Question

The least multiple of 23, which when divided by 18, 21 and 24 leaves remainder 7,10 and 13 respectively is

Answer 1

(1)Take L. C. M of given no 18,21 and 24

2- use the formula

$L.C.M\times M-R=23\times n$

where R is $\frac{sumofreminders}{no.ofterm}$ and M,n are variable

3- use the hit and trial method by putting the values $M$ and check whether the no is divided by 23 or not

we have to find the LCM of the divisors $i.e.18,21and24.$

The LCM of 18, 21 and 24 is $504$

and

$\text{if we subtract 11 from it, we get}493\text{and this is the required number that leaves the remainders 7,}$ $\text{10 and 13 with the divisors 18, 21 and 24 respectively.}$

Now, we have 2 variables and there’s only 1 equation, so obviously we are going to have infinite solutions and we will have to proceed with the trial and error method to obtain the first solution.

If we try with $M$ as 1, we get a number that leaves 10 as the remainder with 23. If we try with $M$ as 2, we get a number that leaves 8 as the remainder with 23. So we expect 0 as the remainder when $M=6.$

If we try with $Mas6$, indeed we get 0 as the remainder with the number $504\times 6–11=3013$

$\text{Hence required number is 3013.}$