Finding cube of 3 digit number using vedic maths.
Lets us consider the number
(153)3
STEPS:
i)Split the given digits into two part
15 3
ii)consider the 1st part i.e. 15 from left to right go on reducing power i.e
153 152 15
iii) Next consider the 2nd part i.e. 3 from right to left go on reducing the power i.e.
153 152.3 15.32 33
iv) Find the values and write in 2nd Row
153 152.3 15.32 33
3375 675 135 27
v) Multiply middle terms by 3 i.e.
153 152.3 15.32 33
3375 675 135 27
×3 ×3
3375 2025 405 27
vi)To get the final answer we are going to keep unit digit as it is rest other digit are carried forward
153 152.3 15.32 33
3375 675 135 27
×3 ×3
3375 2025 405 27
(206) (40) (2) (CARRIED FORWARD)
3581 5 7 7
ANSWER
(153)3 = 35,815,77
To calculate cube root of any Perfect Cube easily, we need to Memorize the cubes, cube of unit digit and its relations from 1 to 10.
| Number | Cube | Cube of unit digit | Relations |
| 13 | 1 | 1 | 1 ⇒1 |
| 23 | 8 | 8 | 8⇒2 |
| 33 | 27 | 7 | 7⇒3 |
| 43 | 64 | 4 | 4⇒4 |
| 53 | 125 | 5 | 5⇒5 |
| 63 | 216 | 6 | 6⇒6 |
| 73 | 343 | 3 | 3⇒7 |
| 83 | 512 | 2 | 2⇒8 |
| 93 | 729 | 9 | 9⇒9 |
| 103 | 1000 | 0 | 0⇒0 |
Mathematics is very interesting and easy to learn when we use tricks to make it simple,now let’s see how easily we can solve cube root of perfects cube by using Vedic method and also by viewing the table above.
Example 1: Find cube root of 19683
Check the last digit number & make a groups of three, three digits
from right side i.e. 19683 Can be written as
19 , 683
STEP 1
- Take the last group which is 683, in this case the last digit of 683 is 3
- Now compare the last digit number i.e. 3 in the above table in the relation we can see that if last digit of perfect cube is 3 then last digit of cube root is 7, 3⇒7
- So we replace 3 by 7 at the right most digit of the cube root
19 , 683
7
STEP 2
- Now check the remaining digit which is left i.e. 19
- We have to check the cube which is less then or equal to the digit 19
- In this case the number 23=8 which is less than 19
- We will replace 19 by 2
i.e. 19 , 683
2 7
Hence the answer for cube root of 19683 is 27
Example 2: Find cube root of 551368
Check the last digit number & make a groups of three, three digits from right side i.e. 19683 Can be written as
551 , 368
STEP 1
- Take the last group which is 368, in this case the last digit of 368 is 8
- Now compare the last digit number i.e. 8 in the above table in the relation we can see that if last digit of perfect cube is 8 then last digit of cube root is 2 i.e 8⇒2
- So we replace 8 by 2 at the right most digit of the cube root
551 , 368
2
STEP 2
- Now check the remaining digit which is left i.e. 551
- We have to check the cube which is less then or equal to the digit 551
- In this case the number 83=512 which is less than 551
- We will replace 551 by 8
i.e. 551 , 368
8 2
Hence the answer for Cube root of 551,368 =82
Example 3: Find cube root of 148,877
Check the last digit number & make a groups of three, three digits from right side i.e. 148,877 Can be written as
148 , 877
STEP 1
- Take the last group which is 877, in this case the last digit of 877 is 7
- Now compare the last digit number i.e. 7 in the above table in the relation we can see that if last digit of perfect cube is 7 then last digit of cube root is 3 i.e 7⇒3
- So we replace 7 by 3 at the right most digit of the cube root
148 , 877
3
STEP 2
- Now check the remaining digit which is left i.e. 148
- We have to check the cube which is less then or equal to the digit 148
- In this case the number 53=125 which is less than 148
- We will replace 148 by 5
i.e 148 , 877
5 3
Hence the answer for Cube root of 19683 =53
Now that we have learned how to find square root of any numbers using Vedic math , we will go ahead and use this technic to find cube of numbers. Finding cube of the number plays a vital role in Mathematics, so it is useful to find quick ways to find the cube of the number. We will learn the Vedic trick which will help us finding the cube of numbers in quick way. To find the cube of numbers we are going to learn four types Vedic Tricks, In the first we are going to find the cube of a number which starts with 1 i.e. 11,12,13……. In second type we are going to find the cube of a number which ends with 1 i.e.21,31,41……In the third type we are going to find the cube of a number having same digit i.e.22,33,44……and Finally in the fourth type we are going to find the cube of a number having different digit.
