Question

The least number which is a perfect square and is divisible by each of the numbers 14, 16, 18 is

• A. 6048
• B. 7056
• C. 1008
• D. 2046

Hint:

Find the LCM first, then multiply by the smallest factor needed to make every prime's power even.

Answer 1

The Correct Option is B

Step 1: List what each divisor demands.
A number divisible by $16 = 2^4$ must carry at least $2^4$. A number divisible by $18 = 2 \times 3^2$ must carry at least $3^2$ (the 2 is already covered). A number divisible by $14 = 2 \times 7$ must carry at least $7^1$.

Step 2: Force every power to be even for a perfect square.
$2^4$ is already even. $3^2$ is already even. $7^1$ is odd, so it must be raised to $7^2$ to keep the number a perfect square.

Step 3: Multiply the smallest required powers.
The smallest such number is $2^4 \times 3^2 \times 7^2 = 16 \times 9 \times 49$. Computing this: $16 \times 9 = 144$, and $144 \times 49 = 7056$.

Final Answer:
This is the same value found by adjusting the LCM, confirming the answer. \[ \boxed{7056} \]