## Math Class 7

an integer is divided by 9 giving a reminder of 6. the resulting quotient when divioded by 4 gives a reminder of 1.the resulting quotient in then divided by 7 giving a quotient of 1 and a reminder of 3. what will be the finale remainder be if the order of the divisor is reversed ?

Dear Parveen,

There are three stages of this question:

1. An integer is divided by 9 giving a remainder of 6.

2. The quotient of above division is divided by 4 and a remainder of 1 is obtained.

3.. The quotient of the above division is divided by 7 and a quotient of 1 and a remainder of 3 is obtained.

Let us move in the reverse order:

3. A number os divided by 7 giving a quotient of 1 and remainder of 3. We know that: Dividend = Divisor * quotient + remainder

The number can hence be found out as:

Dividend = 7 * 1 + 3

= 10

This dividend is the quotient for the second part of the question.

2. The quotient is 10 (as obtained above), divisor is 4 and the remainder is 1.

The number, i.e. the dividend for this division will be equal to:

Dividend = Divisor * quotient + remainder

= 4 * 10 + 1

= 41

This number is the quotient for the first part of the question.

1. The quotient is 41 (as obtained above), the divisor is 9 and the remainder is 6.

The dividend for this division will be the integer we need.

Dividend = Divisor * quotient + remainder

= 9 * 41 + 6

= 369 + 6

= 375

Now if the order of division is reversed, the series of divisions in the question will become:

1. An integer is divided by 7:

We have: 375/7 = 53, remainder = 4

2. The quotient of the above division is divided by 4:

We have 53/4 = 13, remainder = 1

3. The quotient of the above division is divided by 9:

We have: 13/9 = 1, remainder = 4.

Hence we can see that the final remainder we get is equal to 4.

Not explained in detail and proper language

### Respond to this discussion

You need to login to write a comment