By knowing the values below will help us solve the problems easily.
| Number | Square of a number | Cube of a number |
| 1 | 12 = 1 | 13 = 1 |
| 2 | 22 = 4 | 23 = 8 |
| 3 | 32 = 9 | 33 = 27 |
| 4 | 42 = 16 | 43 = 64 |
| 5 | 52= 25 | 53 = 125 |
| 6 | 62=36 | 63 = 216 |
| 7 | 72=49 | 73 = 343 |
| 8 | 82=64 | 83 = 512 |
| 9 | 92=81 | 93 = 729 |
| 10 | 102=100 | 103 = 1000 |
Type 1: Numbers starting with 1
(16)3
1) We consider 1 as 1st term and 6 as 2nd term, we will write the given term as it is i.e. 1 6
2) Square the 2nd term i.e. 62=36 and also cube the 2nd term i.e.=63=216 and write the values in 1st row i.e.
1 6 36 216
3) In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
1 6 36 216
12 72 (Doubled the value)
4) Add them vertically in column. carry forward the 10th places digit to next column
1 6 36 216
+ 12 72
3 12 21 (Carried forward)
4 0 9 6 (Answer)
(16)3=4,096
Type 2: Numbers Ending with 1
(61)3
1) We consider 6 as 1st term and 1 as 2nd term, we will write the given term as it is but in reverse order i.e. 6 1
2) Square the 1st term i.e. 62=36 and also cube the 1st term i.e.=63=216 and write the
values in 1st row i.e.
216 36 6 1
3) In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
216 36 6 1
72 12 (Doubled the value)
4) Add them vertically in column. carry forward the 10th places digit to next column
216 36 6 1
+ 72 12
10 1 (Carried forward)
226 9 8 1 (Answer)
(61)3=226,981
Type 3: Numbers having same digit
(55)3
1) We consider 5 as 1st term and 5 as 2nd term, here both the digit is same so we take any one digit .Cube of 5 =125 and write 4 times
i.e. 125 125 125 125
2) In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
125 125 125 125
250 250 (Doubled the value)
4) Add them vertically in column. carry forward the 10th places digit to next column
125 125 125 125
+ 250 250
41 38 12 (Carried forward)
166 3 7 5 (Answer)
(55)3=166,375
Type 4: Numbers having different digit
(32)3
1) We consider 3 as 1st term and 2 as 2nd term, Cube the 1st term and 2nd term
i.e. (3)3 =27 and (2)3=8
27 8
2) Square the 1st term i.e. 32=9 then multiply by 2nd term i.e.9×2=18
27 18 8
3) Square the 2nd term i.e. 22 =4 then multiply by 1st term i.e.4×3=12
27 18 12 8
4) In 2nd row double the 2 middle terms (. i.e. is 2nd term and 3rd term) & write just
below 2nd & 3rd term.
27 18 12 8
36 24 (Doubled the value)
5) Add them vertically in column. carry forward the 10th places digit to next column
27 18 12 8
+ 36 24
5 3 (Carried forward)
32 7 6 8 (Answer)
(32)3=32,768
Steps for Square Roots Math tricks
For performing square roots, we will have to keep some facts in mind:-
Squares of numbers from 1 to 9 are 1, 4, 9, 16, 25, 36, 49, 64, 81.
Square of a number cannot end with 2, 3, 7, and 8.
We can say that numbers ending with 2, 3, 7, and 8 cannot have a perfect square root.
The square root of a number ending with 1 (1, 81) ends with either 1 or 9
The square root of a number ending with 4 (4, 64) ends with either 2 or 8
The square root of a number ending with 9 (9, 49) ends with either 3 or 7
The square root of a number ending with 6 (16, 36) ends with either 4 or 6
If the number is of ‘n’ digits then the square root will be ‘n/2’ OR ‘(n+1)/2’ digits.
Let us understand an example of finding a square root of 1764.
The number ends with 4. Since it’s a perfect square, square root will end with 2 or 8.
We need to find 2 perfect squares (In Multiples of 10) between which 1764 exists.
