Question

Find the smallest number which when divided by $28$ and $32$, respectively, leaves a remainder, $8$ and $12$ respectively.

Answer 1

Step 1: Find the LCM

Given,

$28-8=20$ and $32-`12=20$ are divisible by the required numbers.

Therefore, the required number will be $20$ less than the LCM of $28$ and $32$

Prime factors of $28$ are

$28=2\times 2\times 7$

Prime factors of $32$ are

$32=2\times 2\times 2\times 2$

$\therefore$LCM of $28$ and $32$ is

LCM$\left(28,32\right)=2\times 2\times 2\times 2\times 7=224$

Step 2: Find the smallest no which gives remainder

To get the smallest number, which gives remainder $8$ and $12$ when divided by $28$ and $32$, we have to subtract $20$ from the LCM of $28$ and $32$

The required the smallest number $=224-20=204$

Hence, the number is $204$.