answer:If there are now 10 birds sitting on the fence after 6 more birds came to join them, we can find out how many birds were initially sitting on the fence by subtracting the 6 birds that joined from the total number of birds now. 10 birds (current) - 6 birds (joined) = 4 birds (initially sitting on the fence) So, initially, there were boxed{4} birds sitting on the fence.

answer:1. **Given Data and Setting Up Variables**: - We are given two distinct points ( A ) and ( B ) in a plane. - Let ( |AB| = 2a ), which implies the distance between ( A ) and ( B ) is ( 2a ). - Let the midpoint of ( AB ) be ( O ). - We need to find the locus of the point ( M ) that satisfies the condition (overrightarrow{MA} cdot overrightarrow{MB} = k^2), where ( k ) is a nonzero constant. 2. **Express Position Vectors**: - The midpoint ( O ) of ( AB ) has the vector relationships: [ overrightarrow{OA} = - overrightarrow{OB} ] This is because ( O ) is the midpoint, and the position vectors balance out. 3. **Translate the Given Condition**: - The condition (overrightarrow{MA} cdot overrightarrow{MB} = k^2) can be translated using vectors starting from ( O ): [ overrightarrow{MA} = overrightarrow{OA} - overrightarrow{OM} ] [ overrightarrow{MB} = overrightarrow{OB} - overrightarrow{OM} ] 4. **Substitute and Simplify**: - We have: [ overrightarrow{OA} = -overrightarrow{OB} ] Therefore: [ overrightarrow{MA} cdot overrightarrow{MB} = (overrightarrow{OA} - overrightarrow{OM}) cdot (overrightarrow{OB} - overrightarrow{OM}) ] Substituting (overrightarrow{OB} = - overrightarrow{OA}): [ = (overrightarrow{OA} - overrightarrow{OM}) cdot (-overrightarrow{OA} - overrightarrow{OM}) ] 5. **Expand and Use Vector Properties**: [ = overrightarrow{OA} cdot (-overrightarrow{OA}) + overrightarrow{OA} cdot (-overrightarrow{OM}) + overrightarrow{OM} cdot (-overrightarrow{OA}) + overrightarrow{OM} cdot (-overrightarrow{OM}) ] Simplifying using the dot products and properties of vectors: [ = -overrightarrow{OA}^2 + overrightarrow{OA} cdot (-overrightarrow{OM}) + (-overrightarrow{OM}) cdot overrightarrow{OA} + (-overrightarrow{OM}) cdot (-overrightarrow{OM}) ] [ = -|overrightarrow{OA}|^2 - 2 overrightarrow{OA} cdot overrightarrow{OM} + |overrightarrow{OM}|^2 ] [ = -a^2 + |overrightarrow{OM}|^2 ] Since (overrightarrow{OA} cdot overrightarrow{OM} = 0) when vectors are perpendicular. 6. **Substitute Back into Condition**: - Given (overrightarrow{MA} cdot overrightarrow{MB} = k^2): [ |overrightarrow{OM}|^2 - a^2 = k^2 ] Solving the equation for ( |overrightarrow{OM}| ): [ |overrightarrow{OM}|^2 = k^2 + a^2 ] [ |overrightarrow{OM}| = sqrt{k^2 + a^2} ] 7. **Conclusion**: Thus, the point ( M ) lies on a circle with center ( O ), and radius (sqrt{k^2 + a^2}). # Conclusion: The locus of the point ( M ) is a circle centered at the midpoint of ( AB ) with a radius of (sqrt{k^2 + a^2}). [ boxed{text{Circle with center } O text{ and radius } sqrt{k^2 + a^2}} ]

