# MSC #29: Using the Ekadhika

Can you solve this problem in less than 3 seconds?

“What are the last four digits in the recurring pattern in the decimal equivalent of 1/19?

A) 6667; B) 0526; C) 8421; D) 9991; E) 3829”

This problem, which was given in the intermediate level of the 1^{st} International Vedic Mathematics
Olympiad cannot be solved using an ordinary calculator because the recurring decimal string is 18 digits long. Manually dividing 1 by 19 will take too much time and very few will actually attempt it.

But by using the Ekadhika, which is the topic of the 29^{th} of our 30 MATH-Inic Specials for Christmas
series, we can know the answer in less than 3 seconds.

Our **29 ^{th} MATH-Inic Special for Christmas** is about a “fun, fast and easy” way to convert a fraction with a prime number denominator into a recurring decimal. In our example here, instead of dividing by 19, we will use its

*,*

**Ekadhika****2**. The

*is also discussed in Chapters 1 and 15 of the “30 Master Strategies in Computing”*

**Ekadhika***of 19 is*

**Ekadhika****2**, which is “one more than the one before” 9 in 19. Because

**1**is less than

**19**, we start with 0 and a decimal point

**1/19 = 0.**Next, we divide 1 by 2 getting a quotient of

**0**and a remainder of

**1**. We place the remainder

**before**the answer figure:

**1/19 = 0.**We then divide our next dividend, 10 by 2 to get the next answer figure, 5

_{1}0**1/19 = 0.**Then we divide 5 by 2 getting an answer of 2 and remainder of 1. Again, we place the remainder in front of the answer.

_{1}0 5.**1/19 = 0.**Repeat the steps, dividing the last answer figure (including the prefixed remainder) by the Ekadhika, 2, and putting down the result as the next answer figure, until we get an answer figure of 1. We stop at this point, since we started with a numerator of 1 and the sequence of the answer figure will just repeat in a cycle.

_{1}0 5_{1}2**1/19 = 0.**

_{1}0 5_{1}2 6 3_{1}1_{1}5_{1}7_{1}8**9**There will be 18 digits in the recurring decimal. This is logical because when dividing by 19 there are only 18 possible remainders. To generalize, the recurring decimal equivalent of 1/n will have

_{1}4 7_{1}3_{1}6 8 4 2 1**at**

**most**(n-1) digits. Once we have completed computing the recurring decimal equivalent of 1/19, we can now use the result to determine the decimal form of any fraction with 19 as the denominator. For example, if we want to know what the value of 12/19 is, we just look at the result of 1/19 and we see that the third digit is

_{1}2 which is actually 12 and the answer is composed of the digits following it:

**0.631578947368421052**. Now we have purposely divided the answer into two lines of 9 digits each. This is to show that there is a further short cut to this method. Take the difference between the denominator and the numerator, 19 – 1= 18. Once we see an

_{1}8 in the answer line, which is the 9

^{th}digit, we can stop computing and just use

*to get the next answer figure. You can see the figures in the first and second lines add up to 9 and if we consider the remainders the total is 19. This can easily be explained by the fact that 1/19 +18/19 = 1 and their decimal equivalents will add up to 0.9999… We can also start computing from the right. Notice that the last answer figure is 1, which is the numerator of the starting fraction. Now instead of dividing by two, we will multiply by 2. From 1, we have 2, 4, 8 and*

**all from 9 and the last from 10**_{1}6. The next answer figure will be 6 x 2 + 1 or

_{1}3. Then we have 3 x 2 + 1 or 7 and so on. This is not possible using the conventional method. Thus, the answer in the IVMO question is 8421.