1. **State the problem:** Find the highest common factor (HCF) of 490, 588, and 882 using the prime factorization method. 2. **Prime factorize each number:** - 490: Divide by 2: 490 \div 2 = 245 Divide 245 by 5: 245 \div 5 = 49 Divide 49 by 7: 49 \div 7 = 7 Divide 7 by 7: 7 \div 7 = 1 So, $490 = 2 \times 5 \times 7 \times 7 = 2 \times 5 \times 7^2$ - 588: Divide by 2: 588 \div 2 = 294 Divide 294 by 2: 294 \div 2 = 147 Divide 147 by 3: 147 \div 3 = 49 Divide 49 by 7: 49 \div 7 = 7 Divide 7 by 7: 7 \div 7 = 1 So, $588 = 2^2 \times 3 \times 7^2$ - 882: Divide by 2: 882 \div 2 = 441 Divide 441 by 3: 441 \div 3 = 147 Divide 147 by 3: 147 \div 3 = 49 Divide 49 by 7: 49 \div 7 = 7 Divide 7 by 7: 7 \div 7 = 1 So, $882 = 2 \times 3^2 \times 7^2$ 3. **Identify common prime factors with the smallest powers:** - Common primes: 2 and 7 - For 2: minimum power is $2^1$ - For 7: minimum power is $7^2$ - 5 and 3 are not common to all three numbers 4. **Calculate the HCF:** $$\text{HCF} = 2^1 \times 7^2 = 2 \times 49 = 98$$ **Final answer:** The HCF of 490, 588, and 882 is $98$.