answer:To solve this problem, we follow these steps: 1. **Calculate the final number of knives Carolyn has:** - Initially, Carolyn has 6 knives. - After trading, she gains 10 knives. - Therefore, the final number of knives is 6 + 10 = boxed{16 text{ knives}}. 2. **Determine the initial number of spoons Carolyn has:** - Since she has three times as many spoons as knives initially, and she starts with 6 knives, the initial number of spoons is 6 times 3 = boxed{18 text{ spoons}}. 3. **Calculate the final number of spoons Carolyn has:** - After trading 6 spoons for 10 knives, the final number of spoons is 18 - 6 = boxed{12 text{ spoons}}. 4. **Find the total final amount of silverware Carolyn has:** - Adding the final numbers of spoons, knives, and forks (remembering she has 12 forks), we get the total as 12 + 16 + 12 = boxed{40 text{ pieces}}. 5. **Calculate the percentage of silverware that is knives:** - To find the percentage, we divide the number of knives by the total number of pieces and multiply by 100%: frac{16}{40} times 100% = boxed{40%}. Therefore, the percentage of Carolyn's silverware that is knives is boxed{40%}.

answer:The number AB_8 in decimal is 8A + B, and the number BA_6 in decimal is 6B + A. Setting these two expressions equal since they represent the same number, we have: [ 8A + B = 6B + A ] Rearranging terms gives: [ 7A = 5B ] [ frac{A}{B} = frac{5}{7} ] Since A and B are digits in their respective bases, and must be integers, check suitable values. For base 8, A can be 0 to 7, and for base 6, B can be 0 to 5. We seek values where frac{A}{B} = frac{5}{7} holds true: [ A = 5, B = 7 times frac{5}{7} = 5 ] Thus, A = 5 and B = 5 are the only values satisfying the equation and the base restrictions. Therefore, the decimal value is: [ 8 times 5 + 5 = 40 + 5 = boxed{45} ]

answer:To determine how many integers Josef could pick such that the quotient is an integer, we need to find the number of positive divisors of 360. 360 can be factored into primes as (360 = 2^3 cdot 3^2 cdot 5^1). The formula for the number of divisors of a number based on its prime factorization is ((e_1 + 1)(e_2 + 1)...(e_k + 1)), where (e_i) are the exponents in the prime factorization of the number. For 360, this becomes: [ (3+1)(2+1)(1+1) = 4 cdot 3 cdot 2 = 24. ] Thus, there are (boxed{24}) integers Josef could pick to make Timothy's number an integer.

answer:1. **Define ( b_k ):** For ( k = 1, 2, ldots, n ), define ( b_k ) as follows: [ begin{aligned} b_k = & a_1 q^{k-1} + a_2 q^{k-2} + cdots + a_{k-1} q + a_k & + a_{k+1} q + a_{k+2} q^2 + cdots + a_n q^{n-k}. end{aligned} ] 2. **Part (a): Show that ( b_k > a_k ):** We need to show that ( b_k > a_k ) for each ( k = 1, 2, ldots, n ). Notice that: [ b_k = a_k + a_1 q^{k-1} + a_2 q^{k-2} + cdots + a_{k-1} q + a_{k+1} q + cdots + a_n q^{n-k}. ] Since all ( a_i ) are positive and ( 0 < q < 1 ), each term ( a_i q^j ) for ( i ne k ) is positive. Therefore, [ b_k > a_k. ] 3. **Part (b): Show the inequality for ( frac{b_{k+1}}{b_k} ):** For ( k = 1, 2, ldots, n-1 ), we need to show that [ q < frac{b_{k+1}}{b_k} < frac{1}{q}. ] Start by considering the difference: [ q b_k - b_{k+1} = left( q^2 - 1 right) left[ a_{k+1} + a_{k+2} q + cdots + a_n q^{n-k-1} right]. ] Since ( q^2 - 1 < 0 ) and all terms within brackets are positive, we have: [ q b_k - b_{k+1} < 0 quad Rightarrow quad q b_k < b_{k+1}. ] This implies: [ q < frac{b_{k+1}}{b_k}. ] Similarly, consider: [ q b_{k+1} - b_k = left( q^2 - 1 right) left[ a_1 q^{k} + a_2 q^{k-1} + cdots + a_{k} right]. ] Since ( q^2 - 1 < 0 ) and all terms within brackets are positive, we have: [ q b_{k+1} - b_k < 0 quad Rightarrow quad q b_{k+1} < b_k. ] This implies: [ frac{b_{k+1}}{b_k} < frac{1}{q}. ] Therefore, we have: [ q < frac{b_{k+1}}{b_k} < frac{1}{q}. ] 4. **Part (c): Show the sum inequality:** Consider the sum: [ b_1 + b_2 + cdots + b_n. ] By substituting the definition of ( b_k ), we get: [ begin{aligned} b_1 + b_2 + cdots + b_n = & a_1 + a_2 q + a_3 q^2 + cdots + a_n q^{n-1} & + a_1 q + a_2 + a_3 q + cdots + a_n q^{n-2} & + a_1 q^2 + a_2 q + a_3 + cdots + a_n q^{n-3} & + cdots + & + a_1 q^{n-1} + a_2 q^{n-2} + cdots + a_n. end{aligned} ] Notice that each ( a_i ) appears multiple times with different powers of ( q ), forming a geometric series sum. Therefore: [ begin{aligned} b_1 + b_2 + cdots + b_n < & (a_1 + a_2 + cdots + a_n) left( 1 + q + q^2 + cdots + q^{n-1} right) & + (a_1 + a_2 + cdots + a_n) left( q + q^2 + cdots + q^{n-1} right). end{aligned} ] This is simplified using the geometric series sum formula: [ begin{aligned} b_1 + b_2 + cdots + b_n < & (a_1 + a_2 + cdots + a_n) left( 1 + 2(q + q^2 + cdots + q^{n-1}) right) = & (a_1 + a_2 + cdots + a_n) left( 1 + 2 frac{q (1 - q^{n-1})}{1 - q} right) < & left( a_1 + a_2 + cdots + a_n right) left( frac{1 + q}{1 - q} right). end{aligned} ] Therefore: [ b_1 + b_2 + cdots + b_n < frac{1 + q}{1 - q} left( a_1 + a_2 + cdots + a_n right). ] **Conclusion:** boxed{}