Question

find the least number between 200 and 500 which leaves a remainder of 3 in each case when divided by 8, 10, 12 ,30

Answer 1

243 is the least no. 200 and 500 which leave remainder 3
solution
we know that we require remainder 3 so the last digit would be three in this case and rest digit subtracting three will be lcm of all four greater than 200 so we get the answer that is 243

Answer 2

Answer:

The least number between 200 and 500 which leaves a remainder of 3 in each case when divided by 8, 10, 12 ,30 is 243.

Step-by-step explanation:

To find : The least number between 200 and 500 which leaves a remainder of 3 in each case when divided by 8, 10, 12 ,30?

Solution :

First we find the LCM of the numbers 8, 10, 12 ,30.

2 | 8  10  12  30

2 | 4   5   6   15

2 | 2   5   3   15

3 | 1   5   3   15

5 | 1   5   1     5

  | 1    1    1    1

$LCM(8,10,12,15)= 2\times2\times2\times3\times5$

$LCM(8,10,12,15)=120$

The number 120 is divisible by 8 ,10,12 and 30. But it is not in between 200 and 500.

So, If we double the number the required number will lie between 200 and 500 i.e. $120\times 2=240$

The required number which is divisible by 8 ,10,12 and 30 and lie between 200 and 500 is 240.

Now, we add 3 in 240, as we need a number which leaves a remainder 3.

$240+3=243$

Therefore, The least number between 200 and 500 which leaves a remainder of 3 in each case when divided by 8, 10, 12 ,30 is 243.