In this article, we shall learn the procedure to divide the large numbers. To make it better understandable, lets go through with the division of polynomials. It will help us to understand the procedure better.
Lets take an example, 12x2-8x-32 by x-2
____________
x-2 | 12x2 - 8x - 32 | 12x + 16
-(12x2 - 24x)
-----------------
16x - 32
-(16x - 32)
-----------------
0
Here, Quotient = 12x + 16 , Remainder = 0
Take one more example, 6x4+13x3+39x2+37x+45 by x2-2x-9
_______________________
x2-2x-9 | 6x4+ 13x3+ 39x2+ 37x+ 45 | 6x2 + 25x + 143
- (6x4 - 12x3- 54x2)
--------------------------
25x3+ 93x2+ 37x
-(25x3- 50x2- 225x)
-------------------------
143x2+ 262x+ 45
-(143x2- 286x -1287)
---------------------------
548x+1332
Here, Quotient = 6x2 + 25x + 143 , Remainder = 548x + 1332
Now, see the above example in different form, which will actually be our procedure for division of large numbers
x2-2x-9 | 6x4+ 13x3+ 39x2 | +37x + 45 |
2 9 12 54
50 225
286 1287
----------------------------------------
6 25 143 | 548 1332
Quotient = 6x2 + 25x + 143 , Remainder = 548x + 1332
Following are the steps that we done above:
Step 1: Divisor ( x2-2x-9 )
The first step is to write the divisor. In second line, write all the coefficients except the first, changing all of those to the opposite sign of coefficients. (Here, opposite signs of '-2' and '-9' are '2' and '9' respectively and first coefficient of x2 i.e. '1' is not being considered )
Step 2: Dividend ( 6x4+13x^3+39x2 +37x + 45 )
Step 3: Procedure
We have to repeat the process up to last digit of the dividend.
IMPORTANT: In this article, we consider only those divisors whose first digit is unity i.e. '1'. The divisors whose first digit is not '1', will be taken in the next article.
The following examples will illustrate this more clearly.
239479 divided by 11203:
1 1 2 0 3 | 2 3 9 4 7 9
-1-2 0-3 | -2 -4 0 -6
-1 -2 0 -3
----------------------
2 1 4 2 1 6
Quotient= 21, Remainder = 4216
1248 divided by 160:
1 6 0 | 1 2 4 8
-6 0 -6 0
24 0
------------------
1-4 28 8
6 28 8
7 12 8 (Remainder is greater than divisor, hence Remainder=288-160=128, Quotient=6+1=7 carry over)
Quotient= 7, Remainder= 128
Another example:
12349 divided by 1133
1 1 3 3 | 1 2 3 4 9
-1-3-3 | -1 -3-3
-1-3-3
------------------
1 1 -1-2+6
1 1 -114
-----------------
1 0 1019
Quotient= 10+1 = 11, Remainder= -100-20+6 = -114 { -1 = -100(hundredth place), -2=-20(tenth place), +6=+6(ones place)}
Since remainder is negative, so we subtract 1 from quotient and subtract the current remainder from divisor
Quotient= 11-1=10, Remainder= 1133-114 = 1019
Quotient= 10, Remainder= 1019
I hope this interesting technique will help in solving large divisions. In next article, we shall discuss about those cases where divisor's first digit is not '1'.
Keep practising and keep sharing your experience with awesome Vedic Math!!
Lets take an example, 12x2-8x-32 by x-2
____________
x-2 | 12x2 - 8x - 32 | 12x + 16
-(12x2 - 24x)
-----------------
16x - 32
-(16x - 32)
-----------------
0
Here, Quotient = 12x + 16 , Remainder = 0
Take one more example, 6x4+13x3+39x2+37x+45 by x2-2x-9
_______________________
x2-2x-9 | 6x4+ 13x3+ 39x2+ 37x+ 45 | 6x2 + 25x + 143
- (6x4 - 12x3- 54x2)
--------------------------
25x3+ 93x2+ 37x
-(25x3- 50x2- 225x)
-------------------------
143x2+ 262x+ 45
-(143x2- 286x -1287)
---------------------------
548x+1332
Here, Quotient = 6x2 + 25x + 143 , Remainder = 548x + 1332
Now, see the above example in different form, which will actually be our procedure for division of large numbers
x2-2x-9 | 6x4+ 13x3+ 39x2 | +37x + 45 |
2 9 12 54
50 225
286 1287
----------------------------------------
6 25 143 | 548 1332
Quotient = 6x2 + 25x + 143 , Remainder = 548x + 1332
Following are the steps that we done above:
Step 1: Divisor ( x2-2x-9 )
The first step is to write the divisor. In second line, write all the coefficients except the first, changing all of those to the opposite sign of coefficients. (Here, opposite signs of '-2' and '-9' are '2' and '9' respectively and first coefficient of x2 i.e. '1' is not being considered )
Step 2: Dividend ( 6x4+13x^3+39x2 +37x + 45 )
The second step is to rearrange the expression of the dividend, leaving space after last two expressions from right. The last two expressions are for the remainder. (Here question arises, how do we know from where we have to split. The answer is : one less than the length of the divisor expression (i.e. number of coefficients + constant). In above expression, length of divisor expression is 3, therefore we shall split after two expressions from right (3 - 1 = 2).
