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.