Question

Find the greatest number of five digits which on dividing by 3, 12 ,18 , 24, 30 and 36 leaves 2, 11, 17, 23, 29 and 35 respectively as remainders.

Answer 1

The LCM of $3,12,18,24,30,36$ is $360$

We have the greatest five digit number : $99999$

When $99999$ is divided by $360$, leaves the remainder $279$

So, $99999-279=99720$

Now here, we can see that

$3-2=1,12-11=1,18-17=1,24-23=1,30-29=1,36-35=1$

The difference of divisors and their corresponding remainders is $1$.

So, the required five digit number will be $1$ less than $99720$.

The required number is $99720-1=99719$