answer:1. Let the n natural numbers be denoted as x_1, x_2, cdots, x_n. 2. Define ( s_n ) as the sum of the ratios of the sums of adjacent pairs of these numbers to their respective numbers: [ s_n = frac{x_1 + x_3}{x_2} + frac{x_2 + x_4}{x_3} + cdots + frac{x_{n-2} + x_n}{x_{n-1}} + frac{x_{n-1} + x_1}{x_n} + frac{x_n + x_2}{x_1} ] 3. By rewriting and combining the terms, it becomes clearer: [ s_n = left( frac{x_1}{x_2} + frac{x_2}{x_1} right) + left( frac{x_3}{x_2} + frac{x_2}{x_3} right) + cdots + left( frac{x_{n-1}}{x_n }+ frac{x_n}{x_{n-1}} right) + left( frac{x_n}{x_1} + frac{x_1}{x_n} right) ] 4. Each fraction pair (left( frac{x_i}{x_j} + frac{x_j}{x_i} right)) is at least 2 by the Arithmetic Mean-Geometric Mean (AM-GM) inequality, since: [ frac{x_i}{x_j} + frac{x_j}{x_i} geq 2 ] 5. Summing these up, we get: [ s_n geq 2n ] 6. To prove the upper bound ( s_n < 3n ) by induction: - Base Case: ( n = 3 ): We have: [ s_3 = frac{x_1 + x_2}{x_3} + frac{x_2 + x_3}{x_1} + frac{x_3 + x_1}{x_2} ] Assume ( x_3 = max {x_1, x_2, x_3} ). Then, ( frac{x_1 + x_2}{x_3} leq 2 ) since ( frac{x_1 + x_2}{x_3} ) is a natural number. Two cases arise: - If ( frac{x_1 + x_2}{x_3} = 2 ): [ Rightarrow x_1 = x_2 = x_3 ] Then: [ s_3 = 3 times 2 = 6 < 9 ] - If ( frac{x_1 + x_2}{x_3} = 1 ): Let ( p = frac{x_2 + x_3}{x_1} ) and ( q = frac{x_3 + x_1}{x_2} ) (both being natural numbers), it follows: [ 2x_2 = (p-1)x_1, quad 2x_1 = (q-1)x_2 ] So we have: [ (p-1)(q-1) = 4 ] This implies: - ( (p-1, q-1) = (4, 1) ) or ( (1, 4) ) - ( (p-1, q-1) = (2, 2) ) Thus: [ s_3 = 1 + p + q = begin{cases} 8 & text{for } (p-1,q-1) = (4, 1) text{ or } (1, 4) 7 & text{for } (p-1, q-1) = (2, 2) end{cases} ] Both are ( < 9 ). Thus in all cases: [ s_3 < 9 ] - Inductive Step: Assume ( s_k < 3k ) for ( n = k ). For ( n = k + 1 ): [ s_{k+1} = frac{x_1 + x_3}{x_2} + frac{x_2 + x_4}{x_3} + cdots + frac{x_{k-1} + x_{k+1}}{x_k} + frac{x_k + x_1}{x_{k+1}} + frac{x_{k+1} + x_2}{x_1} ] Assume ( x_{k+1} = max {x_1, x_2, cdots, x_k, x_{k+1}} ): - If ( frac{x_{k} + x_{1}}{x_{k+1}} = 2 ): [ x_{k+1} = x_1 = x_k ] - So: [ s_{k+1} = 3k - frac{x_{k-1} + x_1}{x_k} - frac{x_k + x_2}{x_1} + frac{x_{k-1} + x_{k+1}}{x_k} + frac{x_k + x_1}{x_{k+1}} + frac{x_{k+1} + x_2}{x_1} ] Which simplifies to: [ s_{k+1} < 3k + 2 < 3(k+1) ] Or if ( frac{x_{k} + x_{1}}{x_{k+1}} = 1 ), ( x_{k+1} = x_1 + x_k ): - And all components naturally numbers: [ s_{k+1} = 3k + 1 + frac{x_{k+1} - x_{1}}{x_k} + frac{x_k + x_1}{x_{k+1}} < 3(k + 1) ] [ s_{k+1} < 3(k + 1) ] 7. Thus, for any natural number ( n geq 3 ): [ 2n leq s_n < 3n ] Therefore: [ boxed{2n leq s_n < 3n} ]

