Question

Find the smallest number which on being added 23 to it is exactly divisible by 32 36 48 and 96

Answer 1

Answer:

x+23 =lcm of 32,36,48and 96 =2^6×3^2=576. x+23 =576. x=576-23=553. Hence, the required smallest number is 553 when added 23 to it exactly divisible by 32,36,48 and 96.

Answer 2

Answer:

553

Step-by-step explanation:

let the required number be x

x+23=lcm of 32,36,48and 96

2⁶*3²=576

x+23=576

x=576-23

x=553

Hence the smallest number is 553 in which is exactly divisible by 32,36,48 and 96.