answer:1. **List and analyze the divisors of each integer from 1 to 5**: - ( 1 ): Divisors are ( {1} ) - ( 2 ): Divisors are ( {1, 2} ) - ( 3 ): Divisors are ( {1, 3} ) - ( 4 ): Divisors are ( {1, 2, 4} ) - ( 5 ): Divisors are ( {1, 5} ) We can observe that the integers 2, 3, and 5 are only divisible by 1 and themselves, which makes them prime. In contrast, the integer 4 is divisible by 1, 2, and 4. 2. **Determine the pairs ((a, b)) such that (1 leqslant a leqslant b leqslant 5)**: The pairs are: [ { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5) } ] There are (5) options for pairs starting with 1, (4) for pairs starting with 2, and so on: [ 5 + 4 + 3 + 2 + 1 = 15 ] 3. **Identify pairs for which ( gcd(a, b) = 1 )**: The greatest common divisor (gcd) is 1 if and only if (a) and (b) share no common divisors other than 1. This is known as being "coprime." Let's examine the pairs: - ( (1, 1), (1, 2), (1, 3), (1, 4), (1, 5) ) - ( (2, 3), (2, 5) ) (Since ( gcd(2, 2) = 2) and (gcd(2, 4) = 2)) - ( (3, 4), (3, 5) ) (Since ( gcd(3, 3) = 3 )) - ( (4, 5) ) (Since ( gcd(4, 4) = 4 )) - ( (5, 5) ) (Since ( gcd(5, 5) = 5 )) 4. **Count the number of pairs for which ( gcd(a, b) neq 1 )**: These pairs are: - ( (2, 2), (2, 4), (3, 3), (4, 4), (5, 5) ) There are 5 such pairs. 5. **Subtract to find pairs where (a) and (b) are coprime**: Since there are 15 total pairs and 5 pairs where ( gcd(a, b) neq 1 ): [ 15 - 5 = 10 ] # Conclusion: There are ( boxed{10} ) pairs ((a, b)) for which (a) and (b) are coprime within the range (1 leqslant a leqslant b leqslant 5).

answer:Let the measures of the angles be 3x, 4x, and 5x. Since these are the angles of a triangle, we have 3x + 4x + 5x = 180^circ. Simplifying, we get: [ 12x = 180^circ ] [ x = 15^circ ] The largest angle, being 5x, is: [ 5x = 5 times 15^circ = boxed{75^circ} ]

answer:To solve this problem, we consider the condition that each community must receive at least one volunteer. There are two scenarios: 1. **One community gets two volunteers, and the other two communities get one volunteer each.** In this case, we first choose one community from the three to receive two volunteers, which can be done in C_3^1 = 3 ways. Then, we choose two volunteers out of the four to go to the selected community, which can be done in C_4^2 = 6 ways. So, there are 3 times 6 = 18 ways in this scenario. 2. **One community gets three volunteers, and one community gets one volunteer.** First, we choose one community from the three to receive three volunteers, which can be done in C_3^1 = 3 ways. Then, we choose three volunteers out of the four to go to the selected community, which can be done in C_4^3 = 4 ways. So, there are 3 times 4 = 12 ways in this scenario. Adding the two scenarios together, we get 18 + 12 = 30 ways. However, this calculation does not consider the unique scenario where one community gets all four volunteers, and the other two communities get none, which is not allowed as each community must receive at least one volunteer. Therefore, the total number of different distribution methods is boxed{30}.

answer:1. Let ( a_{1}, a_{2}, ldots, a_{2015} ) be a sequence of numbers taking values in ( {-1, 0, 1} ), such that sum_{i=1}^{2015} a_{i} = 5 quad text{and} quad sum_{i=1}^{2015} left(a_{i} + 1right)^2 = 3040. 2. Suppose the number of ( -1 )'s is ( x ), the number of ( 0 )'s is ( y ), and the number of ( 1 )'s is ( z ). 3. We have the total number of terms: [ x + y + z = 2015. ] 4. From the first condition ( sum_{i=1}^{2015} a_{i} = 5 ): [ -x + 0 cdot y + z = 5 quad Rightarrow quad z = x + 5. ] 5. For the second condition ( sum_{i=1}^{2015} left( a_{i} + 1 right)^2 = 3040 ), we can write: [ sum_{i=1}^{2015} left(a_{i} + 1 right)^2 = sum_{i=1}^{2015} b_{i} ] where ( b_{i} = (a_{i} + 1)^2 ). We know: - If ( a_{i} = -1 ), then ( b_{i} = ( -1 + 1 )^2 = 0 ), - If ( a_{i} = 0 ), then ( b_{i} = ( 0 + 1 )^2 = 1 ), - If ( a_{i} = 1 ), then ( b_{i} = ( 1 + 1 )^2 = 4 ). 6. Thus: [ x cdot 0 + y cdot 1 + z cdot 4 = 3040. ] 7. Substitute ( z = x + 5 ) into the equation: [ y + 4(x + 5) = 3040. ] 8. Simplifying, we get: [ y + 4x + 20 = 3040 quad Rightarrow quad y + 4x = 3020. ] 9. Consider the equation from the number of terms: [ x + y + z = 2015 quad Rightarrow quad x + y + (x + 5) = 2015 ] or equivalently: [ y + 2x + 5 = 2015 quad Rightarrow quad y + 2x = 2010. ] 10. Now we have the system of linear equations: [ begin{cases} y + 4x = 3020 y + 2x = 2010 end{cases} ] 11. Subtract the second equation from the first: [ (y + 4x) - (y + 2x) = 3020 - 2010 ] which simplifies to: [ 2x = 1010 quad Rightarrow quad x = 505. ] 12. From ( z = x + 5 ): [ z = 505 + 5 = 510. ] # Conclusion: Therefore, the number of ones in the sequence is ( boxed{510} ).