Question

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

Answer 1

Well, that means that it is 3 more than a common multipule of 8,10,12, and 30
find the lcm

8=2*2*2
10=2*5
12=2*2*3
30=2*3*5
lcm=2*2*2*3*5=120
times it by 2 to get it between 200 and 500
240
add 3
243 is da number

Answer 2

Answer:

The required number is 243.

Step-by-step explanation:

Consider the provided numbers.

We need to find a number which leaves a remainder of 3 in each case when divided by 8 ,10,12 and 30.

First find the number which is divisible by 8 ,10,12 and 30,

Find the LCM of the provided numbers as shown below:

8 = 2×2×2

10 = 2×5

12 = 2×2×3

30 = 2×3×5

LCM= 2×2×2×3×5 = 120.

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

If we double the number the required number will lie between 200 and 500.

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

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

240+3=243

Thus, the required number is 243.