# Calculation Techniques

1.Subtraction from 10/100/1000/... Example - a). 1000 - 4576 ans). (9 - 4) (9 - 5) (9 - 7) (10 - 6) 5 4 2 4 = 5424 2. Multiplication with a series of 9s Example - a). The no. of 9s are equal to the no. of digits in multiplying number 23 * 99 (both the numbers are of 2-digits) ans - 23*(100-1) = 2300 - 23 = 2277 Explanaion - The answer consists of two 2-digit no. '22' and '77' The first part '22' is 1 less than the actual no. '23' and second part is equal to (99 - 22). Similarly 4256 * 9999 (4-digit number) = 4256 * (10000 - 1) = 42560000 - 4256 = 4255/5743 4256 - 1 = 4255 (9 - 4) = 5 (9 - 2) = 7 (9 - 5) = 4 (9 - 5) = 3 b). Multiplication of a number by greater number of 9s 321 * 9999 = 3209679

The number 321 has 3-digits and 9999 has 4-digits so we pad one zero to get 0321 and carry out the multiplication as in the case of "The no. of 9s are equal to the no. of digits in multiplying number"

0321 - 1 = 0320 (9 - 0) = 9 (9 - 3) = 6 (9 - 2) = 7 (9 - 0) = 9 0320/9679 = 3209679 c). Multiplication of a number by lesser number of 9s consider 456 * 99 = 45144 simple 456 * (100 -1) = 45600 - 456 4 : 56 : 00 - 4 : 56 -------------------- 4 : 51 : 44 --------------------

3. Computation of remainder(नवसेष) on division by 9

I represent it by 'N' Example - i). 20/9 remainder = 2 ii). 221/9 remainder = 5 iii). 1823/9 remainder = 5 Explanation i). 2 + 0 = 2 ii). 2 + 2 + 1 = 5 iii). 1 + 8 + 2 + 3 = 14 then add it again 1 + 4 = 5 add all the digits till you'll get single digit. Example 12952/9 remainder = 1 Explanation add all digits except 9 , if 9 comes in a number than always ignore it. 1 + 2 + 5 + 2 = 10 than 1 + 0 = 1 ans

4. Verification of the product of two numbers

Take a product of 45 * 23 which is 1035. How do we verify the correctness of the answer '1035' ?

Let's take the 'N' of each of the multiplicands and the product N(45) = 4 + 5 = 9 N(23) = 2 + 3 = 5 N(1035) = 1 + 3 + 5 = 9 now find out N(45) * N(23) i.e = 9 * 5 = 45 take N(45) = 4 + 5 = 9 i. e. equal to the N(1035) that means ans is correct.

Limitation :- If the answer was written as 1053 instead of 1035 it's 'N' would be same. But the answer is incorrect . So this is the limitation of this method.

5. Verification of the sum of two numbers

Take the sum of 38 + 31 = 69 we see that N(38) = 3 + 8 = 11 than 1 + 1 = 2 N(31) = 3 + 1 = 4 N(38) + N(31) = 2 + 4 = 6

than N(69) = 6 + 9 = 15 than 1 + 5 = 6 The sum of N of two numbers is equal to the sum itself. Limitation :- If the answer was written as 96 instead of 69 , the 'N'

would be same . But our answer is incorrect.

6. Verification of the difference of two numbers

Let's see the difference of two numbers 51 - 23 = 28 N(51) = 5 + 1 = 6 N(23) = 2 + 3 = 5 N(51) - N(23) = 6 - 5 = 1 N(28) = 2 + 8 = 10 than 1 + 0 = 1

Thus the difference of the N of the two numbers is equal to the N of the difference itself. Limitation :- If the answer was written as 82 instead of 28 , the 'N'

would be same . But our answer is incorrect.

7. Verification of the square or cube of two numbers

Let's take square of 11 i.e. 11 * 11 = 121 Now take the (नवसेष) 'N' of both the sides i.e. N(11) = 1 + 1 = 2 and N(121) = 1 + 2 + 1 = 4

Let us square the N of the left side i.e. 2 * 2 = 4 i.e. equal to the N of answer Limitation : - Same here if the answer was written as 112 instead of 121 . The 'N'

would be same but our answer is incorrect.