The numbers are 1600 (40) and 2500 (50).Find to whom 1764 is closer. It is closer to 40. Therefore the square root is nearer to 40. Now from Step 2, possibilities are 42 or 48 out of which 42 is closer to 40.
Hence the square root = 42
Steps to find Square root of 3, 4, & 5 digits’ number.
1) Finding Square root of 3 digit
Consider a number
√729
1)Splits the number keeping last 2 digits aside i.e.
7 ǀ 29
2) Check the square digit of number 9 & its unit Square digit is 3& 7
3) Next check Square 7 it lies between 2 & 3 Square (22=4,32=9)
4)Choose the smaller number i.e.2
5)Now there is two possibilities for answer i.e. 23 or 27.
6) The smaller number is 2, consider the next number of 2 is 3 multiply those two number 2×3=6 ,6 is smaller as we compare to given digit i.e. 7 (6˂ 7).
7)We have to consider the greater number i.e.27 so the final answer will be 27.
√729=27
2 )Finding Square root of 4-digit number
Consider the number
√1024
1)Splits the number keeping last 2 digits aside i.e.
10 ǀ 24
2) Check the square digit of number 4 & its unit Square digit is 2& 8
3) Next check Square 10 it lies between 3 & 4 Square (32=9,42=16)
4) Choose the smaller number i.e.3
5) Now there is two possibilities for answer i.e. 32 or 38
6) The smaller number is 3, consider the next number of 3 is 4 multiply these twos’
number 3×4=12 ,12 is greater as we compare to given digit i.e.10 (12> 7)
7) We have to consider the smaller number i.e.32 so the final answer will be 32.
√1024=32
3) Finding square of 5-digit number
√76176
1) Splits the number keeping last 2 digits aside i.e.
7 61 ǀ 76
2) Check the square digit of number 6 & its unit Square digit is 4& 6
3) Next check Square 716 it lies between 27&28 Square (272=729,282=784)
4) Choose the smaller number i.e.27
5) Now there is two possibilities for answer i.e. 274 or 276.
6) The smaller number is 27, consider the next number of 27 is 28, multiply those
two number 27×28=756, is smaller as we compare to given digit i.e.761 (756˂ 761).
7) We have to consider the greater number i.e.276 so the final answer will be 276
√76176=276
Divisibility tests of common numbers
Divisibility of different numbers has been listed below:
- Divisibility by 2 : A number is divisible by 2, if its unit digits is any of 0,2,4,6,8.
In other words, when the unit's place digit is even or zero (0). - Divisibility by 3 : A number is divisible by 3 only when the sum of its digits is divisible by 3. In other words, All numbers with a digit sum of three (3), six (6) or nine (9) are divisible by 3.
- Divisibility by 4 : A number is divisible by four (4), if its last two (2) digits are divisible by 4.
- Divisibility by 5 : A number is divisible by 5 if its unit digit is five (5) or zero (0).
- Divisibility by 6 : A number is divisible by 6 if it is divisible by both two (2) and three (3).
- Divisibility by 8 : A number is divisible by eight (8) only when the number formed by its last three (3) digits is divisible by eight (8).
- Divisibility by 9 : A number is divisible by nine (9) if sum of its digits is divisible by nine (9). In other words, All numbers with a digit sum is nine (9) are divisible by nine (9).
- Divisibility by 10 : A number is divisible by ten (10) only when its unit digit is zero (0).
- Divisibility by 11 : A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either zero (0) or number is divisible by 11. In simple words, Add all the digits in the ODD position and all digits in the EVEN position and subtract the smaller result from the larger result. If we get 0 or 11 or any multiples of 11 , then the number is divisible by 11.
Example : 7352631
Sum of odd digits : 7+5+6+1 = 19
Sum of Even digits : 3+2+3 = 8
19-8 = 11. So, number 7352631 is divisible by 11 - Divisibility by 12 : A number is divisible by 12 if it is divisible by both three (3) and four (4). In other words, All numbers divisible by both 3 and 4 are divisible by 12.
- Divisibility by 15 : A number is divisible by 15 if it is divisible by both three (3) and five (5). In other words, All numbers divisible by both 3 and 5 are divisible by 15.
- Divisibility by 25 : A number is divisible by twenty-five (25) only when the number formed by its last two (2) digits is divisible by twenty-five (25).
