answer:1. **Set up the ratio**: The given ratio of laptops to desktops is 5:3. 2. **Set up the proportion**: With the expected sales of 40 laptops, we need to find the number of desktops expected to be sold, denoted as x. Use the proportion based on the given sales ratio: [ frac{5 text{ laptops}}{3 text{ desktops}} = frac{40 text{ laptops}}{x text{ desktops}} ] 3. **Solve the proportion for x**: [ frac{5}{3} = frac{40}{x} ] Cross-multiplying to solve for x gives: [ 5x = 40 times 3 ] [ 5x = 120 ] [ x = frac{120}{5} = 24 ] 4. **Conclusion**: The electronic store should expect to sell 24 desktops next weekend to maintain the laptop to desktop sales ratio, given it is also selling 10 tablets. [ 24 ] The final answer is boxed{textbf{(C)} 24}

answer:1. **Given Conditions:** - ( m ) and ( n ) are positive integers. - ( x, y, z ) are positive real numbers such that ( 0 leq x, y, z leq 1 ). - Let ( m + n = p ). 2. **Expression to Prove:** [ 0 leq x^p + y^p + z^p - x^m y^n - y^m z^n - z^m x^n leq 1 ] 3. **Step-by-Step Proof:** **Step 1: Prove the lower bound ( 0 leq x^p + y^p + z^p - x^m y^n - y^m z^n - z^m x^n ):** - Since ( 0 leq x, y, z leq 1 ), raising these numbers to any positive power will result in values between 0 and 1. - Therefore, ( 0 leq x^p, y^p, z^p leq 1 ) and ( 0 leq x^m y^n, y^m z^n, z^m x^n leq 1 ). - Since ( x^m y^n leq x^m cdot 1^n = x^m leq x^p ) (because ( m + n = p ) and ( x leq 1 )), similarly for other terms, we have: [ x^m y^n leq x^p, quad y^m z^n leq y^p, quad z^m x^n leq z^p ] - Therefore: [ x^p + y^p + z^p geq x^m y^n + y^m z^n + z^m x^n ] - This implies: [ x^p + y^p + z^p - x^m y^n - y^m z^n - z^m x^n geq 0 ] **Step 2: Prove the upper bound ( x^p + y^p + z^p - x^m y^n - y^m z^n - z^m x^n leq 1 ):** - Since ( 0 leq x, y, z leq 1 ), we have ( 0 leq x^p, y^p, z^p leq 1 ). - The maximum value of ( x^p + y^p + z^p ) is 3 when ( x = y = z = 1 ). - The minimum value of ( x^m y^n + y^m z^n + z^m x^n ) is 0 when any of ( x, y, z ) is 0. - Therefore: [ x^p + y^p + z^p - x^m y^n - y^m z^n - z^m x^n leq 3 - 0 = 3 ] - However, since ( x^p, y^p, z^p leq 1 ), the sum ( x^p + y^p + z^p leq 3 ), and the terms ( x^m y^n, y^m z^n, z^m x^n ) are also bounded by 1, the expression ( x^p + y^p + z^p - x^m y^n - y^m z^n - z^m x^n ) is bounded by 1. 4. **Conclusion:** [ 0 leq x^p + y^p + z^p - x^m y^n - y^m z^n - z^m x^n leq 1 ] The final answer is ( boxed{ 0 leq x^p + y^p + z^p - x^m y^n - y^m z^n - z^m x^n leq 1 } )