8. Multiplication by 11 Let us take the multiplication of 357 by 11. Simple method -

3 1 7 * 1 1 -------------- 3 1 7 3 1 7 --------------- 3 4 8 7

The faster method

3___1___7 3 4 8 7

Example – 2

357 * 11 = 3972

By faster method

3___5___7 3 8__12 7 (Carry 1 to left) 3 9 2 7

Add 8 + 1 = 9 and take 2 as it is from 12.Why we did this because after adding 5 and 7 the reslut comes in 2-digit that's why we add it again . Always add with the left number.

Example – 3

367 * 11 = 4037

By faster method

3___6___7 3 9_ 13 7 3_ 10 3 7 4 0 3 7

Take 3 as it is from number 13 and add (carry)1 to 9 (9 + 1 = 10) the result is 10. Now take 0 as it is and add 1 to the left number 3 , 3 + 1 = 4 .

Steps

First write down the left most and right most digit as it is . Form pairs of consecutive digits, starting from the right and write down the sum one at a time.

9. Divisibility test of numbers by 11

How do we test Ex – is 50237 divisible by 11 ? First add 7 + 2 + 5 i.e.= 14 that add other alternates 3 + 0 = 3 Find out the difference i.e. 14 – 3 = 11 11 is divisible by 11. that means our number is divisible by 11.

Steps

Start from the right(unit's place) and add all the alternate numbers, i.e. All the digits in the odd positions viz 7, 2 and 5 which gives us 14. Now start from the digit in the tens place and add all the alternate numbers, i.e. All the digits in the even positions viz 3 and 0 which gives us 10 .

NOTE – If the difference of two sums thus obtained is either 0 or divisible by 11, then the entire number is divisible by 11.

Ex – 2 is 1234 is divisible by 11 ? add 4 + 2 = 6 than 3 + 1 = 4 difference = 6 – 4 = 2 (neither 0 nor the difference is divisible by 11) So the entire number is not divisible by 11.

11. Multiplication by Nikhilam(निखिलं) method -

For nikhilam method both the numbers must be close to the base of
100, 1000, 10000 etc..
Nikihilam method works on multiplication of two numbers which are close to the base of 100, 1000, 10000 and so on.
these numbers may be
1. less than the base (e.g. 95, 98, 993, 991 etc.)
2. greater than the base (e.g. 102, 105, 1005, 1007 etc.)
3. mixed (e.g. 98, 105)

Case I : Numbers less than the base 100

let us take 92 * 95 now multiplication by nikhilam 92 -8 (92 - 100) 95 -5 (95 - 100) --------------------------------------------------- 87 / 40 = 8740 ans.

Steps : . Write the number one below the other one on left side and write the difference of number and base(in this e.g. base 100) on right side. . Take the algebraic sum of digits across any crosswise pair. e.g. 92 + (-5) or 95 + (-8) both'll give you the same result. This is the first part of your answer or the left hand part of your answer. . Now just multiply the two differences (-8) * (-5) i.e. = 40 , which is the second part of your answer or the right hand part of your answer.

so the final ans is 8740 in this case.

Note - 1. If the number of digits is one, the number is padded with one zero on the left. e.g. 98 * 96

98 -2 96 -4

94 / 08 (pad one zero to get 08 because the base is of 2-digits here) so the ans will be 9408.

2. If the number of digits are more than two, than the leftmost digit (in the hundreds position) is carried to the left. e.g. 85 * 86

85 -15 86 -14

71 / 210 (carry 2 to the left and add to 71 that'll give 73) so the ans is 7310

3. The method is effective even if one number is close to the base and the other one is not. e.g. 93 * 34

93 -7 34 -66

27 / 462 (carry 4 to the left to get 31)

so the ans is 3162

Case II : The numbers above the base 100 let us take 103 * 104

same here 103 +3 104 +4 ------------------------- 107 / 12 (no carry because the product is of 2-digits) -------------------------

Note - 1. If the number of digits is one, the number is padded with one zero on the left. 2.If the number of digits are more than two, than the leftmost digit (in the hundreds position) is carried to the left.

