answer:1. **Analyzing Condition 1**: - If Andrey goes to play hide and seek, then Borya will also go, and Vasya will not go. - This can be written logically as: [ A rightarrow (B land neg V) ] 2. **Analyzing Condition 2**: - If Borya goes to play hide and seek, then Gena or Denis will also go. - This can be written logically as: [ B rightarrow (G lor D) ] 3. **Analyzing Condition 3**: - If Vasya does not go to play hide and seek, then Borya and Denis will also not go. - This can be written logically as: [ neg V rightarrow (neg B land neg D) ] 4. **Analyzing Condition 4**: - If Andrey does not go to play hide and seek, then Borya will go, and Gena will not go. - This can be written logically as: [ neg A rightarrow (B land neg G) ] 5. **Deciphering the Conditions**: - Let's determine whether certain individuals will definitely go or not by combining the logical statements: - From condition 3: [ neg V rightarrow (neg B land neg D) ] This also means: [ B lor D rightarrow V ] - From condition 4: [ neg A rightarrow (B land neg G) ] This allows for two scenarios: - If Andrey does not go (neg A), it's guaranteed that Borya will go (B), and Gena will not go (neg G). 6. **Combining Conditions**: - To see if Borya will go independently of Andrey’s choice, consider condition 3 and 4 together: If Andrey does not go (neg A), Borya will go and Gena will not go. If Andrey does go (A), it's guaranteed Borya will go. Thus, Borya will always go: [ B = text{True} ] 7. **Seeing Vasya's Condition**: - When Borya goes, using condition 3: [ neg B rightarrow V ] Since ( B ) is True, Vasya must go by elimination (otherwise, if B were False, Vasya must not go). 8. **Checking Denis and Gena's Positions**: - From above: [ B rightarrow (G lor D) ] Borya still goes; thus, either Gena or Denis must go. - Considering earlier that Gena does not participate if Andrey does not and since Vasya will play (by condition 1 if A were true), Denis should go. Therefore, the conclusions are: - Borya (B), Vasya (V), and Denis (D) will play. # Conclusion: [ boxed{text{With Borya, Vasya, and Denis.}} ]

answer:1. Firstly, let's prove that there are infinitely many prime numbers which, when divided by ( n ), leave a remainder different from ( 1 ). - Assume for contradiction that there are only finitely many such primes, and multiply these primes together. Let's denote this product by ( P ). - Multiply ( P ) by ( n ) and subtract ( 1 ), giving us ( Q = nP - 1 ). Since ( n > 2 ), ( Q ) does not leave a remainder of ( 1 ) when divided by ( n ). - None of the finitely many primes divide ( Q ), as ( Q ) leaves a different remainder when divided by ( n ). - Thus, the prime factors of ( Q ) must be of the form ( 1 + nm ). However, the product of such numbers remains of this form, but ( Q ) contradicts this, so we have a contradiction. - Therefore, there are infinitely many primes that, when divided by ( n ), leave a remainder different from ( 1 ). 2. Since there are infinitely many primes and only finitely many residue classes mod ( n ), by the pigeonhole principle, there must be at least ( 2(n+1) ) primes in some residue class mod ( n ). - Let these primes be ( p_1, p_2, ldots, p_{2(n+1)} ), all of which yield the same remainder ( a ) when divided by ( n ). 3. We use the auxiliary theorem on the exponentiation: - According to the pigeonhole principle, among the powers ( a^1, a^2, ldots, a^{n+1} ), there exist two that are congruent mod ( n ), say ( a^{alpha_1} cong a^{alpha_2} mod n ) with ( alpha_1 > alpha_2 ). - Therefore, ( n mid a^{alpha_2} (a^{alpha_1 - alpha_2} - 1) ). - Since ( a ) is coprime to ( n ), it implies ( n mid a^{alpha_1 - alpha_2} - 1 ), meaning ( a^{alpha} equiv 1 mod n ) for some smallest integer ( alpha leq n ). 4. Let the smallest such ( alpha ) be ( r ): - We can form the polynomial ( P = p_{i_1} p_{i_2} cdots p_{i_r} equiv a^r equiv 1 mod n ), hence ( P in V_n ). - ( P ) must be irreducible in ( V_n ) because if it were reducible, any divisor ( D ) would also need to belong to ( V_n ), but such ( D neq 1 ) and ( D neq P ) would imply ( D = p_{j_1} p_{j_2} cdots p_{j_k} equiv a^k neq 1 mod n ), which contradicts the minimality of ( r ). - Hence, ( P ) is irreducible and ( r > 1 ) since ( a notequiv 1 mod n ). 5. Choosing two irreducible elements ( P_1 ) and ( Q_1 ) in ( V_n ): - ( P_1 = p_1 p_2 ldots p_r ) and ( Q_1 = q_1 q_2 ldots q_r ) where ( P_1, Q_1 in V_n ). - The product ( P_1 Q_1 = p_1 p_2 ldots p_r q_1 q_2 ldots q_r = q_1 q_2 ldots q_r p_1 p_2 ldots p_r = P_1' Q_1' ), with ( P_1' neq P_1 ) and ( Q_1' neq Q_1 ) due to ( r > 1 ), also in ( V_n ). - ( P_1' ) and ( Q_1' ) remain irreducible and different from ( P_1 ) and ( Q_1 ), confirming the assertion. Hence, we have shown that such ( r in V_n ) exists, which can be factored in more than one way using irreducible elements of ( V_n ). (boxed{})

