This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Here are the solutions to the problems. 4) Step 1: Factorize each expression. x^2 + x = x(x+1) x^2 - 1 = (x-1)(x+1) x^2 - x = x(x-1) Step 2: Identify all unique factors and their highest powers. The unique factors are x, (x+1), and (x-1). Each appears with a highest power of 1. Step 3: Multiply these highest powers together to find the L.C.M. L.C.M = x(x+1)(x-1) This can also be written as: L.C.M = x(x^2-1) = x^3-x x(x^2-1) or x^3-x 5) Step 1: Find the prime factorization of each number. 9 = 3^2 15 = 3 × 5 20 = 2^2 × 5 24 = 2^3 × 3 Step 2: Identify all unique prime factors and their highest powers. • Highest power of 2: 2^3 • Highest power of 3: 3^2 • Highest power of 5: 5^1 Step 3: Multiply these highest powers together to find the L.C.M. L.C.M = 2^3 × 3^2 × 5^1 = 8 × 9 × 5 = 72 × 5 = 360 The least number of sweets is 360. 360 6) Step 1: Find the prime factorization of each number. 27 = 3^3 30 = 2 × 3 × 5 45 = 3^2 × 5 Step 2: Identify all unique prime factors and their highest powers. • Highest power of 2: 2^1 • Highest power of 3: 3^3 • Highest power of 5: 5^1 Step 3: Multiply these highest powers together to find the L.C.M. L.C.M = 2^1 × 3^3 × 5^1 = 2 × 27 × 5 = 54 × 5 = 270 Step 4: Add the remainder to the L.C.M. to find n. Since the remainder is always 3, n = L.C.M + 3. n = 270 + 3 = 273 273 7) Step 1: Find the prime factorization of each number (20, 24, 28). 20 = 2^2 × 5 24 = 2^3 × 3 28 = 2^2 × 7 Step 2: Identify all unique prime factors and their highest powers. • Highest power of 2: 2^3 • Highest power of 3: 3^1 • Highest power of 5: