1. **Problem Statement:** Find the prime factorisation of the numbers 72, 756, 187, and 630, expressing each answer in index notation. 2. **Formula and Rules:** Prime factorisation means expressing a number as a product of prime numbers raised to their powers. For example, if a number $n = p_1^{a} \times p_2^{b} \times \cdots$, where $p_i$ are primes and $a,b,\ldots$ are their exponents. 3. **Prime Factorisation Steps:** (a) For 72: - Divide by smallest prime 2: $72 \div 2 = 36$ - Divide 36 by 2: $36 \div 2 = 18$ - Divide 18 by 2: $18 \div 2 = 9$ - 9 is divisible by 3: $9 \div 3 = 3$ - 3 is prime. So, $72 = 2^3 \times 3^2$. (b) For 756: - Divide by 2: $756 \div 2 = 378$ - Divide 378 by 2: $378 \div 2 = 189$ - 189 divisible by 3: $189 \div 3 = 63$ - 63 divisible by 3: $63 \div 3 = 21$ - 21 divisible by 3: $21 \div 3 = 7$ - 7 is prime. So, $756 = 2^2 \times 3^3 \times 7^1$. (c) For 187: - Check divisibility by 2 (no), 3 (no), 5 (no), 7 (no), 11 (no), 13 (yes): $187 \div 11 = 17$ (actually 187 ÷ 11 = 17 is incorrect, check 187 ÷ 11 = 17? 11*17=187 yes) - 17 is prime. So, $187 = 11^1 \times 17^1$. (d) For 630: - Divide by 2: $630 \div 2 = 315$ - 315 divisible by 3: $315 \div 3 = 105$ - 105 divisible by 3: $105 \div 3 = 35$ - 35 divisible by 5: $35 \div 5 = 7$ - 7 is prime. So, $630 = 2^1 \times 3^2 \times 5^1 \times 7^1$. **Final answers:** - (a) $72 = 2^3 \times 3^2$ - (b) $756 = 2^2 \times 3^3 \times 7$ - (c) $187 = 11 \times 17$ - (d) $630 = 2 \times 3^2 \times 5 \times 7$