Vedic Mathematics Series Part 2

We initially looked at Vedic Mathematics’  Ek?dhikena P?rvena formula to help us find the square of numbers ending in 5 and then extended it to other numbers near the same by combining the formula with some other properties. Now we shall use the same formula in a different context.

Find recurring fractions whose denominators are numbers are ending in ‘9’:

This includes all numbers of the form 1/X9 where x ε {1, 2…..9}. To find the decimal representation of such numbers we can use our Ekadhika Purvena. There are two methods in this which we shall now look one by one.
Division Method:  To illustrate the same let’s find the value of Value of 1/29.

Now the decimal representation will be recurring since the denominator cannot be factorized into prime factors of 2 and 5. The numbers of decimal places before the recurring starts is the difference between the numerator and denominator which in this case is 29 -1=28 digits.

Here for application of the formula the purva (previous) for the denominator is the 10’s digit which is 2. Hence Ekadhikena purva (one more than the previous) is 2+ 1 = 3.

The formula is to be applied in a different context for this problem which is shown as below:

1.    Start with 1/ Ekadhikena purva = 1/3  =  remainder is 1 and quotient is  0 represented as (0.₁0)
2.    Now read ₁0 as 10 and divide this by Ekadhikena purva which is 10/3 with remainder 1 and quotient 3 represented as (0.₁0₁3)
3.    Repeat the same for 13 read as 13 leading to (.₁0₁3₁4)
4.    Repeating (.₁0₁3₁4₂4)
5.    Resulting in (.₁0₁3₁4₂4₀8₂2₁7₂5₁8₀6₀2₂0..) = 0.03448275860..
Note that we can go on getting the complete decimal representation only by using 3’s division which is far simpler than trying to divide by 29

Doing the same for say 17/19

Purva for the denominator = 1

Ekadhikena purva (one more than the previous) is 1+ 1 = 2

1.    17/2 = 0.₁8
2.    18/2=0.₁8₀9
3.    09/2=0.₁8₀9₁4
4.    14/2=0.₁8₀9₁4₀7
5.    … =₁8₀9₁4₀7₁3₁6₀8₀4₀2₀1₁0₀5₁2₀6₀3₁1₁5₁7₁8₀9₁4… => 0.894736842105263157..

Hence we can do this for any 1/X9 and most AB/X9 fraction.

Multiplication Method: Same thing can be done in different way using only multiplication too
Say to find the Value of 1 / 19

Here the again the format is 1/X9 where x ε {1, 2…..9}. Here the last digit is 9. The purva (previous) for the denominator is the 10’s digit which is 1. Hence Ekadhikena purva (one more than the previous) is 1+ 1 = 2.
In the multiplication this Ekadhikena purva is the multiplier for the calculation rather than the divider. In this the process happens leftward.

1.    Start with the last digit in the numerator which is ‘1’
2.    Multiply by 2 and put it to the left = (21)
3.    Multiply by 2 and put it to the left = (421)
4.    Multiply by 2 and put it to the left = (8421)
5.    Multiply by 2 and put it to the left with a carryover of 1 represented as = (¹68421)
6.    Multiply by 2, add the carry over = 6x2 + 1 =13 represented as (¹3¹68421)
7.    Multiply by 2, add the carry over = 3x2 + 1 =7 represented as (⁰7¹3¹68421)
8.    Similarly (⁰7¹3¹68421)
9.    ⁰1¹0⁰5¹2⁰6⁰3¹1¹5¹7¹8⁰9¹4⁰7¹3¹6⁰8⁰4⁰2⁰1 =0.1052631578947368421
10.    From this onwards as per the rule (19-1 =18 digits) the decimal recurs.

Here we get the decimal without any division and without any calculations involving 19 and just 1 and 2

Similarly for 11/19

1.    Start with ¹1
2.    Multiply by 2 and add carry over = 1x2+1 = 3 => ⁰3¹1
3.    Similarly 5¹7¹8⁰9¹4⁰7¹3¹6⁰8⁰4⁰2⁰1¹0⁰5¹2⁰6⁰3¹1 = 0.578947368421052631

Insights for objective type questions

1.    For any 1/X9 type question the answer is a recurring decimal that recurs only after X9-1 decimals. Any recurrence before that can help to eliminate that answer
2.    If the choices presented are differing towards the starting digit start with the division method and do just till you have enough digits to eliminate last but one choices
3.    If the choices presented are differing towards the ending digits start with the multiplication method and do just till you have enough digits to eliminate last but one choices
4.    Another peculiarity say for 11/19 0.578947368421052631 = 578947368 || 421052631 the numbers in the 1st set are 9’s compliment of the 2nd set of numbers. You can hence find the answers by finding half of the digits.
5.    You can save further time by looking for this 9’s complement pattern in the choices in the questions without any calculation.
Hence using this technique any division of the form 1/X9 (and most of form AB/X9) can be easily done and further by applying the insights, seemingly time consuming problems can be done within 10 seconds or less by elimination.

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