 Let us solve more problems related to Nicollum method of division where the base is greater than divisor Particular problem that I want to share with you which will solve more concepts for us is going to be 147 Divided by seven. So let's just try to solve this using Nicollum method So first of all, we will have to find out the supplement divisor And the supplement divisor is going to be 10 minus 7 is equal to 3 Now the first thing that we do is we create three parts of the page We write our divisor seven On the left hand side and below that we write the Supplement divisor as three now we will write 147 like this one Four and seven because the rightmost part gets only one digit because the base that we have chosen has only one zero in it Now let's put this horizontal line here and then bring this one down and then we'll multiply this one by three So what we're doing here one times three will write the result below four and the result is Seven now we will again use this seven and multiply with three. So let's multiply Seven with the three and we get 21. We write this result here like this And the result we get is 28 and we already know that we cannot have two digits in the last column So we will cancel this two out from here and we'll write two here and Now because we have gotten this two in the central part We will again multiply this two with the supplement divisor three So two times three which is six. We will write the result here on the right hand side and on the right hand side We get 14 and on the central part we get 19 Again this 14 is two digit numbers We cannot have this one and we have to move this to the central part and because now we have got one as A new member in the central part. We will have to multiply this one with three again so we get one times three is equal to three and on the right hand side we have Seven as a remainder and 20 on the central part. We can see that the remainder is seven We can see that seven is equal to the divisor. We still need to divide that by seven Now because we are stuck in a problem like this There is something called Compliment of the remainder now how we can find this compliment of the remainder We write this seven that we've got as zero and seven and then we think about the base we borrow one from the left-hand side and Subtract seven from the base the base is going to be ten So we write one below this dotted zero and once we subtract that seven from ten we get three but we put a Sign like this a small cap on the head of three which also means minus three So you could also read this as ten minus three basically So we are writing the seven as thirteen with a negative sign for three So basically is ten minus three. So we just replace that seven with something like this which is This and now we can again continue cancelling this one out putting it on the central part and Then multiplying this one with the supplement divisor Which is three again and this gives me three as a result and I write the result On the right-hand side and so the remainder that I get finally because the cap three is minus three We get zero as a remainder and 21 as the quotient so that's why here quotient is 21 and remainder is is zero