Question

Find the greatest number of $5$ digits which on dividing by $10,12,16,20$, and $24$ leaves in each case $3$ as the remainder.

Answer 1

First, we will find the greatest number of $5$ digits that is divided completely by each$10,12,16,20$ and $24$
If the number is divided by $10,12,16,20$, and $24$, it must be divided by their L.C.M. also.
L.C.M. of $10,12,16,20$ and $24$

L.C.M. = $2\times 2\times 2\times 2\times 3\times 5$
= $240$
The greatest number of $5$ digits which is completely divisible by $10,12,16,20$ and $24$ must be a multiple of $240$
To find the greatest number of $5$ digits we follow the long division method

The $5$ digit greatest number which is completely divisible by $240$ or with $10,12,16,20$ and $24$ = $99999-159=99,840$
We want the that gives remainder 3 in each case i.e. = $99840+3=99843$
So, the number is $99843$