1. **Problem statement:** (a) Write $14M$ as a product of its prime factors in index form. (b) Given $M = 2^7 \times 3^3 \times 5^2$ and $R$ an integer such that $\frac{M}{R}$ is a cube number, find the smallest possible value of $R$. 2. **Step (a): Express $14M$ in prime factors.** - First, factorize 14 into primes: $14 = 2 \times 7$. - Given $M = 2^7 \times 3^3 \times 5^2$. - Multiply $14$ and $M$: $$14M = (2 \times 7)(2^7 \times 3^3 \times 5^2) = 2^{1+7} \times 3^3 \times 5^2 \times 7^1 = 2^8 \times 3^3 \times 5^2 \times 7^1$$ 3. **Step (b): Find smallest $R$ such that $\frac{M}{R}$ is a cube number.** - A cube number has all prime exponents multiples of 3. - $M = 2^7 \times 3^3 \times 5^2$. - Let $R = 2^a \times 3^b \times 5^c$ where $a,b,c \geq 0$. - Then $\frac{M}{R} = 2^{7 - a} \times 3^{3 - b} \times 5^{2 - c}$ must have exponents divisible by 3. Check each prime: - For 2: $7 - a \equiv 0 \pmod{3}$. - $7 \equiv 1 \pmod{3}$, so $a \equiv 1 \pmod{3}$. - Smallest $a$ is 1. - For 3: $3 - b \equiv 0 \pmod{3}$. - $3 \equiv 0 \pmod{3}$, so $b \equiv 0$. - Smallest $b$ is 0. - For 5: $2 - c \equiv 0 \pmod{3}$. - $2 \equiv 2 \pmod{3}$, so $c \equiv 2 \pmod{3}$. - Smallest $c$ is 2. Therefore, $R = 2^1 \times 3^0 \times 5^2 = 2 \times 25 = 50$. **Final answers:** - (a) $14M = 2^8 \times 3^3 \times 5^2 \times 7^1$ - (b) Smallest $R = 50$