Multiply the numbers close to the power of 10s in just 5 seconds! - Part 1

Multiplication of numbers that are close to the power of 10s (i.e 99999 x 99997 ; 99982 99992) can be easily done in just 5 seconds. Tough to believe it right, but it is possible and all that you need to know is basic subtraction, addition and multiplication tables till 5. Just, follow the below examples with their explanation and by the end of the article you would be all set.

Part 1: Multiplying numbers below to the power of 10s

Example 1: 9 X 6

Step 1: Identify the base (power of 10) to which the number is close. Here, 9 and 6 are close to 10. Therefore, base is 10

Step 2: Identify by how much both the numbers are less than the base. Here, 9 is less by 1 and 6 is less by 4.

Step 3: Write the original numbers along with the number by which it is less than 10 as below.

9 - 1

6 - 4

Step 4: Now, Cross subtract the original number and the deficit number. That is subtract 9 by 4 or subtract 6 by 1. So, we get

9 - 1

X

6 - 4

-----------

5

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficit numbers are 1 and 4. Therefore, we have

9 - 1

X |

6 - 4

-----------

5 / 4 = 54 (since base is 10 we can have 1 digit on the right hand side which is 4)

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Therefore we have 9 X 6 = 54

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Example 2: 7 X 6

Step 1: Identify the base (power of 10) to which the number is close. Here, 7 and 6 are close to 10. Therefore, base is 10

Step 2: Identify by how much both the numbers are less than the base. Here, 7 is less by 3 and 6 is less by 4.

Step 3: Write the original numbers along with the number by which it is less than 10 as below.

7 - 3

6 - 4

Step 4: Now, Cross subtract the original number and the deficit number. That is subtract 7 by 4 or subtract 6 by 3. So, we get

7 - 3

X

6 - 4

-----------

3

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficiet numbers are 3 and 4. Therefore, we have

7 - 3

X  |

6 - 4

-----------

3 / 2

1

-----------

Step 6: Now, since the base is 10, the right hand most digit is units and can have only one digit. Therefore, keep the 2 of the 12 on the right hand side and carry the 1 over the left and change the 3 into 4.

7 - 3

X |

6 - 4

-----------

3 / 2 = 42

1

-----------

Therefore we have 7 X 6 = 42

Note: If the number of digits are more than the number of zeroes in the base, the excess digit or digits are to be added to left hand side of the answer.

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Example 3: 91 X 99

Step 1: Identify the base (power of 10) to which the number is close. Here, 91 and 99 are close to 100. Therefore, base is 100

Step 2: Identify by how much both the numbers are less than the base. Here, 91 is less by 9 and 99 is less by 1.

Step 3: Write the original numbers along with the number by which it is less than 100 as below.

91 - 9

99 - 1

Step 4: Now, Cross subtract the original number and the deficient number. That is subtract 99 by 9 or subtract 91 by 1. So, we get

91 - 9

X

99 - 1

-----------

90

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficiet numbers are 9 and 1. Therefore, we have

91 - 9

X  |

99 - 1

-----------

90 / 9

-----------

Step 6: Now, since the base is 100, the right hand most digit is tens and units and should have two digit. Therefore, fill the vacancie by adding zero. thus

91 - 9

X  |

99 - 1

-----------

90 / 09 = 9009

-----------

Therefore we have 91 X 99 = 9009

Note: If the Right hand side contains less number of digits than the number of zeros in the base, the remaining digits are filled up by giving zero or zeroes on the left side of the right hand side digit.

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Example 4: 93 X 97

Step 1: Identify the base (power of 10) to which the number is close. Here, 93 and 97 are close to 100. Therefore, base is 100

Step 2: Identify by how much both the numbers are less than the base. Here, 93 is less by 7 and 97 is less by 3.

Step 3: Write the original numbers along with the number by which it is less than 100 as below.

93 - 7

97 - 3

Step 4: Now, Cross subtract the original number and the deficient number. That is subtract 93 by 3 or subtract 97 by 7. So, we get

93 - 7

X

97 - 3

-----------

90

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficiet numbers are 7 and 3. Therefore, we have

93 - 7

   X |

97 - 3

-----------

90 / 21 = 9021

-----------

Therefore we have 93 X 97 = 9021

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Example 5: 883 X 997

Step 1: Identify the base (power of 10) to which the number is close. Here, 883 and 997 are close to 1000. Therefore, base is 1000.

Step 2: Identify by how much both the numbers are less than the base. Here, 883 is less by 117 and 997 is less by 3.

Step 3: Write the original numbers along with the number by which it is less than 100 as below.

883 - 117

997 - 003

Step 4: Now, Cross subtract the original number and the deficient number. That is subtract 883 by 3 or subtract 117 by 7. So, we get

883 - 117

X

997 - 003

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

880

Step 5: Now, vertically multiply the two deficit numbers. Here, the deficiet numbers are 117 and 3. Therefore, we have

883 - 117

X   |

997 - 003

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

880 / 351 = 880351 (since base is 1000 we can have 3 digits on the right hand side)

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Therefore we have 883 X 997 = 880351

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This method is known as "Nikhilam Navatashcaramam Dashatah - All from 9 and the last from 10" in Vedic Mathematics.

Note:

1. This method can be easily applied to the numbers that are near to the base 10s i.e., 10,100, 1000....

2. If one number is near base of 10s and the other is not, then also it can easily be applied - Ex 88 X 99 or 25 X 98 and above Example 5.

3. The deficit numbers can easily written by following the rule that all the digits (of the original number) are to be subtracted from 9 but the last (right hand most one) one should be subtracted from 10. Ex deficit of 63 is 37 (6-9 =3 / 10-3=7).

With little bit of practice by following the above method, I am sure you can do multiplication of the numbers near to power of 10s in just 5 seconds. Wondering that above method is good for the numbers that are below base 10s but what about numbers which are above it? Don't worry, Vedic mathematics has a solution for it too, I shall discuss the same in my next article, till then practice this one and let me know if you have any questions.