- Multiply the first coefficient of the dividend (i.e. '6' in above example) by each digit of the rewritten divisor (i.e. 2 & 9) on the left (we get, 6x2=12, 6x9=54). Write each digit one after the other leaving the first place.
- Now see across the row, add the digits(i.e. 13+12= 25). Now repeat the same as done in Step3(i), multiply '25' by rewritten divisor (i.e. 2 & 9) we get, 25x2=50, 25x9=225. Write each digit one after the other leaving the first two places.
- Repeating the same as done in last step, now add (39+54+50=143). Now multiply 143 by (2 & 9) and we get (286 & 1287).
- Now add remaining columns which is the remainder, add (37+225+286=548, 45+1287=1332) and Quotient is (6 25 143)expression.
We have to repeat the process up to last digit of the dividend.
IMPORTANT: In this article, we consider only those divisors whose first digit is unity i.e. '1'. The divisors whose first digit is not '1', will be taken in the next article.
The following examples will illustrate this more clearly.
239479 divided by 11203:
1 1 2 0 3 | 2 3 9 4 7 9
-1-2 0-3 | -2 -4 0 -6
-1 -2 0 -3
----------------------
2 1 4 2 1 6
Quotient= 21, Remainder = 4216
- Start with writing the new divisor in next line, which is -1 -2 0 -3.
- Rearrange the dividend leaving space after 4 digits from right.(One less than the length of the divisor, 5-1=4 )
- Multiply (-1 -2 -0 -3) by 2 (the first digit of the dividend) and write the answer (-2 -4 0 -6) down in the next row (Row 2), from column 2 onwards.
- Add the numbers in column 2 to get the next multiplication factor which is (3 + (-2) = 1) in this example.
- Multiply the new divisor by this multiplication factor and write the answer (-1 -2 0 -3) to the right, i.e Row 3, Columns 3 onwards.
- Add all the columns to get the answer. The quotient is on the left and remainder on the right. (Note: Here the remainder is in last 4 columns)
- Remainder = 4216 (i.e. 4000+200+10+6, '4' on thousandth place, '2' on hundredth place, '1' on tenth place & '6' on ones place) Similarly, Quotient = 21 (i.e. 20+1, '2' on tenth place & '1' on ones place)
1248 divided by 160:
1 6 0 | 1 2 4 8
-6 0 -6 0
24 0
------------------
1-4 28 8
6 28 8
7 12 8 (Remainder is greater than divisor, hence Remainder=288-160=128, Quotient=6+1=7 carry over)
Quotient= 7, Remainder= 128
- Start with writing down the new divisor which is -6 -0.
- Rearrange the dividend leaving space after 2 digits from right.(One less than the length of the divisor, 3-1=2 )
- Multiply this (-6 0) by 1 (the first digit of the dividend) and write the answer (-6 0) in the next row (Row 2), in column 2 and 3.
- Add the numbers in column 2 to get the next multiplication factor which is (2 + (-6)=-4 in this example.
- Multiply the new divisor (-6 0) by this multiplication factor (-4) and write the answer (24 0) to the right, i.e. Row 3, Columns 3 and 4.
- Add all the columns to get the answer. The quotient is on the left and remainder on the right. (Note: Here the remainder is in last 2 columns). So the left side is 10-4=6. ('1' on tenth place & '-4' on ones place) The right side is (280+8)=288. ('28' on tenth place & '8' on ones place) Remainder is bigger than 160 so we subtract the divisor from it and carry over 1 to quotient.
Another example:
12349 divided by 1133
1 1 3 3 | 1 2 3 4 9
-1-3-3 | -1 -3-3
-1-3-3
------------------
1 1 -1-2+6
1 1 -114
-----------------
1 0 1019
Quotient= 10+1 = 11, Remainder= -100-20+6 = -114 { -1 = -100(hundredth place), -2=-20(tenth place), +6=+6(ones place)}
Since remainder is negative, so we subtract 1 from quotient and subtract the current remainder from divisor
Quotient= 11-1=10, Remainder= 1133-114 = 1019
Quotient= 10, Remainder= 1019
I hope this interesting technique will help in solving large divisions. In next article, we shall discuss about those cases where divisor's first digit is not '1'.
Keep practising and keep sharing your experience with awesome Vedic Math!!