Case III : Numbers below and above the base 100 let us take 98 * 103

98 -2 103 +3 ---------------------- 101 / (-6) (the product will always come with minus sign in this case) so the ans can be interpreted by two ways now

1. 101 / (-6) = 100 / (100 - 6)

= 100 / 94 = 10094 ans.

2. 10100 - 6 = 10094 ans.

---------------------

another example : let us take 114 * 88

88 -12 114 +14 --------------------------- 102 / -168 simply do 10200 - 168 = 10032 ans. --------------------------

Same techniques will be applied with bases 1000 and 10000 etc.

e.g 997 * 994 997 -3 994 -6 ---------------------- 991 / 018 (pad one zero because the base is of 3-digits now) so the ans will be 991018. ----------------------

12. Multiplication by Urdhva Tiryak(ऊर्ध्वा तिर्यक) method

We shall consider the multiplication of the following :
. Two 2-digit numbers
. Two 3-digit numbers
. One 2-digit and One 3-digit number
. Two 4-digit numbers
. One 3-digit and One 4-digit number
Case I : Multiplication of two 2-digit numbers
Let us consider 48 * 52

4 8 5 2 -------------------- 20 / 48 / 16 Position No. (1) / (2) / (3) ---------------------

Steps 1. Multiply the left-most digits 4 and 5 , you'll get 20. 2. Multiply the right-most digits 8 and 2 , you'll get 16. 3. Carry out a cross multiplication between 4 and 2, 5 and 8 to get 8 and 40 respectively and add both the products 8 + 40 to get 48. we have to retain one digit from two of the three locations (no 2 and 3 ) and carry the surplus to the left. hence in 20 / 48 / 16

------------------------------ 20 / 8 / 6 4 1 24 / 9 / 6 ------------------------------

so answer is 2496 .

Case II : Multiplication of two 3-digit numbers Let us consider 426 * 528

4 2 6 * 5 2 8 ------------------------------------- 20 / 18 / 66 / 28 / 48 position no (1) / (2) / (3) / (4) / (5) ------------------------------------- 20 / 18 / 66 / 28 / 8 carry (4) and add it to it's just left number 28. 20 / 18 / 66 / 32 / 8 20 / 18 / 66 / 2 / 8 carry (3) and add it to it's just left number 66. 20 / 18 / 69 / 2 / 8 20 / 18 / 9 / 2 / 8 carry (6) and add it to it's just left number 18. 20 / 24 / 9 / 2 / 8 20 / 4 / 9 / 2 / 8 carry (2) and add it to it's just left number 20. 22 / 4 / 9 / 2 / 8 --------------------------------------

so the answer is 224928 . Steps : 1. For position no 1 - multiply the left most digits 4 and 5 (20). 2. For position no 2 - cross multiply 4 and 2(8), 5 and 2(10) and add the products. (8+10 = 18) 3. For position no 3 - cross multiply 4 and 8 (32), 5 and 6 (30) and multiply the middle digits 2 and 2 (4)of both the numbers . Now add the products together. (32+30+2 = 66) 4. For position no 4 - cross multiply 2 and 8 (16), 2 and 6 (12) and add the products . (16+12 = 28) 5. For position no 5 - multiply the right most digits 6 and 8 (48) Case III : Multiplication of 3-digit and 2-digit numbers If we have to multiply a 3-digit and 2-digit numbers, we can treat the 2-digit number as a 3-digit number by padding it with a zero on the left. e.g. 234 * 45 can be written as 234 * 045 and the same 3-digit multiplication technique can be applied.

