The smallest 5 digit number which when divided by 7,11 and 21 leaves the remainder 3 in each case is

We know that -

Dividend=divisor*quotient+remainder

For our question dividend must be of 5 digit.Let dividend be x.

Divisor=LCM(7,11,21)=231.

Let quotient be y.

Remainder is 3.

Substitute above values in equation-

x=231*y+3

Now we need to substitute different value of y(y=1,2,3…) in order to find smallest 5 digit x.

When y=1 then x=234

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When y=43 then x=9936

When y=44 then x=10167

When y=45 then x=10398

So least 5 digit number(10167) is obtained at y=44.

The 5 digit number when divided by 7,11,21 which gives remainder 3 is 10167.

So,Option B is correct choice.