Question

What will be the highest three digit number which when divided by 3, 7 and 21 leaves remainder 2?

Answer 1

Answer: 989

Step-by-step explanation:

(Given)

The highest three digit number which when divided by 3, 7 and 21 leaves reminder 2

According to the question and as we know highest three digit number is 999

and the L.C.M of 3, 7, and 21 is 21

then we have to divide 999 with 21 = $\frac{999}{21}$  

then we got remainder is 12

then we have to substract 12 from 999 then

999-12=987

then as per the question it has to leave remainder 2 and we have to add 2 with 987

then required number is 989

The highest three digit number is 989

Answer 2

Given:

Remainder=2

Divided by 3, 7, and 21

To find:

The highest 3-digit number is divisible by 3, 7, 21, and leaves the remainder as 2

Solution:

The highest 3-digit number that is divisible by 3, 7, 21, and leaves the remainder as 2 is 989.

We can find the number by following the given steps-

We know that for a number to be divisible by the given three numbers, it has to be divisible by its LCM.

The three given numbers are 3, 7, 21.

The LCM of 3, 7, 21 is 21.

So, the highest 3-digit number should be divisible by 21.

The highest 3-digit number=999

Now, the number closest to 999 which is also divisible by the LCM of 3, 7, and 21 can be obtained by dividing 999 by the LCM obtained.

On dividing, we get the remainder as 12.

So, 999-12=987

The highest three-digit number divisible by 21 is 987.

But we have to find the number which also leaves 2 as the remainder.

So, we will add the required remainder to the number obtained.

The required number=987+2=989

Therefore, the highest 3-digit number that is divisible by 3, 7, 21, and leaves the remainder as 2 is 989.