- Divisibility by 27 : A number is divisible by twenty-seven(27) if sum of its digits is divisible by twenty-seven(27).
- Divisibility by 125 : A number is divisible by one hundred and twenty-five (125) only when the number formed by its last three (3) digits is divisible by one hundred and twenty-five (125).
Multiply 11 to 19 In Your Head
It is assumed that you know your multiplication table reasonably well up to 10x10.
Example 1: Multiply 14 by 17
- Always place the larger number of the two on top in your mind.
- Add right portion of lower number. 17 + 4 = 21
- Add a zero behind it (multiply by 10) to get 210.
- Multiply right portions of both numbers (4x7 = 28)
- Add 210 + 28 = 238.
Example 2: Multiply 15 by 18
- Place the larger number of the two on top in your mind.
- First add 18 + 5 = 23
- Add a zero behind it (multiply by 10) to get 230.
- Multiply the covered lower 5 x the single digit above it the "8" (5x8 = 40)
- Add 230 + 40 = 270.
That is It! Wasn't that easy? Practice it on paper first!
Note: This Method works only numbers between 11 and 19.
When both the numbers are more than the same base. This method is extension of the above method i.e.
we are going to use same sutra here and applying it to larger numbers.
Step 1: Here base is 10
12 + 2 [12 is 2 more than 10 also called surplus]
14 + 4 [14 is 4 more than 10also called surplus]
Step 2: Cross add: 12 + 4 =16 or 14 + 2 = 16,(both same) which gives first part of answer = 16
Step 3: Vertical multiplication: 2 × 4 = 8
So, 12 + 2
14 +4
16 / 8So, 12 × 14 = 168
(14 + 2 = 12 + 4)
Step1: Here base is 100
105 + 05 [105 is 5 more than 100 or 5 is surplus]
107 + 07 [107 is 7 more than 100 or 7 is surplus]
Base here is 100 so we will write 05 in place of 5and 07 in place of 7
Step 2: Cross add: 105 + 7 = 112 or 107 + 5 = 112 which gives first part of the answer = 112
Step 3: Vertical multiplication: 05 × 07 = 35 (two digits are allowed)
As the base in this problem is 100 so two digits are allowed in the second part.
So, 105 × 107 = 11235
Step 1: Here base is 100
112 + 12 [2 more than 100 i.e. 12 is surplus]
115 + 15 [15 more than 100 i.e. 15 is surplus]
Step 2: Cross add: 112 + 15 = 127 = 115 + 12 to get first part of answer
i.e.127
Step 3: Vertical multiplication 12 × 15 = ? Oh, my god!It’s such a big number. How to get product of this? Again use the same method to get the product.
12 + 2
15 + 5
12 + 5 = 15 + 2 = 17/ (1) 0, 17 + 1 / 0 = 180 i.e. 12 × 15 = 180
But only two digits are allowed here, so 1 is added to 127 and we get (127 + 1) = 128
So, 112 × 115 = 128, 80
Same sutra applied to bigger numbers which are less than the same base.
Step 1: Check the base: Here base is 100 so we are allowed to have two digits on the right hand side.
99 – 01 (1 less than 100 ) i.e. 01 deficiency
98 – 02 (2 less than 100) i.e. 0 2 deficiency
Step 2: Cross – subtract: 99 – 02 = 97 = 98 – 01 both same so first part of answer is 97
Step3: Multiply vertically – 01 × – 02 = 02 (As base is 100 so two digits are allowed in second part
So, 99 × 98 = 9702
Step1: Here base is 100
So, 89 – 11 (i.e. deficiency = 11)
88 – 12 (i.e. deficiency = 12)
Step2: Cross subtract: 89 – 12 = 77 = 88 – 11(both same)
So, first part of answer can be 77
Step 3:Multiply vertically – 11 × – 12
Again to multiply 11 × 12 apply same rule
11 + 1 (10 + 1)
12 + 2 (10 + 2)
11 + 2 = 13 = 12 + 1 / 1 × 2 = 12 so, 11 × 12 = (1) 32 as only two digits are allowed on right hand
side so add 1to L.H.S.
