You would get the quotient first and count the digits.
All fractions ever meant was simply division. 1/8 = 1 divided by 8. Perform this division by hand, without using remainders, and you'll see that the quotient of 1 and 8 is 0.125.
No, because a quotient requires two numbers. Given the two numbers it is quite easy to work out the number of digits in the quotient.
647.25
7
You can't tell anything about the quotient until you know whatthe divisor is going to be.-- If I divide your 4,796 by 4, the quotient is 1,199 . . . 4 digits.-- And if I divide it by 2,398, the quotient is 2 . . . . only 1 digit.
A divisor is a number that divides another number, also known as the dividend, without leaving a remainder. The quotient, on the other hand, is the result of dividing the dividend by the divisor. In other words, the quotient is the answer to a division problem. The divisor and the quotient are related in that the divisor is used to divide the dividend and obtain the quotient.
All fractions ever meant was simply division. 1/8 = 1 divided by 8. Perform this division by hand, without using remainders, and you'll see that the quotient of 1 and 8 is 0.125.
int dividend,divisor,remainder; int division(int p,int q){ int quotient=1; /*if divisor and diviend are equal then quotient=1*/ if(p==q){ remainder=0; return 1; } /*if dividend is smaller than divisor then remainder=dividend*/ if(p<q){ remainder=p; return 0; } /*shift left till divisor > dividend*/ while(p>=q){ q<<=1; quotient<<=1; } /*shift right for one time so that divisor become smaller than dividend*/ q>>=1; quotient>>=1; /*again call division recurcively*/ quotient+=division(p-q,divisor); return quotient; } int main(){ cout<<"\nEnter dividend:"; cin>>dividend; cout<<"\nEnter divisor:"; cin>>divisor; cout<<"\nQuotient:"<<division(dividend,divisor); cout<<"\nRemainder:"<<remainder; //system("pause"); return 0; }
No, because a quotient requires two numbers. Given the two numbers it is quite easy to work out the number of digits in the quotient.
It is the amount of times 0.5 goes into 22.4, without the remainder. So the quotient is 44.
647.25
7
You can't tell anything about the quotient until you know whatthe divisor is going to be.-- If I divide your 4,796 by 4, the quotient is 1,199 . . . 4 digits.-- And if I divide it by 2,398, the quotient is 2 . . . . only 1 digit.
(235 × 25) + (1625 ÷ 65) = 5,900Product is multiplication. Quotient is division.Note: 235 × 25 + 1625 ÷ 65 = 5,900 is the same result but without the brackets. This is because the multiplication and division have the same order of precedence and would both be calculated first before the addition.
well, if u dont want the airthematic operator to be used for division and if the divisor is 2 here is the possible solution. shift the dividend to right by one bit. for eg: if the diviedend is 127 and divisor is 2, when 127 is shifted by one bit, it is 63 by normal division we get the same quotient(i.e int divided by int 127/2 =63).
Without going into the intricacies of long division........ Division is successive subtraction as opposed to multiplication which is successive addition. Let's say you want to divide 19 by 3, in successive subtraction you would first see if you can take away 3 from 19. The answer is yes. So you take away 3 and create a variable called quotient (which initially has a value of 0). Since you were able to successfully take away 3 from 19 during this first attempt, increment the quotient by 1. Since you took away 3 from 19 and accounted for it in the quotient (which is the number of times you are able to successfully take away 3 - the divisor) see what is left in the original number. 19 is now 16. Can you take away 3 from 16. The answer is yes. Increment the quotient - now it should be 2 and 16 will become 13. Keep doing this. You will see that you can do this six times in all (the quotient will have incremented to 6) and then you will be left with 1 from which you cannot take away 3. If you are limited to just integer division the process ends here. Of course if one knows multiplication tables, then this problem can be solved in one step. One would know that 3 can be taken away from 19 six times with a remainder of 1 at the end. So to illustrate this further, if we started with 19 pencils and the teacher wanted us to make bundles of 3 pencils, division or successive subtraction tells us that we can make six bundles with 1 pencil left over.
Impossible