Question

The least numbers divisible by 2,3,7 and 9 is :

• A. 126
• B. 256
• C. 251
• D. 189

Hint:

The "least number divisible by" a set of numbers is their LCM (Least Common Multiple). Numbers: 2, 3, 7, 9. Notice that if a number is divisible by 9, it's automatically divisible by 3. So, we need the LCM of 2, 7, and 9. Prime factors: \(2 = 2\) \(7 = 7\) \(9 = 3^2\) LCM = Product of the highest powers of all prime factors involved = \(2^1 \times 3^2 \times 7^1 = 2 \times 9 \times 7 = 126\).

Answer 1

The Correct Option is A

Concept: The least number divisible by a set of given numbers is their Least Common Multiple (LCM). Step 1: List the numbers The given numbers are 2, 3, 7, and 9. Step 2: Find the prime factorization of each number
\(2 = 2^1\)
\(3 = 3^1\)
\(7 = 7^1\)
\(9 = 3^2\) Step 3: Calculate the LCM To find the LCM, take the highest power of each prime factor present in any of the numbers. The prime factors involved are 2, 3, and 7.
Highest power of 2: \(2^1 = 2\)
Highest power of 3: \(3^2 = 9\) (from the number 9, as \(3^2>3^1\))
Highest power of 7: \(7^1 = 7\) LCM = \(2^1 \times 3^2 \times 7^1\) LCM = \(2 \times 9 \times 7\) LCM = \(18 \times 7\) Calculate \(18 \times 7\): \(18 \times 7 = (10+8) \times 7 = 70 + 56 = 126\). So, the LCM is 126. The least number divisible by 2, 3, 7, and 9 is 126. Note: Since 9 is a multiple of 3, any number divisible by 9 is automatically divisible by 3. So, we effectively need the LCM of 2, 7, and 9. LCM(2, 7, 9) = LCM(2, 7, \(3^2\)). Since 2, 7, and 9 are pairwise coprime (no common factors other than 1), their LCM is their product: \(2 \times 7 \times 9 = 14 \times 9 = 126\).