Question

The smallest natural number which is divisible by 24, 58, 48 and 12 is:

• A. 1465
• B. 1392
• C. 1441
• D. 1312

Answer 1

Answer:

B. 1392

To find the smallest natural number divisible by 24, 58, 48, and 12, we need to find their LCM (Least Common Multiple).

Prime factorization

• 24 = ( 2^3 \times 3 )

• 58 = ( 2 \times 29 )

• 48 = ( 2^4 \times 3 )

• 12 = ( 2^2 \times 3 )

  Take highest powers of each prime

• Highest power of 2 = ( 2^4 )

• Highest power of 3 = ( 3^1 )

• Highest power of 29 = ( 29^1 )

Multiply

$LCM = 2^4 \times 3 \times 29$

$= 16 \times 3 \times 29$
$= 48 \times 29$
$= 1392$

The smallest natural number divisible by 24, 58, 48, and 12 is 1392.