answer:1. **Calculate 7 diamond 24:** [ 7 diamond 24 = sqrt{7^2 + 24^2} = sqrt{49 + 576} = sqrt{625} = 25 ] 2. **Calculate (-24) diamond (-7):** [ (-24) diamond (-7) = sqrt{(-24)^2 + (-7)^2} = sqrt{576 + 49} = sqrt{625} = 25 ] 3. **Apply the operation diamond to the results from steps 1 and 2:** [ (7 diamond 24) diamond ((-24) diamond (-7)) = 25 diamond 25 = sqrt{25^2 + 25^2} ] 4. **Simplify the expression:** [ sqrt{25^2 + 25^2} = sqrt{625 + 625} = sqrt{1250} ] 5. **Recognize that sqrt{1250} can be rewritten using sqrt{2}:** [ sqrt{1250} = sqrt{2 times 625} = sqrt{2} times sqrt{625} = 25sqrt{2} ] 6. **Conclude with the final answer:** [ 25sqrt{2} ] The final answer is boxed{textbf{(C) } 25sqrt{2}}

answer:1. **Understanding the Problem**: Egor Sergeevich received train tickets between the following pairs of cities: - Saint Petersburg and Tver - Yaroslavl and Nizhny Novgorod - Moscow and Kazan - Nizhny Novgorod and Kazan - Moscow and Tver - Moscow and Nizhny Novgorod 2. **Graph Representation**: We represent the cities and the routes Egor has using a graph: - The vertices (cities) are: Saint Petersburg, Tver, Yaroslavl, Nizhny Novgorod, Moscow, and Kazan. - The edges (routes) between them are: - Saint Petersburg - Tver - Yaroslavl - Nizhny Novgorod - Moscow - Kazan - Nizhny Novgorod - Kazan - Moscow - Tver - Moscow - Nizhny Novgorod 3. **Observation**: For Egor to visit each of these six cities exactly once, starting and ending at different cities, it implies that the starting and ending points must have an odd degree in the graph. This is because they must be endpoints of a path that passes through all vertices exactly once (Eulerian path criteria for non-cyclic path). 4. **Degree Calculation**: - **Saint Petersburg**: connected to Tver (degree = 1) - **Tver**: connected to Saint Petersburg and Moscow (degree = 2) - **Yaroslavl**: connected to Nizhny Novgorod (degree = 1) - **Nizhny Novgorod**: connected to Yaroslavl, Kazan, and Moscow (degree = 3) - **Moscow**: connected to Tver, Kazan, and Nizhny Novgorod (degree = 3) - **Kazan**: connected to Moscow and Nizhny Novgorod (degree = 2) 5. **Identifying Candidate Starting Points**: According to the degree calculation: - Saint Petersburg (degree = 1) - Yaroslavl (degree = 1) Only these cities have odd degrees, making them the only possible starting or ending points. 6. **Conclusion**: Egor's journey could have started or ended in either Saint Petersburg or Yaroslavl, given the structure of the graph and the Eulerian path property. Therefore, the possible starting cities for Egor's journey are: [ boxed{text{Saint Petersburg or Yaroslavl}} ]

