answer:Let f(r) = r^{13} - r^5 + 1. According to the Remainder Theorem, the remainder when f(r) is divided by r - 1 is f(1) since r - 1 = 0 when r = 1: f(1) = 1^{13} - 1^5 + 1 = 1 - 1 + 1 = 1. Thus, the remainder when r^{13} - r^5 + 1 is divided by r-1 is boxed{1}.

answer:Solution: (1) Transposing terms, we get: 3y=4-2x, To make the coefficient 1, we get: y= frac {4-2x}{3}; (2) Transposing terms, we get: 2x=4-3y, To make the coefficient 1, we get: x= frac {4-3y}{2} . To write the equation 2x+3y-4=0 in the form where y is represented by an algebraic expression containing x, we need to move the terms containing y to one side of the equation, and the other terms to the other side, then make the coefficient 1. This gives us the form: y= frac {4-2x}{3}; To write it in the form where x is represented by an algebraic expression containing y, we need to move the terms containing x to one side of the equation, and the other terms to the other side, then make the coefficient 1. This gives us the form: x= frac {4-3y}{2} . This question tests basic equation operation skills, such as transposing, combining like terms, and making the coefficient 1. To represent someone, you should put them on one side of the equation, move the other terms to the other side, then combine like terms and make the coefficient 1. This allows you to represent y using an algebraic expression containing x, or represent x using an algebraic expression containing y. The final answers are: boxed{y= frac {4-2x}{3}} and boxed{x= frac {4-3y}{2}} .

answer:To find the distance from the center of the circle to the chord, let's denote the following: - ( R = 13 ) is the radius of the circle. - The length of the chord is ( 10 ). 1. **Identify the relevant geometry:** - The center of the circle is denoted by ( O ). - The chord is denoted by ( AB ). - Let ( M ) be the midpoint of the chord ( AB ). Since ( M ) is the midpoint, ( AM = MB = frac{10}{2} = 5 ). According to the Perpendicular Bisector Theorem: > The line segment ( OM ) from the center of the circle to the midpoint of a chord is perpendicular to the chord and bisects it. 2. **Formulate a right triangle:** - The triangle ( OMA ) is right-angled at ( M ). 3. **Using the Pythagorean theorem:** - ( OM ) is the distance from the center ( O ) to the chord ( AB ) (this is what we need to find). - ( OA ) is the radius of the circle: ( OA = 13 ). Using Pythagoras’ theorem in triangle ( OMA ): [ OA^2 = AM^2 + OM^2 ] 4. **Substitute the known values:** - ( OA = 13 ) - ( AM = 5 ) [ 13^2 = 5^2 + OM^2 ] 5. **Solve for ( OM ):** [ 169 = 25 + OM^2 ] [ OM^2 = 169 - 25 ] [ OM^2 = 144 ] [ OM = sqrt{144} = 12 ] # Conclusion: The distance from the center of the circle to the chord is ( boxed{12} ).

answer:1. **Determine the size and sum of the set ( S ):** Given: [ S = {1994, 1997, 2000, ldots, 1994 + 3k} ] The set ( S ) is an arithmetic sequence with the first term ( a = 1994 ) and common difference ( d = 3 ). The number of terms in ( S ), denoted by ( |S| ), is ( k + 1 ) since it starts from ( 1994 ) and increments by ( 3 ) up to ( 1994 + 3k ). The sum of all terms in ( S ), denoted by ( M ), is calculated as follows: [ M = sum_{j=0}^{k} (1994 + 3j) ] Simplifying the sum: [ M = sum_{j=0}^{k} 1994 + sum_{j=0}^{k} 3j ] The first sum: [ sum_{j=0}^{k} 1994 = 1994(k + 1) ] The second sum, which is the sum of the first ( k+1 ) natural numbers: [ sum_{j=0}^{k} 3j = 3 sum_{j=0}^{k} j = 3 left(frac{k(k+1)}{2}right)= frac{3k(k+1)}{2} ] Thus: [ M = 1994(k + 1) + frac{3k(k+1)}{2} ] Factor out (k+1): [ M = (k+1)left(1994 + frac{3k}{2}right) ] 2. **Equation for ( x ) in terms of ( k ):** Suppose sets ( A ) and ( B ) are such that each contains numbers from set ( S ), with sums equal to ( 9 ) times each other. Let the sum of numbers in ( B ) be ( x ). Consequently, the sum of numbers in ( A ) would be ( 9x ). Therefore: [ 10x = M = (k+1)left(1994 + frac{3k}{2}right) ] Solving for ( x ): [ x = frac{(k+1)(1994 + frac{3k}{2})}{10} = frac{(k+1)(19940 + 3k)}{20} ] 3. **Finding ( k ) so that ( x ) is an integer:** For ( x ) to be an integer, ( frac{(k+1)(3k+8)}{20} ) must be an integer. This implies: [ 20 mid (k+1)(3k+8) ] This condition can be broken down into two parts: [ 4 mid (k+1)(3k+8) quad text{and} quad 5 mid (k+1)(3k+8) ] 4. **Condition ( 5 mid (k+1)(3k+8) ):** Since ( 5 mid (k+1)(3k+8) ): [ 5 mid (k+1) ] Denote ( k+1 ) as ( 5m ), where ( m in mathbb{N} ): [ k = 5m - 1 ] Substitute back: [ k+1 = 5m ] The equation now becomes: [ 20 mid 5m cdot (3(5m-1)+8) = 5m cdot (15m-3+8) = 5m cdot (15m+5) ] For this to hold true under modulo 4: [ 4 mid 5m cdot (3m+1) ] This results in ( m ) having to be either ( 4t ) or ( 4t+1 ): [ m = 4t quad text{or} quad m = 4t+1 ] 5. **Analyzing the cases:** - **For ( m = 4t ):** Substituting ( m = 4t ): [ k = 5(4t) - 1 = 20t - 1 ] Which implies ( |S| = 20t ). Arranging the numbers of ( S ) into groups of 20, the largest and smallest can form ( B ) while others form ( A ). - **For ( m = 4t + 1 ):** Substituting ( m = 4t + 1 ) into ( k ): [ k = 5(4t + 1) - 1 = 20t + 4 ] 6. **Conclusion:** By iterating these cases and analyzing the sums, we conclude the two possible forms for ( k ): [ boxed{k = 20t - 1 text{ where } t in {1, 2, 3, ldots} text{ or } k = 20t + 4 text{ where } t in {8, 9, 10, ldots}} ]