 8{,}520{,}000 , text{rub.} ] 5. **Calculate Taxes for Different Regimes:** **Option (a):** [ text{Income Tax} = 150{,}000{,}000 times 6% = 9{,}000{,}000 , text{rub.} ] **Insurance Contributions Deduction:** [ text{Annual Insurance Contributions} = 1{,}380{,}000 times 12 = 16{,}560{,}000 , text{rub.} ] However, the deduction can only be up to 50% of the tax amount: [ text{Deduction Limit} = frac{9{,}000{,}000}{2} = 4{,}500{,}000 , text{rub.} ] Thus, the final tax: [ text{Income Tax Payable} = 9{,}000{,}000 - 4{,}500{,}000 = 4{,}500{,}000 , text{rub.} ] **Option (b):** [ text{Tax on Income minus Expenses} = 8{,}520{,}000 times 15% = 1{,}278{,}000 , text{rub.} ] Compare with Minimum Tax (1% of total income): [ text{Minimum Tax} = 150{,}000{,}000 times 1% = 1{,}500{,}000 , text{rub.} ] Since the minimum tax is higher, the tax payable will be the minimum tax: [ text{Tax Payable} = 1{,}500{,}000 , text{rub.} ] 6. **Analyze the Results:** By comparing the final tax liabilities: [ text{Option (a) Tax Payable} = 4{,}500{,}000 , text{rub.} ] [ text{Option (b) Tax Payable} = 1{,}500{,}000 , text{rub.} ] **Conclusion:** Under the Unified Simplified Taxation System (USN), the more favorable option for Artur and Timur is the "income minus expenses" regime at 15% with a minimum tax of 1% of income: [ boxed{text{Option (b)}} ]

answer:At the start of the race, each car has 2 passengers and 1 driver, making a total of 3 people per car. With 20 cars, that's 20 * 3 = 60 people. When the cars go past the halfway point, each car gains another passenger. That means each car now has 3 passengers and 1 driver, making a total of 4 people per car. With 20 cars, that's 20 * 4 = 80 people. Therefore, by the end of the race, there are boxed{80} people in the cars.

answer:1. **Definition**: A natural number is called an almost square if it is equal to the product of two consecutive natural numbers. Let's denote an almost square as ( n(n+1) ), where ( n ) is a natural number. 2. **Expression as a Fraction**: We want to prove that any almost square can be expressed as a quotient of two almost squares. Consider the almost square ( n(n+1) ). We can rewrite it as follows: [ n(n+1) = frac{n(n+1)(n+2)}{(n+2)} ] 3. **Recognize Patterns**: The denominator ( n+2 ) and the numerator ( n(n+1)(n+2) ) both need to be structured in such a way that they can be expressed using almost squares. 4. **Expanding and Simplifying**: Let's expand the numerator ( n(n+1)(n+2) ): [ n(n+1)(n+2) = n(n+1)^2 + n(n+1) cdot 2 = n(n^2 + 2n + 1) + 2n(n+1) ] This gives: [ n(n+1)(n+2) = n^3 + 2n^2 + n + 2n^2 + 2n = n^3 + 4n^2 + 3n ] Rearrange and factor the terms: [ n(n+1)(n+2) = (n^2(n+1)) + (2n(n+1)) ] 5. **Denominator**: The denominator ( (n+1)(n+2) ) also needs to be detailed as: [ (n+1)(n+2) = n^2 + 3n + 2 ] 6. **Combine and Simplify Expression**: Now, we combine the expressions together: [ frac{n(n+1)(n+2)}{(n+2)} = frac{(n^2 + 2n)(n^2 + 2n + 1)}{(n+1)(n+2)} ] 7. **Conclusion**: In the numerator of the fraction, it's clear that we also have almost squares. Hence, we have shown that any almost square ( n(n+1) ) can indeed be expressed as the quotient of two almost squares. [ boxed{} ]

answer:1. Since gcd(x, y) divides both x and y, it must also divide x + y = 780. 2. The goal is to find the largest divisor of 780 that can be used as the greatest common divisor. 3. The divisors of 780 are 1, 2, 3, 4, 5, 6, 10, 12, 13, 15, 20, 26, 30, 39, 52, 60, 65, 78, 130, 156, 195, 260, and 390. The largest divisor less than 780 is 390. 4. Let x = 390 and y = 780 - 390 = 390. Then gcd(x, y) = gcd(390, 390) = 390. Therefore, the largest possible value of gcd(x, y) when their sum is 780 is boxed{390}.