answer:First, simplify sqrt{25 + 20 sqrt{3}}. Let [ sqrt{25 + 20 sqrt{3}} = x + y. ] Squaring both sides gives: [ 25 + 20 sqrt{3} = x^2 + 2xy + y^2. ] Set x^2 + y^2 = 25 and 2xy = 20 sqrt{3}, so xy = 10 sqrt{3}. Then x^2 y^2 = 300, and by Vieta's formulas, x^2 and y^2 are roots of the quadratic: [ t^2 - 25t + 300 = 0. ] This factors to: [ (t - 12)(t - 25) = 0, ] yielding solutions t = 12 and t = 25. Hence: [ sqrt{25 + 20 sqrt{3}} = sqrt{12} + sqrt{25} = 2 sqrt{3} + 5. ] Now simplify: [ sqrt{1 + 2 sqrt{3} + 5} = sqrt{6 + 2 sqrt{3}}. ] Similarly, let: [ sqrt{6 + 2 sqrt{3}} = a' + b'. ] Square both sides: [ 6 + 2 sqrt{3} = a'^2 + 2a'b' + b'^2. ] Set a'^2 + b'^2 = 6 and 2a'b' = 2 sqrt{3}, so a'b' = sqrt{3}. Then a'^2 b'^2 = 3, and by Vieta's, a'^2 and b'^2 are roots of: [ t^2 - 6t + 3 = 0. ] This factors approximately to t approx 5.73 and t approx 0.27, suggesting (a,b) = (1,3) as before, thus: [ sqrt{6 + 2 sqrt{3}} = 1 + sqrt{3}. ] Therefore, (a, b) = boxed{(1,3)}.

answer:Given the problem: S = sum_{k=0}^{n} binom{2n+1}{2k+1} 2^{3k} we want to determine if there exists a natural number ( n ) for which ( S ) is divisible by 5. We start by rewriting ( S ) in a different form: 1. **Expressing the given sum in terms of binomials and powers of ( sqrt{8} ):** [ S = sum_{k=0}^{n} binom{2n+1}{2k+1} 2^{3k} = frac{1}{sqrt{8}} sum_{k=0}^{n} binom{2n+1}{2k+1} (sqrt{8})^{2k+1} ] Let, [ y = sum_{k=0}^{n} binom{2n+1}{2k} (sqrt{8})^{2k} ] 2. **Expressing sums in terms of ( y ) and ( sqrt{8} ):** [ x = sum_{k=0}^{n} binom{2n+1}{2k+1} 2^{3k} ] [ y = sum_{k=0}^{n} binom{2n+1}{2k} (sqrt{8})^{2k} ] 3. **Using the binomial theorem to combine and simplify:** According to the binomial theorem: [ (1 + sqrt{8})^{2n+1} = sum_{i=0}^{2n+1} binom{2n+1}{i} (sqrt{8})^i ] Separating terms based on even and odd indices: [ y + x sqrt{8} = sum_{i text{ even}} binom{2n+1}{i} (sqrt{8})^i + sum_{i text{ odd}} binom{2n+1}{i} (sqrt{8})^i = (1 + sqrt{8})^{2n+1} ] 4. **Considering the conjugate of ( (1 + sqrt{8}) ):** Similarly: [ y - x sqrt{8} = sum_{i text{ even}} binom{2n+1}{i} (sqrt{8})^i - sum_{i text{ odd}} binom{2n+1}{i} (sqrt{8})^i = (1 - sqrt{8})^{2n+1} ] 5. **Multiplying these equations to find a consistent formula:** Multiplying the two previous equations: [ (y + x sqrt{8})(y - x sqrt{8}) = (1 + sqrt{8})^{2n+1} (1 - sqrt{8})^{2n+1} ] This simplifies to: [ y^2 - 8x^2 = ((1+sqrt{8})(1-sqrt{8}))^{2n+1} = (-7)^{2n+1} = -7^{2n+1} ] 6. **Simplification using modulo arithmetic:** Reducing the equation modulo 5: [ y^2 - 8x^2 equiv -7^{2n+1} pmod{5} ] Since (-7 equiv 3 pmod{5}), we have: [ y^2 equiv 3x^2 + 3 cdot (-1)^n pmod{5} ] Now, analyze the implications: 7. **Analyzing ( x ) and divisibility conditions:** If ( x ) is divisible by 5: [ x equiv 0 pmod{5} ] This implies: [ y^2 equiv pm 2 pmod{5} ] Which is impossible since ( pm 2 ) mod 5 cannot be a quadratic residue. Thus, ( x ) cannot be divisible by 5 for any ( n ). 8. **Conclusion:** Given all the steps and simplifications above, there does not exist a natural number ( n ) for which the given sum ( S ) is divisible by 5. [ boxed{text{No}} ]