answer:To prove that the product (xyz) is divisible by (5) and (55), we will proceed in two parts. # Part (a): Prove that (xyz) is divisible by (5) 1. **Lemma**: If (x, y, z) form a Pythagorean triple, then at least one of (x, y, z) is divisible by (5). **Proof**: Consider the reduced residue classes modulo (5), i.e., ({0, 1, -1}). We need to show that no permutation of (pm 1) and (pm 1) can be congruent to (1) or (-1). - The possible quadratic residues modulo (5) are (0, 1, 4). Note that (4 equiv -1 pmod{5}). - Therefore, the possible values for (x^2, y^2, z^2) modulo (5) are ({0, 1, -1}). 2. **Case Analysis**: - Suppose (x^2 + y^2 = a^2), (y^2 + z^2 = b^2), and (z^2 + x^2 = c^2). - If (x, y, z) are not divisible by (5), then (x^2, y^2, z^2) must be congruent to (1) or (-1) modulo (5). 3. **Contradiction**: - Assume (y^2 equiv -1 pmod{5}). Then (x^2 equiv 1 pmod{5}) and (z^2 equiv -1 pmod{5}). - From (y^2 + z^2 = b^2), we get (-1 + (-1) equiv b^2 pmod{5}), which implies (b^2 equiv -2 pmod{5}). This is a contradiction because (-2) is not a quadratic residue modulo (5). - Similarly, if (y^2 equiv 1 pmod{5}), then (x^2 equiv -1 pmod{5}) and (z^2 equiv -1 pmod{5}). From (z^2 + x^2 = c^2), we get (-1 + (-1) equiv c^2 pmod{5}), which implies (c^2 equiv -2 pmod{5}). This is also a contradiction. 4. **Conclusion**: - Therefore, at least one of (x, y, z) must be divisible by (5). # Part (b): Prove that (xyz) is divisible by (55) 1. **Divisibility by (5)**: - From part (a), we have already established that at least one of (x, y, z) is divisible by (5). 2. **Divisibility by (11)**: - Consider the equations modulo (11). The quadratic residues modulo (11) are ({0, 1, 3, 4, 5, 9}). - Similar to the argument for (5), we need to show that at least one of (x, y, z) is divisible by (11). 3. **Case Analysis**: - Suppose (x, y, z) are not divisible by (11), then (x^2, y^2, z^2) must be one of the quadratic residues modulo (11). 4. **Contradiction**: - Assume (y^2 equiv -1 pmod{11}). Then (x^2 equiv 1 pmod{11}) and (z^2 equiv -1 pmod{11}). - From (y^2 + z^2 = b^2), we get (-1 + (-1) equiv b^2 pmod{11}), which implies (b^2 equiv -2 pmod{11}). This is a contradiction because (-2) is not a quadratic residue modulo (11). - Similarly, if (y^2 equiv 1 pmod{11}), then (x^2 equiv -1 pmod{11}) and (z^2 equiv -1 pmod{11}). From (z^2 + x^2 = c^2), we get (-1 + (-1) equiv c^2 pmod{11}), which implies (c^2 equiv -2 pmod{11}). This is also a contradiction. 5. **Conclusion**: - Therefore, at least one of (x, y, z) must be divisible by (11). Since (xyz) is divisible by both (5) and (11), it must be divisible by (55). The final answer is ( boxed{ xyz } ) is divisible by (55).

answer:To solve this problem, we will reverse the transformations and find the original coordinates (a, b). Step 1: Reflect the point about the line y=x The final position of the transformed point is (4,1). Reflecting this point about the line y=x involves swapping the coordinates. Thus, the point (4, 1) reflects to (1, 4). Step 2: Reverse the 90^circ counterclockwise rotation around (2,3) After reflection, the point before rotation is (1, 4). To find the original position ((a, b)) before the rotation, we apply the reverse of a 90^circ clockwise rotation about the point (2,3). [ (x', y') = (x_c + (y - y_c), y_c - (x - x_c)) ] Plugging in: [ x' = 2 + (4 - 3) = 2 + 1 = 3 ] [ y' = 3 - (1 - 2) = 3 - (-1) = 4 ] Thus, the original coordinates of P are (3, 4). Step 3: Calculate b - a [ b - a = 4 - 3 = 1 ] 1 The final answer is boxed{- **A) 1**}

answer:To find out how many apples Joan picked from the orchard, we need to subtract the number of apples Melanie gave her from the total number of apples Joan has now. Joan has 70 apples now. Melanie gave Joan 27 apples. So, the number of apples Joan picked from the orchard is: 70 (total apples) - 27 (apples given by Melanie) = 43 apples Joan picked boxed{43} apples from the orchard.

answer:First, find the sets A and B. Set A = {x in mathbb{N} mid -1 < log_2{x} leq 2} = {x in mathbb{N} mid frac{1}{2} < x leq 4} = {1, 2, 3, 4}. Set B = {x mid frac{5 - 2x}{x - 6} > 0} = {x mid frac{5}{2} < x < 6}. Now, find the intersection of sets A and B: A cap B = {3, 4}. Next, find the number of subsets of A cap B. The number of subsets of any set with n elements is 2^n. In this case, n = 2, so there are 2^2 = 4 subsets. Therefore, Z(A cap B) = boxed{4}. The answer is (B).