Division when divisor is near to the Base of 10 power
This division method work if divisor is near to the base like 10, 100, 1000 etc. Here Nikhilam sutra of vedic maths is useful.
We can also call it Base method of division. Because we apply it when the divisor is near the base of power 10.
I explained it by taking lots of different sums, its really a very easy & interesting way of doing division.
Case 1: When divisor is near to base 10
Here we solve the sums when divisor is near to the base 10 like, 8, 7, 6 etc. We can apply same method for divisor 9 also. But I already publish a blog that explain the shortcut division trick for divisor 9.
Now onward we use abbreviations,
R = Remainder
Q = Quotient
Diff. = Difference
Ex i) 23 ÷ 8
Step 1: Identify Base & Difference
As divisor is 8, which is near to 10, so Base = 10
Diff. = base – number = 10 – 8 = 2
Step 2: Split the dividend in to two parts (Q & R) in such a way that Number of digit in remainder side is equal to number of zero in base.
Base: 10
Divisor = 8
diff. = 2
As here Base 10 have 1 zero so in remainder side, we take 1 digit
Step 3:Take 2 down as it is at quotient place. This is our first digit of Q.
Step 4: Now multiply this 2 with diff. 2.
So, 2 x 2 = 4 & add this 4 in next digit of dividend i.e. 3 & write down the total at R place.
Answer, Q = 2 & R = 7
Ex. ii) 112 ÷ 7
Base 10
Div. = 7
Diff. = 3
As remainder side answer is 14, which is greater than 7(divisor) is not advisable. So, we re-divide the R.
14 ÷ 7, Q = 2, R = 0
Addition of both Q is our final Q.
Q = 14 + 2 = 16, & R = 0
Case ii)When divisor is near to the base 100, 1000, 10000 etc.
Ex. iii) 1324 ÷ 98
Base: 100 (As 98 is close to 100)
Divisor: 98
Diff. =100-98 =02
Step 1: Divide the dividend in two parts Q & R. In R side we take 2 digit as base is 100. We take 1 as it is in Q place.
Step2: Multiply 1 with diff. 02 & add this in next two digits
1 x 02 = 02
As divisor have 2- digits so, we consider 2 digits to add (02 instead of 2)
We add here second digit i.e. 3+0=3
Step 3: Now multiply this 3 with diff. 02 = 06
Add this result from third digit of dividend. Now all digits are cover, so we add up the numbers.
In remainder last digit is 4 + 6 = 10, so we carry-over 1 in left hand side digit i.e. 4 + 1 = 5, so R = 50.
Answer, Q = 13 & R = 50.
Ex. iv) 711 ÷ 96
Base = 100
Divisor = 96
Diff. = 04
Answer, Q = 7 & R = 39
Ex. v) 1431 ÷ 88
Base = 100
Divisor = 88
Diff. = 12
Here, Q = 15 & R = 111 > divisor(88)
So, we re-divide it, 111 ÷ 88, Q = 1, R = 23
So, Final Q = 15 + 1 = 16 & R = 23
Ex. vi) 210021 ÷ 8888
Base = 10000
Divisor = 8888
Diff. = 1112
Q = 23, R = 5597
Case iii) When divisor is greater than base
Ex. vii) 1962 ÷ 112
Step 1: As we do in previous sum find out diff.
Base = 100 (As divisor is close to 100)
Divisor = 112
Diff. = -(12) (Base – divisor)
Step 2: Divide the dividend in two parts (Q part & R part). In R side we take 2 digit as base is 100.
Step 3: First digit 1 we take in Q place as it is & for next step, we multiply it with diff. –(12) & add the result from second digit of dividend.
Step 4: Now multiply second digit of Q (8) with diff. 8 x -(12) = -(96) & add them from third digit of dividend.
Here, Q = 18 & R = -(54)
But -ve remainder is not advisable, so we reduce Q by 1 & subtract the remainder from the divisor.
Q = 18 -1 =17 & R = 112 – 54 =58
Answer Q = 17 & R = 58
Ex. viii) 120887 ÷ 1
Base =1000
Divisor = 1212
Diff. = -(212)
Q = 100 + 00 -1 = 99 (By using place value) & Q = 899
Answer, Q = 99 & R = 899
Ex. ix) 1129 ÷ 108
Base = 100
Divisor = 108
Diff. = -(08)
Q = 11, R = -60 + 1 = -59
But -ve remainder is not advisable, so we reduce Q by 1 & subtract the remainder from the divisor.
Q = 11 -1 = 10, R = 108 – 59 = 49
Find this links to study more:
1. Division trick when divisor is 9
2. Multiply 99 x 96, 9998 x 9982… in 3 seconds by Base method
3. Find the cube & cube root of perfect cube number by very easy way
4. What is Vedic Mathematics & 16- Sutras
5. Find the square & square – root of any number by very easy way
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