Question

Find the least multiple of $23$ which when divided by $18,21$ and $24$ leaves remainder $7,10$ and $13$ respectively

Answer 1

From given data

When a number is divided by $18$, it leaves a remainder of $7$.

When the same number is divided by $21$, it leaves a remainder of $10$.

When the same number is divided by $24$, it leaves a remainder of $13$.

Observing the constant difference of $11$ between divisor and remainder:

We can conclude that if $11$ is subtracted from the LCM of $18,21$ and $24$, then it will leave a remainder of $7$ when divided by $18$, a remainder of $10$ when divided by $21$ and a remainder of $13$ when divided by $24$.

When $11$ is subtracted from any multiple of LCM, it will give the same result.

Now,

$18=2\times 3\times 3$

$21=3\times 7$

$24=2\times 2\times 2\times 3$

∴ LCM of $18,21$ and $24=2^3\times 3^2\times 7=504$

So, any number of the form $(504n–11)$ will give the required remainders, when $n$ is an integer.

But, this number also has to be divisible by $23$.

Using trial and error method, for $n=6$:

We find that, $504n–11=504\times 6–11=3024–11=3013$

Thus, the required number which satisfies all the conditions is $3013$.