answer:1. Since f(-1)=a-b+1=0quad① And for any real number x in the domain, f(x)geqslant 0 holds, and a > 0 Therefore, triangle =b^{2}-4aleqslant 0quad② Solving equations ① and ② simultaneously, we get (a-1)^{2}leqslant 0, hence a=1, b=2 Thus, f(x)=x^{2}+2x+1 Hence, F(x)= begin{cases} (x+1)^{2},x > 0 -(x+1)^{2},x < 0 end{cases} 2. Since xin[-2,2], g(x)=f(x)-kx=x^{2}+(2-k)x+1 is a monotonic function Also, the symmetry axis of the function g(x) is x= dfrac {k-2}{2} Therefore, dfrac {k-2}{2}geqslant 2 or dfrac {k-2}{2}leqslant -2 Hence, boxed{kgeqslant 6 text{ or } kleqslant -2}

answer:Given that (a, b, c,) and (d) are positive real numbers such that (abcd = 1), we aim to show that: [a^{2} + b^{2} + c^{2} + d^{2} + ab + ac + ad + bc + bd + cd geq 10.] 1. **Identify the sum and the number of terms**: We are given a sum of 10 terms: [S = a^{2} + b^{2} + c^{2} + d^{2} + ab + ac + ad + bc + bd + cd.] 2. **Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality)**: The AM-GM Inequality states that for any non-negative real numbers (x_1, x_2, ldots, x_n): [ frac{x_1 + x_2 + cdots + x_n}{n} geq sqrt[n]{x_1 x_2 cdots x_n}. ] Multiplying through by (n), we obtain: [ x_1 + x_2 + cdots + x_n geq n sqrt[n]{x_1 x_2 cdots x_n}. ] 3. **Apply AM-GM to our specific sum**: For our sum (S) composed of 10 terms: [ S = a^{2} + b^{2} + c^{2} + d^{2} + ab + ac + ad + bc + bd + cd, ] applying AM-GM we get: [ a^{2} + b^{2} + c^{2} + d^{2} + ab + ac + ad + bc + bd + cd geq 10 sqrt[10]{a^{2} times b^{2} times c^{2} times d^{2} times ab times ac times ad times bc times bd times cd}. ] 4. **Simplify the product inside the root**: [ a^{2} times b^{2} times c^{2} times d^{2} times ab times ac times ad times bc times bd times cd = (abcd)^2 times (abcd)^3 = (abcd)^{5}. ] Given (abcd = 1), we have: [ (abcd)^5 = 1^{5} = 1. ] 5. **Conclusion**: Substituting this back into our inequality: [ a^{2} + b^{2} + c^{2} + d^{2} + ab + ac + ad + bc + bd + cd geq 10 sqrt[10]{1} = 10. ] Thus, we have shown that: [ a^{2} + b^{2} + c^{2} + d^{2} + ab + ac + ad + bc + bd + cd geq 10. ] [boxed{10}]