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30MSC 2023 #19: Using The Ekadhika

30MSC 2023 #19: Using the Ekadhika

A fraction is usually converted to its decimal form by dividing the numerator by the denominator. This would be easy if the denominator is small, or the fraction is terminating. However, our featured example states that the fraction 1/49 has a recurring decimal string and we are required to give the last five digits. To get the answer using the conventional method, continuous division by 49 is needed, until we reach the end of the repeating string which is 42 digits long. This would be tedious and be prone to errors.

We can use the Ekadhika (the number one more) of the divisor. Here the Ekadhika is the number one more than the number before 9. Therefore, the Ekadhika of 49 is 5. We can use the 5 as the divisor instead of 49 to get the repeating decimal string equivalent of 1/49. In the following computations the remainder is enclosed in parentheses and placed before the quotient to be part of the next dividend but not part of the decimal string.

1/49 = 0.0…                 Let us start with 0 and a decimal point

1/49 = 0.(1)0…               1 divided by 5 is 0 remainder 1. The remainder is placed before the answer digit

1/49 = 0.(1)02…             10 divided by 5 is 2

1/49 = 0.(1)02(2)0…         2 divided by 5 is 0 remainder 2

1/49 = 0.(1)02(2)04…      20 divided by 5 is 4 

1/49 = 0.0204…                                           

But as we said before, the recurring string is 42 digits long. And it would take a while even while using 5 as the divisor to know the complete string. The last digit of this string, however, indicates that the next digits will repeat the previous pattern. This means that the last digit of the string is the original numerator of this fraction, 1.

We can then start with 1 and reverse the operation: to get the next digit to the left, we multiply by 5.

1/49 =                        …1       Start with 1

1/49 =                      …51     1 multiplied by 5 is 5

1/49 =               …(2)551     5 multiplied by 5 is 25

1/49 =       … (2)7(2)551     5 multiplied by 5 is 25 plus 2 is 27

1/49 =  …(3)7(2)7(2)551    7 multiplied by 5 is 35 plus 2 is 37.

Therefore, the last five digits of the recurring decimal string of 1/49 is 77551

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