So, L.H.S. = 77 + 1 = 78
Hence 89 × 88 = 7832
Step 1: As the numbers are near 1000 so the base here is 1000 and hence three digits allowed on the
right hand side
988 – 012 (012 less than 1000) i.e. deficiency = 0 12
999 – 001 (001 less than 1000) i.e. deficiency = 00 1
Step 2: Cross – subtraction: 988 – 001 = 987 = 999 – 012 = 987
So first part of answer can be 987
Step 3: Multiply vertically: –012 xs – 001 = 012 (three digits allowed)
988 × 999 = 987012
As, 99 = 0 Product (L.H.S.) = 0 × 8 = 0 [taking 9 = 0]
98 = 8
R.H.S. = 9702 = 7 + 2 = 9 = 0 9702 = 9 both are same
As both the sides are equal answer may be correct.
89 = 8
88 = 8 + 8 = 16 = 1 + 6 = 7 (add the digits)
L.H.S. = 8 × 7 = 56 = 5 + 6 = 11 = 2 (1 + 1)
R.H.S. = 7832 = 8 + 3 = 11 = 1 + 1 = 2
As both the sides are equal, so answer is correct
988 = 8 + 8 = 16 = 1 + 6 =7
999 = 0
As 0 × 7 =0 = LHS
987012 = 0 (As 7 + 2 = 9 = 0 , 8 + 1 = 9 = 0 also 9 = 0 )
RHS = 0
As LHS = RHS So, answer is correct.
(i) 97 × 99
(xv) 8987 × 9996.
Step1: Base here is 100000 so five digits are allowed in R.H.S.
87798 – 12202 (12202 less than 100000) deficiency is 12202
99995 – 00005 (00005 less than100000) deficiency is 5
Step 2: Cross – subtraction: 87798 -00005 =87793
Also 99995 – 12202 = 87793 (both same)
So first part of answer can be 87793
Step 2 : Multiply vertically: –12202 × – 00005 = + 61010
87798 × 99995 = 8779361010
Checking:
87798 total 8 + 7 + 7 + 8 = 30 = 3 (single digit)
99995 total = 5
LHS = 3 x 5 =15 total = 1 + 5 = 6
RHS = product = 8779361010 total = 15 = 1 + 5 = 6
L.H.S = R.H.S. So, correct answer
Step 1: Base have is 100000 so five digits are allowed in R.H.S.
88777 – 11223 i.e. deficiency is 11223
99997 – 00003 i.e. deficiency is 3
Step 2: Cross subtraction: 88777 – 00003 = 88774 = 99997 – 11223
So first part of answer is 88774
Step 3: Multiply vertically: – 11223 × – 00003 = + 33669
88777 × 99997 = 8877433669
Checking: 88777 total 8 + 8 + 7 + 7 + 7 = 37 = + 10 = 1
99997 total = 7
LHS = 1 × 7 = 7
RHS = 8877433669 =8 + 8 + 7 + 7 + 4 = 34 = 3 + 4 = 7
i.e. LHS = RHS So, correct answer
(i) 999995 × 739984
Step 1: Here base is 10000 so four digits are allowed
10021 + 0021 (Surplus)
10003 + 0003 (Surplus)
Step 2: Cross – addition 10021 + 0003 = 10024 = 10003 + 0021 (both same)
First part of the answer may be 10024
Step 3: Multiply vertically: 10021 × 0003 = 0063 which form second part of the answer
10021 × 10002 = 100240063
Checking:
10021 = 1+ 2 + 1 + 1 = 4
10003 = 1 + 3 = 4
LHS = 4 × 4 = 16 = 1 + 6 = 7
RHS = 100240063 = 1 + 2 + 4 = 7
As LHS = RHS So, answer is correct
Step 1: Here base is 10000 so four digits are allowed in RHS
11123 + 1123 (surplus)
10003 + 0003 (surplus)
Step 2: Cross – addition: 11123 + 0003 = 11126 = 10003 + 1123 (both equal)
First part of answer is 11126
Step 3: Multiply vertically: 1123 × 0003 = 3369 which form second part of answer
11123 × 10003 = 111263369
Checking:
11123 = 1 + 1 + 1 + 2 + 3 = 8
10003 = 1 + 3 = 4 and 4 × 8 = 32 = 3 + 2 =5
LHS = 5
R.H.S = 111263369 = 1 + 1 + 1 + 2 = 5
As L.H.S = R.H.S So, answer is correct
(i) 10004 × 11113
Step 1: 98 Here base is 100 deficiency=02
9 Base is 10 deficiency = 1
98 – 02 Numbers of digits permitted on R.H.S is 1 (digits in lower base )
Step 2: Cross subtraction: 98-188
It is important to line the numbers as shown because 1 is not subtracted from 8 as usual but from
9 so as to get 88 as first part of answer.
