answer:To demonstrate that any segment can be divided into black and white sub-segments such that the integrals of any polynomial of degree at most (n) over these black and white sub-segments are equal, we proceed by induction. 1. **Base Case (n=0):** - For (n=0), the polynomial of degree (0) is a constant function. The integral of a constant over any segment is the product of this constant and the length of the segment. - Hence, it suffices to divide the segment into two equal parts. 2. **Induction Step:** - Assume that for a segment ([a ; b]) and any polynomial of degree at most (n-1), the segment can be divided into black and white sub-segments such that the integrals over these sub-segments are equal. - Let (c) be the midpoint of ([a ; b]). - According to our induction hypothesis, in the interval ([a ; c]), we can divide it into sub-segments ( B_{1}, ldots, B_{r} ) (black) and ( W_{1}, ldots, W_{s} ) (white) such that the integrals of any polynomial of degree at most ( n-1 ) are equal. - Let ( f(x) ) be any polynomial of degree ( n ). - We need to prove that this property holds for ([a ; b]). 3. **Re-labeling Integrals:** - Denote the integral of ( f(x) ) over ( B_{i} ) as: [ int_{B_{i}} f(x) d x ] - Similarly, denote the integral over ( W_{i} ). 4. **Translation and Coloring:** - Move each black segment ( B_{i} ) from ([a ; c]) to the interval ([c; b]) and color them white. Denote them as ( W_{s+i} ). - Move each white segment ( W_{i} ) similarly and denote them as ( B_{r+i} ). 5. **Combining Results:** - The segments ( B_{1}, ldots, B_{r+s} ) (black) and ( W_{1}, ldots, W_{r+s} ) (white) now form the division of ([a ; b]). 6. **Proving Equality of Integrals:** - For black segments and white segments after translation, we have: [ int_{W_{r+i}} f(x) d x = int_{B_{i}} f(y+c) d y ] This follows from the substitution: [ y = x - c, quad dy = dx ] - Thus: [ int_{W_{r+i}} f(x) d x - int_{B_{i}} f(x) d x = int_{B_{i}} (f(x+c) - f(x)) d x ] - By adding similar results: [ sum_{i=1}^{r+s} int_{W_{i}} f(x) d x - sum_{i=1}^{r+s} int_{B_{i}} f(x) d x = sum_{i=1}^{r} int_{B_{i}} (f(x+c) - f(x)) d x - sum_{i=1}^{s} int_{W_{i}} (f(x+c) - f(x)) d x ] - Here, if ( f(x) ) is a polynomial of degree at most ( n ), ( f(x+c) - f(x) ) is a polynomial of degree at most ( n-1 ). 7. **Applying Induction Hypothesis:** - By the induction hypothesis, the integral of ( f(x+c) - f(x) ) over black and white segments in ([a ; c]) are equal. Hence, the right-hand side sums to zero: [ = 0 ] - Therefore, the left-hand side must also be zero, proving our hypothesis for degree ( n ): [ sum_{i=1}^{r+s} int_{W_{i}} f(x) d x = sum_{i=1}^{r+s} int_{B_{i}} f(x) d x ] This completes the proof by induction. blacksquare

answer:1. **Define the Triangular Number:** Let ( T(m) ) be the ( m )-th triangular number, defined as: [ T(m) = frac{m(m+1)}{2} ] Triangular numbers represent the sum of the first ( m ) natural numbers. 2. **Existence of ( m_0 ):** For any given positive integer ( n ), there exists a unique ( m_0 ) such that: [ T(m_0) < n leq T(m_0 + 1) ] This is because triangular numbers increase monotonically, and for any ( n ), there is a unique interval ([T(m_0), T(m_0 + 1)]) that contains ( n ). 3. **Express ( n ) in terms of ( m_0 ):** Let: [ a + b - 2 = m_0 ] and [ a = n - T(m_0) ] We need to show that ( a ) and ( b ) are positive integers and that they form a unique ordered pair. 4. **Positivity of ( a ):** Since ( T(m_0) < n ), we have: [ a = n - T(m_0) > 0 ] Thus, ( a ) is a positive integer. 5. **Expression for ( b ):** From ( a + b - 2 = m_0 ), we can solve for ( b ): [ b = m_0 - a + 2 ] Substitute ( a = n - T(m_0) ) into the equation for ( b ): [ b = m_0 - (n - T(m_0)) + 2 = T(m_0) + m_0 + 2 - n ] Using the definition of the next triangular number: [ T(m_0 + 1) = T(m_0) + m_0 + 1 ] we get: [ b = (T(m_0) + m_0 + 1) - n + 1 = T(m_0 + 1) - n + 1 ] 6. **Positivity of ( b ):** Since ( n leq T(m_0 + 1) ), we have: [ T(m_0 + 1) - n geq 0 ] Therefore: [ b = T(m_0 + 1) - n + 1 geq 1 ] Thus, ( b ) is a positive integer. 7. **Uniqueness of ( (a, b) ):** The uniqueness of ( m_0 ) ensures the uniqueness of the pair ( (a, b) ). Since ( m_0 ) is uniquely determined by ( n ), and ( a ) and ( b ) are uniquely determined by ( m_0 ), the pair ( (a, b) ) is unique. (blacksquare)

answer:First, consider the function [ f(x, y, z) = |x + y + z| + |x + y - z| + |x - y + z| + |-x + y + z|. ] Using symmetry arguments, (f(x, y, z)) is symmetric about the planes (xy), (xz), and (yz). Since we are looking for the volume within specific bounds, consider the situation in the first octant where (x geq y) and (x geq z). Here, [ f(x, y, z) = 3x + y + z + |-x + y + z| geq 3x + y + z + (y + z - x) = 2x + 2y + 2z. ] Given (f(x, y, z) leq 6), we deduce (x + y + z leq 3). Now, restrict the region to the bounds (0 leq x leq 2), (0 leq y leq 1), (0 leq z leq 1). The additional constraint (x + y + z leq 3) does not affect the given bounds. The region of interest is a rectangular prism defined by the maximum values of (x), (y), and (z). The volume of the rectangular prism is [ 2 times 1 times 1 = 2. ] Therefore, the volume of the region defined by the inequality and the bounds is [ boxed{2}. ]

answer:Let's denote the number that was added as ( x ). According to the problem, we have the following sequence of operations: 1. A certain number (62.5) is added to ( x ), giving us ( 62.5 + x ). 2. The sum is multiplied by 2, giving us ( 2(62.5 + x) ). 3. The product is divided by 5, giving us ( frac{2(62.5 + x)}{5} ). 4. 5 is subtracted from the quotient, giving us ( frac{2(62.5 + x)}{5} - 5 ). 5. The remainder left is half of 44, which is 22. So, we have the equation: [ frac{2(62.5 + x)}{5} - 5 = 22 ] Now, let's solve for ( x ): [ frac{2(62.5 + x)}{5} = 22 + 5 ] [ frac{2(62.5 + x)}{5} = 27 ] [ 2(62.5 + x) = 27 times 5 ] [ 2(62.5 + x) = 135 ] [ 62.5 + x = frac{135}{2} ] [ 62.5 + x = 67.5 ] [ x = 67.5 - 62.5 ] [ x = 5 ] The number that was added is boxed{5} .