answer:Answer: 11998001Solution: In the equation f(x+2)-f(x)=3 x+2, we will substitute for x the numbers 0,2,4, ldots, 3998. We get:begin{aligned}& f(2)-f(0)=3 cdot 0+2 & f(4)-f(2)=3 cdot 2+2end{aligned}f(4000)-f(3998)=3 cdot 3998+2Adding the equations, we get: f(4000)-f(0)=3 cdot(0+2+4+cdots+3998)+2 cdot 2000= 3 cdot frac{3998 cdot 2000}{2}+4000=11998000. Then f(4000)=11998001.

answer:1. Initial Casework: We start by determining the values of ( f(1) ), ( f(2) ), and ( f(3) ). - For ( a = b = c = 1 ): [ f(1) + f(1) + f(1) - 1 cdot 1 - 1 cdot 1 - 1 cdot 1 mid 1 cdot f(1) + 1 cdot f(1) + 1 cdot f(1) - 3 cdot 1 cdot 1 cdot 1 ] [ 3f(1) - 3 mid 3f(1) - 3 ] This is always true, so no new information is gained. - For ( a = 2, b = c = 1 ): [ f(2) + f(1) + f(1) - 2 cdot 1 - 1 cdot 1 - 1 cdot 2 mid 2f(2) + f(1) + f(1) - 3 cdot 2 cdot 1 cdot 1 ] [ f(2) + 2f(1) - 4 mid 2f(2) + 2f(1) - 6 ] Let ( f(1) = 1 ): [ f(2) + 2 - 4 mid 2f(2) + 2 - 6 ] [ f(2) - 2 mid 2f(2) - 4 ] This implies ( f(2) = 4 ).2. Using ( P(a,1,1) ) and ( P(a,2,2) ): - For ( P(a,1,1) ): [ f(a) + f(1) + f(1) - a cdot 1 - 1 cdot 1 - 1 cdot a mid af(a) + f(1) + f(1) - 3a ] [ f(a) + 2 - 2a mid af(a) + 2 - 3a ] [ f(a) - 2a + 2 mid af(a) - 3a + 2 ] Simplifying, we get: [ f(a) - 2a + 2 mid (a-1)f(a) - 3a + 2 ] - For ( P(a,2,2) ): [ f(a) + f(2) + f(2) - a cdot 2 - 2 cdot 2 - 2 cdot a mid af(a) + 2f(2) - 6a ] [ f(a) + 8 - 4a mid af(a) + 8 - 6a ] [ f(a) - 4a + 8 mid af(a) - 6a + 8 ]3. Analyzing the Divisibility Conditions: - From ( f(a) - 2a + 2 mid 2(a-1)^2 ) - From ( f(a) - 4a + 8 mid 4(a-2)^2 )4. Considering ( a = p+1 ) for some prime ( p ): - From the first condition, if ( a = p+1 ): [ f(a) in {a^2, 4a-3, 2a+1, 2a, 3a-2, a, 2a-2, 1, 2a^2-2a+1} ]5. Using Euclidean Algorithm: - From the second condition, ( f(a) in {a^2, 4a-3} ) if ( p > 61 ).6. Using ( P(a,a,1) ): - We use the Euclidean algorithm to eliminate the ( f(a) = 4a-3 ) case. - Conclusion: For ( p > 61 ) a prime, we have ( f(p+1) = (p+1)^2 ).7. Using ( P(a,p+1,p+1) ) for large primes ( p ): - We get: [ (f(a) - a^2) + (p+1-a)^2 mid 2(p+1)(p+1-a)^2 ] - Thus for all large ( p ): [ (f(a) - a^2) + (p+1-a)^2 mid 2(p+1)(a^2 - f(a)) ] - If ( f(a) neq a^2 ), the last part causes a contradiction when ( p ) becomes sufficiently large in comparison to ( a ).Thus, ( f(a) = a^2 ) which works clearly. Proved.The final answer is ( boxed{ f(a) = a^2 } )

answer:【Answer】Solution: Largest number: thousand's place maximum 2, hundred's place 0, ten's place maximum 2, unit's place 0, 2020,Second: thousand's place maximum 2, hundred's place 0, ten's place maximum 1, unit's place 1, 2011,Third: thousand's place maximum 2, hundred's place 0, ten's place maximum 1, unit's place 0, 2010, ..If arranged from smallest to largest, the second to last number is 2011, so the answer is: 2011.

answer:Solution. From the condition : n=4.Rewrite the inequalities as:left{begin{aligned} & 16^{3} text{, then for } n=5 text{ the given system is already inconsistent: the intervals } [sqrt[3]{3}, sqrt[4]{4}] text{ and } [sqrt[5]{2}, sqrt[5]{6}] text{ do not intersect.} & text{For } n=4 text{, it is not difficult to find a value of } x text{ that satisfies all four inequalities (for example, } x=1.45 text{).} end{aligned}right.

answer:Just due to the increase in the dollar exchange rate, Petya received a 9.5% profit. In addition, he received a 10 % profit from the bank and 10 cdot 0.095=0.95 % profit from converting this profit into rubles. In total, he received 20.45 % more profit than the 20 % that Vasya received. Therefore, Vasya's initial capital was larger.## AnswerVasya's.

answer:Initially, it is possible to register 5 times 3 times 4= 60 bicycles. Let's analyze the two possible situations:- We increase two letters in one of the sets. With this, we can haveSuggestion: Calculate the initial number of plates that can be made with the elements of sets mathcal{A}, mathcal{B}, and mathcal{C}, and then recalculate by analyzing the various possibilities of increasing the elements of the sets by 1 or 2.| mathcal{A} times mathcal{B} times mathcal{C} | Number of Plates || :---: | :---: || 7 times 3 times 4 | 84 || 5 times 5 times 4 | 100 || 5 times 3 times 6 | 90 |Thus, with the modification shown, the maximum number of new plates is 100-60=40.- Increase one letter in two of the sets. With this, we can have| mathcal{A} times mathcal{B} times mathcal{C} | Number of Plates || :---: | :---: || 6 times 4 times 4 | 96 || 6 times 3 times 5 | 90 || 5 times 4 times 5 | 100 |In this case, the maximum number of new plates is also 40.Suggestion: In item (b), consider the players who are eliminated instead of those who advance to the next rounds.