Step 3: Vertical multiplication: (-02) x (-1) = 2 (one digits allowed )
Second part = 2
98 × 9 = 882
(Through 9 – check method)
98 = 8 , 9 = 0, LHS = 98 × 9 = 8 × 0 = 0
RHS = 882 = 8 + 8 + 2 = 18 = 1 + 8 = 9 = 0
As LHS = RHS So, correct answer
Step 1: 993 base is 1000 and deficiency is 007
97 base is 100 and deficiency is 03
993 – 007 (digits in lower base = 2 So, 2 digits are permitted on
× 97 – 03 RHS or second part of answer)
Step 2: Cross subtraction:
993– 03
963
Again line the number as shown because 03 is subtracted from 99 and not from 93 so as to get 963
which from first part of the answer.
Step 3: Vertical multiplication: (–007) – (–03) = 21 only two digits are allowed in the second part
of answer So, second part = 21
993 × 97 = 96321
Checking: (through 9 – check method)
993 = 3 97 = 7
L.H.S. = 3 × 7 = 21 = 2 + 1 = 3
R.H.S. = 96321 = 2 + 1 = 3
As LHS =RHS so, answer is correct
988 base is 1000 and deficiency is 012
9996 – 0004 (digits in the lower base are 3 so,3digits
× 988 – 012 permitted on RHS or second part of answer)
Step 2 : Cross – subtraction:
9996– 012
9876
Well, again take care to line the numbers while subtraction so as to get 9876 as the first part of the answer.
Step3 : Vertical multiplication: (–0004) × (–012) = 048
9996 × 988 = 9876048
Checking:(9 – check method)
9996 = 6, 988 = 8 + 8 + = 16 = 1 + 6 = 7
LHS = 6 × 7 = 42 = 4 + 2 = 6
RHS = 9876045 = 8 + 7 = 15 = 1 + 5 = 6
As, LHS =RHS so, answer is correct
Example 1: 105 × 12
Step 1: 105 base is 100 and surplus is 5
12 base is 10 and surplus is 2
105 + 05 (digits in the lower base is 1 so, 1 digit is permitted in the second part of answer )
12 + 2
Step 2: Cross – addition:
105
+ 2
125 (again take care to line the numbers properly so as to get 125 )
First part of answer may be 125
Step 3: Vertical multiplication : 05 × 2 = (1)0 but only 1 digit is permitted in the second part so 1
is shifted to first part and added to 125 so as to get 126
105 × 12 = 1260
Checking:
105 = 1 + 5 = 6 , 12 = 1 + 2 = 3
LHS = 6 × 3 = 18 = 1 + 8 = 9 = 0
RHS = 1260 = 1 + 2 + 6 = 9=0
Step1: 1122 – base is 1000 and surplus is 122
104 – base is 100 and surplus is 4
1122 + 122
104 + 04 (digits in lower base are 2 so, 2-digits are permitted in the second part of answer )
Step 2: Cross – addition
1122
+ 04 (again take care to line the nos. properly so as to get 1162)
1162
40
First part of answer may be 1162
Step 3: Vertical multiplication: 122 × 04 = 4, 88
But only 2 – digits are permitted in the second part, so, 4 is shifted to first part and added to 1162
to get 1166 ( 1162 + 4 = 1166 )
1122 × 104 = 116688
Can be visualised as: 1122 + 122
104 + 04
1162 / (4) 88 = 116688
+ 4 /
Checking:
1122 = 1 + 1 + 2 + 2 + = 6, 104 = 1 + 4 =5
LHS = 6 × 5 = 30 = 3
RHS = 116688 = 6 + 6 = 12 = 1 + 2 = 3
As LHS = RHS So, answer is correct
Now doing the question directly
10007 + 0007 base = 10000
× 1003 + 003 base = 1000
10037 / 021 (three digits per method in this part)
10007 × 10003 = 10037021
Checking : 10007 = 1 + 7 = 8 , 1003 = 1 + 3 = 4
LHS = 8 × 4 = 32 = 3 + 2 = 5
RHS = 10037 021 = 1 + 3 + 1 = 5
As LHS = RHS so, answer is correct
(i) 1015 × 103
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