1. **State the problem:** We are given that the least common multiple (LCM) of 15, 18, and a third number $x$ is 1260. We need to find the value of $x$. 2. **Recall the formula and rules:** The LCM of multiple numbers is the smallest number divisible by all of them. For two numbers $a$ and $b$, $\text{LCM}(a,b) = \frac{|a \times b|}{\text{GCD}(a,b)}$. For three numbers, the LCM can be found stepwise: $\text{LCM}(a,b,c) = \text{LCM}(\text{LCM}(a,b), c)$. 3. **Calculate LCM of 15 and 18:** - Prime factorization: $15 = 3 \times 5$, $18 = 2 \times 3^2$ - LCM takes highest powers: $2^1, 3^2, 5^1$ - So, $\text{LCM}(15,18) = 2 \times 3^2 \times 5 = 2 \times 9 \times 5 = 90$ 4. **Use the LCM with the third number $x$:** We know $\text{LCM}(15,18,x) = 1260$, so $$\text{LCM}(90, x) = 1260$$ 5. **Find $x$ using the LCM formula:** $$\text{LCM}(90, x) = \frac{90 \times x}{\text{GCD}(90, x)} = 1260$$ 6. **Rearrange to find $x$:** $$1260 = \frac{90 \times x}{\text{GCD}(90, x)} \implies 1260 \times \text{GCD}(90, x) = 90 \times x$$ $$x = \frac{1260 \times \text{GCD}(90, x)}{90} = 14 \times \text{GCD}(90, x)$$ 7. **Determine possible values of $x$:** Since $x = 14 \times \text{GCD}(90, x)$ and $\text{GCD}(90, x)$ divides 90, possible divisors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Check which $x$ divides 1260: - For $\text{GCD}(90,x) = 5$, $x = 14 \times 5 = 70$ - Check if LCM(90,70) = 1260: - $\text{GCD}(90,70) = 10$ - $\text{LCM}(90,70) = \frac{90 \times 70}{10} = 630$ - Not 1260, so discard. Try $\text{GCD}(90,x) = 10$: - $x = 14 \times 10 = 140$ - $\text{GCD}(90,140) = 10$ - $\text{LCM}(90,140) = \frac{90 \times 140}{10} = 1260$ This matches the required LCM. **Final answer:** $$\boxed{140}$$