answer:To solve this problem, we need to find the number of 11-digit positive integers ( K ) that satisfy the following conditions: 1. ( K ) is divisible by ( 2 ), ( 3 ), and ( 5 ). 2. There exists some positive integer ( K' ) such that ( K' sim K ) and ( K' ) is divisible by ( 7 ), ( 11 ), ( 13 ), ( 17 ), ( 101 ), and ( 9901 ). Let's break down the problem step by step. 1. **Divisibility by 2, 3, and 5:** - For ( K ) to be divisible by ( 2 ), its last digit must be even. - For ( K ) to be divisible by ( 3 ), the sum of its digits must be divisible by ( 3 ). - For ( K ) to be divisible by ( 5 ), its last digit must be ( 0 ) or ( 5 ). Since ( K ) is an 11-digit number, the last digit must be ( 0 ) to satisfy both divisibility by ( 2 ) and ( 5 ). Therefore, ( K ) ends in ( 0 ). 2. **Divisibility by 7, 11, 13, 17, 101, and 9901:** - We need to find a rearrangement ( K' ) of ( K ) such that ( K' ) is divisible by these numbers. - Notice that ( 7 times 11 times 13 = 1001 ) and ( 101 times 9901 = 1000001 ). Therefore, ( K' ) must be divisible by ( 1001 ) and ( 1000001 ). 3. **Properties of multiples of 11 and 101:** - For ( K ) to be divisible by ( 11 ), the alternating sum of its digits must be divisible by ( 11 ). - For ( K ) to be divisible by ( 101 ), the sum of every second digit must be divisible by ( 101 ). 4. **Counting the number of valid ( K ):** - Since ( K ) ends in ( 0 ), we need to arrange the remaining 10 digits such that the sum of the digits is divisible by ( 3 ) and the alternating sum is divisible by ( 11 ). Let's consider the digits ( 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 ) and one additional digit to make it 11 digits. The sum of these digits is ( 45 ). To make the sum divisible by ( 3 ), we can add ( 0 ) (since ( 45 ) is already divisible by ( 3 )). Now, we need to ensure that the alternating sum is divisible by ( 11 ). This can be achieved by arranging the digits appropriately. 5. **Calculating the number of valid permutations:** - There are ( 10! ) ways to arrange the 10 digits ( 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 ). - Since the last digit is fixed as ( 0 ), we have ( 10! ) valid permutations. Therefore, the number of valid 11-digit integers ( K ) is ( 10! = 3628800 ). The final answer is ( boxed{3628800} ).

answer:Given overrightarrow {a}=(1,0,2) and overrightarrow {b}=(0,1,2), let's find overrightarrow {a}-2overrightarrow {b}: begin{align*} overrightarrow {a}-2overrightarrow {b} &= (1,0,2) - 2 cdot (0,1,2) &= (1,0,2) - (0,2,4) &= (1-0, 0-2, 2-4) &= (1, -2, -2). end{align*} Next, we compute the magnitude of overrightarrow {a}-2overrightarrow {b}: begin{align*} |overrightarrow {a}-2overrightarrow {b}| &= sqrt{(1)^2 + (-2)^2 + (-2)^2} &= sqrt{1 + 4 + 4} &= sqrt{9} &= boxed{3}. end{align*}

answer:1. **Calculate the projection of mathbf{v} onto mathbf{w}:** [ text{Proj}_{mathbf{w}} mathbf{v} = frac{mathbf{v} cdot mathbf{w}}{mathbf{w} cdot mathbf{w}} mathbf{w} = frac{begin{pmatrix} 1 -2 -2 end{pmatrix} cdot begin{pmatrix} 1 1 1 end{pmatrix}}{begin{pmatrix} 1 1 1 end{pmatrix} cdot begin{pmatrix} 1 1 1 end{pmatrix}} begin{pmatrix} 1 1 1 end{pmatrix} = frac{1 - 2 - 2}{1 + 1 + 1} begin{pmatrix} 1 1 1 end{pmatrix} = -1 begin{pmatrix} 1 1 1 end{pmatrix} ] [ boxed{begin{pmatrix} -1 -1 -1 end{pmatrix}} ] 2. **Calculate the projection of mathbf{u} onto mathbf{w}:** [ text{Proj}_{mathbf{w}} mathbf{u} = frac{mathbf{u} cdot mathbf{w}}{mathbf{w} cdot mathbf{w}} mathbf{w} = frac{begin{pmatrix} 2 -1 3 end{pmatrix} cdot begin{pmatrix} 1 1 1 end{pmatrix}}{begin{pmatrix} 1 1 1 end{pmatrix} cdot begin{pmatrix} 1 1 1 end{pmatrix}} begin{pmatrix} 1 1 1 end{pmatrix} = frac{2 - 1 + 3}{3} begin{pmatrix} 1 1 1 end{pmatrix} = 4/3 begin{pmatrix} 1 1 1 end{pmatrix} ] [ boxed{begin{pmatrix} 4/3 4/3 4/3 end{pmatrix}} ] 3. **Sum the projections of mathbf{v} and mathbf{u} onto mathbf{w}:** [ begin{pmatrix} -1 -1 -1 end{pmatrix} + begin{pmatrix} 4/3 4/3 4/3 end{pmatrix} = begin{pmatrix} -1 + 4/3 -1 + 4/3 -1 + 4/3 end{pmatrix} = begin{pmatrix} 1/3 1/3 1/3 end{pmatrix} ] [ boxed{begin{pmatrix} 1/3 1/3 1/3 end{pmatrix}} ]