Shortcut to find the Cube of a number
Very often we have to find the cube, i.e. third power of 2 digit numbers. Cubes of very large numbers are rarely used.
Cubes of all the single digits should be memorized. Find below the table of cubes of first ten natural numbers -
13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125,
63 = 216, 73 = 343, 83 = 512, 93 = 729, 103 = 1000
To find the cube of any 2 digit number, we have to take the following steps
First Step: The first thing we have to do is to put down the cube of the tens-digit in a row of 4 figures. The other three numbers in the row of answer should be written in a geometrical ratio in the exact proportion which is there between the digits of the given number.
Second Step: The second step is to put down, under the second and third numbers, just two times of second and third number. Then add up the two rows.
Finding the cube of 12
Or, 123 = ?
First Step: Digit in ten’s place is 1, so we write the cube of 1. And also as the ratio between 1 and 2 is 1:2, the next digits will be double the previous one. So, the first row is
1 2 4 8
Step II: In the above row our 2nd and 3rd digits (from right) are 4 and 2 respectively. So, we write down 8 and 4 below 4 and 2 respectively. Then add up the two rows.
Ex 2: 163 = ?
Soln:
Explanations: 13 (from 16) = 1. So, 1 is our first digit in the first row. Digits of 16 are in the ratio 1:6, hence our other digits should be 1×6 = 6, 6×6 = 36, 36×6 = 216. In the second row, double the 2nd and 3rd number is written. In the third row, we have to write down only one digit below each column (except under the last column which may have more than one digit). So, after putting down the unit-digit, we carry over the rest to add up with the left-hand column. Here,
i) Write down 6 of 216 and carry over 21.
ii) 36 + 72 + 21 (carried) = 129, write down 9 and carry over 12.
iii) 6 + 12 + 12 (carried) = 30, write down 0 and carry over 3.
iv) 1 + 3 (carried) = 4, write down 4.
Popularity: 54%
Divisibility Rule of 7, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47
You might have seen divisibility rules of various numbers. But most of them very conveniently skip the ones which are very difficult and a divisibility rule for which is very much required. Today I’ll show you the divisibility rule for 7, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.
|
Number |
Method |
Example |
|
7 |
Subtract 2 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 7, the original number is also divisible by 7 | Check for 945: : 94-(2*5)=84. Since 84 is divisible by 7, the original no. 945 is also divisible |
|
13 |
Add 4 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 13, the original number is also divisible by 13 | Check for 3146:: 314+ (4*6) = 338:: 33+(4*8) = 65. Since 65 is divisible by 7, the original no. 3146 is also divisible |
|
17 |
Subtract 5 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 17, the original number is also divisible by 17 | Check for 2278:: 227-(5*8)=187. Since 187 is divisible by 17, the original number 2278 is also divisible. |
|
19 |
Add 2 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 19, the original number is also divisible by 19 | Check for 11343:: 1134+(2*3)= 1140. (Ignore the 0):: 11+(2*4) = 19. Since 19 is divisible by 19, original no. 11343 is also divisible |
|
23 |
Add 7 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 23, the original number is also divisible by 23 | Check for 53935:: 5393+(7*5) = 5428 :: 542+(7*8)= 598:: 59+ (7*8)=115, which is 5 times 23. Hence 53935 is divisible by 23 |
|
29 |
Add 3 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 29, the original number is also divisible by 29 | Check for 12528:: 1252+(3*8)= 1276 :: 127+(3*6)= 145:: 14+ (3*5)=29, which is divisible by 29. So 12528 is divisible by 23 |
|
31 |
Subtract 3 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 31, the original number is also divisible by 31 | Check for 49507:: 4950-(3*7)=4929. Since 492-(3*9) is divisible by 465:: 46-(3*5)=31. Hence 49507 is divisible by 31 |
|
37 |
Subtract 11 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 37, the original number is also divisible by 37 | Check for 11026:: 1102 - (11*6) =1036. Since 103 - (11*6) =37 is divisible by 37. Hence 11026 is divisible by 31 |
|
41 |
Subtract 4 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 41, the original number is also divisible by 41 | Check for 14145:: 1414 - (4*5) =1394. Since 139 - (4*4) =123 is divisible by 41. Hence 14145 is divisible by 41 |
|
43 |
Add 13 times the last digit to the remaining truncated number. Repeat the step as necessary. If the result is divisible by 43, the original number is also divisible by 43.
*This process becomes difficult for most of the people because of multiplication with 13. |
Check for 11739:: 1173+(13*9)= 1290:: 129 is divisible by 43. 0 is ignored. So 11739 is divisible by 43 |
|
47 |
Subtract 14 times the last digit from remaining truncated number. Repeat the step as necessary. If the result is divisible by 47, the original number is also divisible by 47. This too is difficult to operate for people who are not comfortable with table of 14. | Check for 45026:: 4502 - (14*6) =4418. Since 441 - (14*8) =329, which is 7 times 47. Hence 45026 is divisible by 47 |
Notes: In applying the above methods, stop repeating the step of adding or subtracting (as the case may be) from the remaining truncated number as soon as you realize that the truncated number is divisible by the given number.
Popularity: 14%
Vedic Multiplication by 9, 99, 999 and so on
When any number has to be multiplied by a series of 9s, like 9, 99, 999, 9999 and so on than we can apply this very simple vedic maths technique to increase your speed of calculation.
Multiplication with 9/ 99 / 999 and so on.
we know, 789 × 999 = 788,211
You will get the answers in two parts,
- The left hand side of the answer: subtract 1 from 789, which is 788
- The right hand side of the answer subtract 789 from 1000 = 1000-789= 211
Thus, 999 x 789 = 789-1 | 1000-789 = 788, 211 (answer)
{for the right hand side of the answer, 789 should be subtracted from (999+1)}
or, 99999 x 78 = 78-1 | 100000 - 78
= 7799922
{78 should be subtracted from (99999+1)}
Another example:
1203579 × 9999999 = 1203579-1 | 10000000- 1203579
=120357887964 21
Number in red is 1 less than 1203579. Number in blue is (10000000-1203579). Hence the answer.
This method has to be altered a little bit when number of 9s are lessers than the number of digit in the divisor.
1432 x 9 = 1432 (10 – 1) = 14320 – 1432 = 12888
So for multiplication with 9, put a zero after that number and subtract the number itself from that.
Likewise for 99 put two zeroes after that number .
3256 x 99 = 325600 – 3256 = 322344
Popularity: 23%




