Question

What will be the least number which when doubled will be exactly divisible by 10, 18, 21, and 27?

• A. 1820
• B. 1830
• C. 945
• D. 1890

Hint:

To find the least number divisible by several numbers, find the LCM of those numbers. Then, divide the LCM by 2 if the number must be doubled.

Answer 1

The Correct Option is C

Step 1: Find the LCM of 10, 18, 21, and 27. The prime factorizations are: \[ 10 = 2 \times 5, \quad 18 = 2 \times 3^2, \quad 21 = 3 \times 7, \quad 27 = 3^3. \] The LCM is taken by selecting the highest powers of all prime factors: \[ \text{LCM} = 2^1 \times 3^3 \times 5^1 \times 7^1 = 2 \times 27 \times 5 \times 7 = 1890. \] Step 2: Find the least number such that when doubled it is divisible by 1890. Let the number be \( x \). When doubled, \( 2x \) must be divisible by 1890, so: \[ 2x = 1890 \quad \Rightarrow \quad x = \frac{1890}{2} = 945. \] So, the least number is 945.