What will be the least number which when doubled becomes exactly divisible by 9, 15, 21, and 30?

To find the least number which when doubled will be exactly divisible by 9, 15, 21, and 30, we need to find the least common multiple (LCM) of these numbers.

Step 1: Prime factorization of the numbers

• 9 = 3^2
• 15 = 3 * 5
• 21 = 3 * 7
• 30 = 2 * 3 * 5

Step 2: Find the LCM The LCM is found by multiplying the highest power of all the prime numbers that appear in the factorization of each number. So, LCM = 2^1 * 3^2 * 5^1 * 7^1 = 2 * 9 * 5 * 7 = 630

Step 3: Find the least number Since we want the number that when doubled is divisible by the LCM, we divide the LCM by 2. So, the least number = 630 / 2 = 315

Therefore, 315 is the least number which when doubled will be exactly divisible by 9, 15, 21, and 30.