answer:Given the sequence {a_n}, we have: [ a_1 = -1, quad a_2 = 1, quad a_3 = -2 ] with the condition for all n in mathbb{N}_{+}: [ a_n a_{n+1} a_{n+2} a_{n+3} = a_n + a_{n+1} + a_{n+2} + a_{n+3} quad text{and} quad a_{n+1} a_{n+2} a_{n+3} neq 1 ] Step-by-Step Calculation: 1. **Finding (a_4):** For (n = 1): [ a_1 a_2 a_3 a_4 = a_1 + a_2 + a_3 + a_4 ] Substituting the known values: [ (-1)(1)(-2)a_4 = -1 + 1 - 2 + a_4 ] Therefore: [ 2a_4 = -2 + a_4 ] Solving for (a_4): [ 2a_4 - a_4 = -2 ] [ a_4 = -2 ] 2. **Finding (a_5):** For (n = 2): [ a_2 a_3 a_4 a_5 = a_2 + a_3 + a_4 + a_5 ] Substituting the known values: [ (1)(-2)(-2)a_5 = 1 - 2 - 2 + a_5 ] Therefore: [ 4a_5 = -3 + a_5 ] Solving for (a_5): [ 4a_5 - a_5 = -3 ] [ 3a_5 = -3 ] [ a_5 = -1 ] 3. **Finding (a_6):** For (n = 3): [ a_3 a_4 a_5 a_6 = a_3 + a_4 + a_5 + a_6 ] Substituting the known values: [ (-2)(-2)(-1)a_6 = -2 - 2 - 1 + a_6 ] Therefore: [ -4a_6 = -5 + a_6 ] Solving for (a_6): [ -4a_6 - a_6 = -5 ] [ -5a_6 = -5 ] [ a_6 = 1 ] 4. **Observing the pattern:** We notice that: [ a_1 = -1, quad a_2 = 1, quad a_3 = -2, quad a_4 = -2, quad a_5 = -1, quad a_6 = 1 ] We deduced: [ a_7 = -2, quad a_8 = -2, quad ldots ] The sequence repeats every 4 terms: (a_{n+4} = a_n). Hence, it's a periodic sequence with a period of 4. 5. **Sum of the first 4321 terms:** Since the sequence is periodic with a period of 4 and the sum of one period is: [ a_1 + a_2 + a_3 + a_4 = -1 + 1 - 2 - 2 = -4 ] Number of complete periods in the first 4321 terms: [ leftlfloor frac{4321}{4} rightrfloor = 1080 quad text{with a remainder of 1} ] The sum of the first 4320 terms is: [ 1080 times (-4) = -4320 ] Adding the first term of the next period ((a_1)): [ S_{4321} = -4320 - 1 = -4321 ] # Conclusion: [ boxed{-4321} ]

answer:1. We begin by noting the set of numbers on the blackboard: ( 2^0, 2^1, 2^2, ldots, 2^{15}, 2^{16} ). There are ( 17 ) numbers in total. 2. When we choose any two numbers, ( a ) and ( b ), to subtract (assuming ( a > b )), the result of ( a - b ) will then be written back on the board in place of ( a ) and ( b ). 3. We aim to determine the largest possible number that can remain on the board after repeatedly performing such subtractions until only one number is left. **Observation:** - Suppose we take two numbers and subtract the smaller from the larger. - If we reverse the order of terms in the final subtraction (i.e., subtract the smaller from the larger), the final number will be negated. - Hence, the possible final numbers come in pairs with opposite signs. 4. Therefore, the largest possible number is the negative of the smallest possible number that can be on the board at the end. 5. To find the smallest possible number that can remain, consider the smallest number on the board initially, which is ( 1 ) (i.e., ( 2^0 )). - We can subtract all other numbers from this smallest number. 6. Summing up all other numbers, we have: [ sum_{k=1}^{16} 2^k = 2^1 + 2^2 + 2^3 + ldots + 2^{16} ] This sum is a geometric series: [ sum_{k=1}^{16} 2^k = 2(1 + 2 + 4 + ldots + 2^{15}) ] The sum of the first ( n ) terms of a geometric series ( a r^k ) is given by: [ S_n = a frac{r^n - 1}{r-1} ] where ( r = 2 ), ( a = 2 ), and ( n = 16 ). Therefore: [ sum_{k=0}^{15} 2^k = frac{2^{16} - 1}{2 - 1} = 2^{16} - 1 ] Hence: [ sum_{k=1}^{16} 2^k = 2^{16} - 1 = 65536 - 1 = 65535 ] 7. The smallest possible result when subtracting the sum of all other numbers from ( 2^0 ): [ 2^0 - (2^1 + 2^2 + 2^3 + ldots + 2^{16}) = 1 - 65535 = -65535 ] 8. The negative of the smallest possible number is then: [ -(-65535) = 65535 ] 9. Therefore, the largest possible number that can remain when only one number is left on the blackboard is: [ boxed{65535} ]