Question

Find the least number of five-digits which when divided by 12, 15, 18 and 27 leaves a remainder 3 in each case. ​(please give statement)

Answer 1

Answer:

543

Step-by-step explanation:

12 = 3*2*2

15 = 3*5

18 = 3*2*3

27 = 3*3*3

therefore, its L.c.m = 3*2*3*2*5*3 = 540

but the least no. leaving 3 as remainder when divided is = 540+3 = 543

required answer is 540

Answer 2

Answer:

The least five-digit number which when divided by $12,15,18$ and $27$ leaves a remainder $3$ is $10263$.

Step-by-step explanation:

We need the least five-digit number which when divided by $12,15,18$ and $27$ leaves a remainder $3$.

First let us find the smallest number that when divided by these numbers leaves a remainder zero.

We need to find the LCM.

The LCM of $12,15,18$ and $27$ is

$12=2\times2\times3\\15=3\times5\\18=2\times3\times3\\27=3\times3\times3$

$LCM=2\times2\times3\times3\times3\times5=540$

This is the smallest number that when divided by these numbers leaves a remainder zero.

But we need a five-digit number.

So we need a smallest five-digit number which is a multiple of $540$.

$18\times540=9720$

$19\times540=10260$

So the smallest five-digit number which when divided by these numbers leaves a remainder zero is $10260$.

But we need a remainder of $3$. So, we simply add $3$ to this result giving a final answer of $10263$.