answer:1. **Define the sum ( S )**: Let ( S = x_1 + x_2 + cdots + x_n ). 2. **Bound the first product term**: For any ( 1 leq i leq n ), we have: [ 1 + frac{1}{x_1 + x_2 + cdots + x_i} ] Notice that [ 1 + frac{1}{x_1 + x_2 + cdots + x_i} leq 1 + frac{1}{i} ] because ( x_1 + x_2 + cdots + x_i geq i ). This follows from (x_i geq 1) for each (i). 3. **Bound the second product term**: For any ( 1 leq i leq n ), consider: [ 1 - frac{1}{x_i + x_{i+1} + cdots + x_n} ] Notice that [ 1 - frac{1}{x_i + x_{i+1} + cdots + x_n} leq 1 - frac{1}{S - (i-1)} ] because ( x_i + x_{i+1} + cdots + x_n leq S - (i - 1) ). 4. **Simplify the bounds**: Hence, [ 1 - frac{1}{S - (i-1)} = frac{S - (i-1) - 1}{S - (i-1)} = frac{S - i}{S + 1 - i} ] 5. **Bound the product of the first expression**: Consider the product of the first term: [ prod_{i=1}^n left(1 + frac{1}{x_1 + x_2 + cdots + x_i}right) leq prod_{i=1}^n left(1 + frac{1}{i}right) = prod_{i=1}^n frac{i + 1}{i} ] Therefore, [ prod_{i=1}^n frac{i + 1}{i} = frac{2}{1} cdot frac{3}{2} cdot cdots cdot frac{n+1}{n} = frac{n+1}{1} = n+1 ] 6. **Bound the product of the second expression**: Consider the product of the second term: [ prod_{i=1}^n left(1 - frac{1}{x_i + x_{i+1} + cdots + x_n}right) leq prod_{i=1}^n frac{S - i}{S + 1 - i} ] Now, [ prod_{i=1}^n frac{S - i}{S + 1 - i} ] is a telescoping product: [ frac{S - 1}{S} cdot frac{S - 2}{S - 1} cdot frac{S - 3}{S - 2} cdot cdots cdot frac{S - (n-1)}{S - (n-2)} cdot frac{S - n}{S - (n-1)} = frac{S - n}{S} ] 7. **Combine both parts**: Now we combine the two parts: [ prod_{i=1}^n left(1 + frac{1}{x_1 + x_2 + cdots + x_i}right) cdot prod_{i=1}^n left(1 - frac{1}{x_i + x_{i+1} + cdots + x_n}right) ] So, [ n + 1 + frac{S - n}{S} leq n + 1 ] 8. **Conclusion**: Therefore, we have shown that: [ prod_{i=1}^n left(1 + frac{1}{x_1 + x_2 + cdots + x_i}right) + prod_{i=1}^n left(1 - frac{1}{x_i + x_{i+1} + cdots + x_n}right) leq n + 1 ] [ boxed{n + 1} ]

answer:1. **Prime factorization of new LCMs**: - 120 = 2^3 cdot 3 cdot 5 - 1000 = 2^3 cdot 5^3 - 480 = 2^5 cdot 3 cdot 5 2. **Constraints on x, y, z**: - x must include at least the highest power of primes found in both 120 and 480, which is 2^5 and 3, but no 5^3. - y must include at least 3 and 5, with up to 2^3 but no higher as 2^5 nmid 120. - z must include at least 5^3 and can include up to 2^3. No 3 is necessary due to 3 nmid 1000. 3. **Finding possible values**: - Possible values for x: multiples of 32 that may include 3 (e.g., 32, 96). - Possible values for y: multiples of 15 that may include higher powers of 2 up to 2^3 (e.g., 15, 30, 60, 120). - Possible values for z: multiples of 125 that may include higher powers of 2 up to 2^3 (e.g., 125, 250, 500, 1000). 4. **Combining the constraints**: - Check each combination of (x, y, z) to ensure all three LCM conditions are met. - Valid triples are found by ensuring x and z share the necessary least powers while combining appropriately with y. 5. **Counting valid triples**: - (32, 120, 1000) - (96, 60, 250) - (96, 120, 250) The number of ordered triples (x, y, z) that satisfy the given conditions is 3. The final answer is boxed{textbf{(B)} 3}