answer:Let's denote the number of bears Jane makes per week without an assistant as B and the number of hours she works per week without an assistant as H. Her rate of making bears without an assistant is then B/H bears per hour. When Jane works with an assistant, she makes some percent more bears, let's call this increase P percent, so she makes (1 + P/100) * B bears per week. She also works 10 percent fewer hours, so she works (1 - 10/100) * H = 0.9 * H hours per week. The problem states that having an assistant increases Jane's output of toy bears per hour by 100 percent. This means her new rate of making bears with an assistant is 2 * (B/H) bears per hour because a 100 percent increase doubles the rate. Now, we can set up an equation using the new rate and the new number of hours to find the number of bears made with an assistant: 2 * (B/H) * 0.9 * H = (1 + P/100) * B Simplifying the left side of the equation by canceling out H, we get: 1.8 * B = (1 + P/100) * B Now, we can solve for P: 1.8 * B = B + (P/100) * B 1.8 * B = B + (P * B)/100 1.8 = 1 + P/100 0.8 = P/100 P = 0.8 * 100 P = 80 So, the percentage increase in bears per week when she works with an assistant is boxed{80} percent.

answer:1. Consider the sum of the form sum_{x=1}^{1989 cdot 1990} left( lfloor sqrt{x} rfloor + lfloor -sqrt{x} rfloor right). 2. We need to analyze the behavior of lfloor sqrt{x} rfloor + lfloor -sqrt{x} rfloor separately for square and non-square values of x. 3. **Case 1: (x) is a perfect square** - For (x = k^2), where (k) is an integer, we have: [ lfloor sqrt{k^2} rfloor = k quad text{and} quad lfloor -sqrt{k^2} rfloor = -k ] - So, [ lfloor sqrt{k^2} rfloor + lfloor -sqrt{k^2} rfloor = k + (-k) = 0 ] 4. **Case 2: (x) is not a perfect square** - For (x) such that (x neq k^2), we have: [ k < sqrt{x} < k+1 quad text{for some integer } k ] - Thus, [ lfloor sqrt{x} rfloor = k quad text{and} quad lfloor -sqrt{x} rfloor = - (k+1) ] - So, [ lfloor sqrt{x} rfloor + lfloor -sqrt{x} rfloor = k - (k+1) = -1 ] 5. Next, we count the number of perfect squares and non-perfect squares in the range from 1 to (1989 cdot 1990). 6. **Counting perfect squares**: - The number of perfect squares in the range (1) to (1989 cdot 1990) is given by the largest integer (k) such that (k^2 leq 1989 cdot 1990). - Consider sqrt{1989 cdot 1990} approx sqrt{1989^2 cdot 1990} = sqrt{1989^2 cdot (1989+1)} approx 1989). - Hence, there are (1989) perfect squares in the range. 7. **Counting non-perfect squares**: - The total number of integers from 1 to (1989 cdot 1990) is (1989 cdot 1990). - Thus, the number of non-perfect squares is: [ 1989 cdot 1990 - 1989 = 1989(1990 - 1) = 1989 cdot 1989 = 1989^2 ] 8. Summing up contributions: - Sum of contributions from perfect squares: [ sum_{k=1}^{1989} left( lfloor sqrt{k^2} rfloor + lfloor -sqrt{k^2} rfloor right) = 1989 cdot 0 = 0 ] - Sum of contributions from non-perfect squares: [ sum_{k neq text{perfect square}} left( lfloor sqrt{k} rfloor + lfloor -sqrt{k} rfloor right) = 1989^2 cdot (-1) = -1989^2 ] 9. Therefore, the total sum is: [ 0 + (-1989^2) = -1989^2 ] 10. Evaluate (1989^2): [ 1989^2 = 1989 cdot 1989 = (2000 - 11)^2 = 2000^2 - 2 cdot 2000 cdot 11 + 11^2 ] [ = 4000000 - 44000 + 121 = 3956121 ] **Conclusion:** [ boxed{-3956121} ]

answer:Given that the radius of the Earth is R, the radius of the latitude circle at 45°N is frac{sqrt{2}}{2}R. Since the longitudes of the two cities are 30°E and 120°E respectively, the length of the chord L connecting the two cities is frac{sqrt{2}}{2}R cdot sqrt{2} = R. Therefore, the angle angle AOB formed by the lines connecting cities A and B to the center of the Earth, O, is frac{pi}{3}. Hence, the distance between cities A and B is frac{pi}{3}R. So, the answer is: boxed{frac{pi}{3}R}.

answer:1. Let's start with the given number ( 12008 ). 2. The problem states we need to insert some number of '3's between the two zeros of 12008. For example, if we insert one '3', we get the number 12308. We need to show that the number is divisible by 19 for any number of 3s inserted. 3. For this proof, observe any possible number formed when inserting any number of 3s between the two zeros. Let ( n ) be a non-negative integer representing the number of '3's inserted. The general form of the number is thus ( 12underbrace{3 cdots 3}_{n text{ times}})08. We express this number as ( 12000 + 3 times 10^n + 8 ). 4. We now need to demonstrate that this number is divisible by 19 for every ( n ). Write the number formally as: N = 12000 + 3 times 10^n + 8. 5. Notice that 12000 mod 19 can be calculated: begin{align*} 12000 div 19 &= 631.578947 12000 - 631 cdot 19 &= 12000 - 11989 = 11. end{align*} Thus, ( 12000 equiv 11 pmod{19} ). 6. Next, compute ( 3 times 10^n mod 19 ). - Since ( 10 mod 19 = 10 ), we have ( 10^n mod 19 ). - Using Fermat’s Little Theorem, ( a^{p-1} equiv 1 pmod{p} ), where ( a = 10 ) and prime ( p = 19 ), gives: [ 10^{18} equiv 1 pmod{19}. ] - Hence, the computation ( 3 times 10^n mod 19 ) is cyclic with maximum cycle length being 18. 7. Compute 8 mod 19, which is straightforward: 8 mod 19 = 8. 8. Combine these to show that ( 12000 + 3 times 10^n + 8 ) is divisible by 19: begin{align*} 12008 + 3 times 10^n &equiv 11 + 8 equiv 19 equiv 0 pmod{19}. end{align*} 9. Therefore, any such number ( 12000 + 3 times 10^n + 8 ), with any ( n ), is divisible by 19. Conclusion. (blacksquare)