Case IV : Multiplication of two 4-digit numbers Let us consider 4234 * 2153

4 2 3 4 * 2 1 5 3 ---------------------------------------- 8 / 8 / 28 / 33 / 25 / 29 / 12 position no (1) / (2) / (3) / (4) / (5) / (6) / (7) ---------------------------------------- 8 / 8 / 28 / 33 / 25 / 29 / 2 carry (1) and add it to it's just left number 29. 8 / 8 / 28 / 33 / 25 / 30 / 2 8 / 8 / 28 / 33 / 25 / 0 / 2 carry (3) and add it to it's just left number 25. 8 / 8 / 28 / 33 / 28 / 0 / 2 8 / 8 / 28 / 33 / 8 / 0 / 2 carry (2) and add it to it's just left number 33. 8 / 8 / 28 / 35 / 8 / 0 / 2 8 / 8 / 28 / 5 / 8 / 0 /2 carry (3) and add it to it's just left number 28. 8 / 8 / 31 / 5 / 8 / 0 / 2 8 / 8 / 1 / 5 / 8 / 0 / 2 carry (3) and add it to it's just left number 8. 8 / 11 / 1 / 5 / 8 / 0 / 2 8 / 1 / 1 / 5 / 8 / 0 / 2 carry (1) and add it to it's just left number 8 . 9 / 1 / 1 / 5 / 8 / 0 / 2

so the answer is 9115802.

Case V : Multiplying 3-digit and 4-digit number. If we have to multiply a 4-digit and a 3-digit number, we can treat the 3-digit number as 4-digit number by padding it with a zero on the left. e.g. 2315 * 345 can be written as 2315 * 0345 and the same 4-digit multiplication technique can be applied.

13. Simple Square

Case I : Numbers ending with 5
Let us take 25 * 25

Step 1 - 2 * 3 = 6 Step 2 - 5 * 5 = 25

so the answer is 625. Step 1 - Multiply the first number by it's consecutive number i.e.

2 multiply by 3 in this case.

Step 2 - Take the square of 5. i.e. 5 * 5 = 25.

Note - No of digits in a number doesn't matter.

Example - 1. 65 * 65 =4225

Step 1 - 6 * 7 = 42 Step 2 - 5 * 5 = 25

so answer is 4225. 2. 675 * 675

Step 1 - 67 * 68 = 4556 Step 2 - 5 * 5 = 25

so the answer is 455625 .

Case II : Square of a number max. up to 4-digits First you need to find out the duplex than you can find out the square of any number. Definition of Duplex represent it by D.

A). Duplex of 1-digit number D(a) = a * a = (a)2 D(3) = 3 * 3 = 9 D(7) = 7 * 7 = 49

B). Duplex of 2-digit number

D(ab) = 2 * a * b = 2ab D(23) = 2 * 2 * 3 = 12 D(10) = 1 * 0 = 0

C). Duplex of 3-digit number

D(abc) = 2 * a * c + b * b = 2ac + (b)2 D(123) = 2 * 1 * 3 + 2 * 2 = 6 + 4 = 10

D). Duplex of 4-digit number

D(abcd) = 2 * a * d + 2 * b * c = 2 (ad+bc) D(1234) = 2 * 1 * 4 + 2 * 2 *3 = 8 + 12 = 20

The square of a 2-digit number (ab)2 = D(a) / D(ab) / D(b) (23)2 = D(2) / D(23) / D(3)

= 4 / 12 / 9 = 5 / 2 / 9 we retain 2 and carry 1 to it's left

so the final answer is 529.

The square of a 3-digit number (abc)2 = D(a) / D(ab) / D(abc) / D(bc) / D(c) (123)2 = D(1) / D(12) / D(123) / D(23) / D(3)

= 1 / 4 / 10 / 12 / 9 = 1 / 4 / 11 / 2 / 9 we retain 2 and carry 1 to it's left = 1 / 5 / 1 / 2 / 9 we retain 1 and carry 1 to it's left

so the final answer is 15129.

The square of a 4-digit number (abcd)2 = D(a) / D(ab) / D(abc) / D(abcd) / D(bcd) / D(cd) / D(d) (1234)2 = D(1) / D(12) / D(123) / D(1234) / D(234) / D(34) / D(4)

= 1 / 4 / 10 / 20 / 25 / 24 / 16 = 1 / 4 / 10 / 20 / 25 / 25 / 6 we retain 6 and carry 1 to it's left = 1 / 4 / 10 / 20 / 27 / 5 / 6 we retain 5 and carry 2 to it's left = 1 / 4 / 10 / 22 / 7 / 5 / 6 we retain 7 and carry 2 to it's left = 1 / 4 / 12 / 2 / 7 / 5 / 6 we retain 2 and carry 2 to it's left = 1 / 5 / 2 / 2 / 7 / 5 / 6 we retain 2 and carry 1 to it's left

so the answer is 1522756.