Question

What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24?

• A. 264
• B. 259
• C. 269
• D. None of these

Hint:

When a number increased by a constant is divisible by several numbers, take the LCM of those divisors and subtract the constant.

Answer 1

The Correct Option is B

We are told that $(\text{number} + 5)$ is divisible by 8, 11, and 24.
First, find the least common multiple (LCM) of 8, 11, and 24.
Prime factors: $8 = 2^3$, $11$ is prime, $24 = 2^3 \times 3$.
LCM = $2^3 \times 3 \times 11 = 264$.
Therefore, the number + 5 must be a multiple of 264.
The smallest positive number satisfying this is when: Number + 5 = 264 $\Rightarrow$ Number = $264 - 5 = 259$.
Thus, the smallest number is $\boxed{259}$.